Problem solving questions have five definite options, that is, “cannot be determined” and “data not sufficient” are not given as options. So this means that in all cases, data is sufficient for us to answer the question. So as long as the data we assume conforms to all the data given in the question, we are free to assume and make the problem simpler for ourselves. The concept is not new – you have been already doing it all along – every time you assume the total to be 100 in percentage questions or the value of n to be 0 or 1, you are assuming that as long as your assumed data conforms to the data given, the relation should hold for every value of the unknown. So the relation should be the same when n is 0 and also the same when n is 1.

Now all you have to do is go a step further and, using the same concept, assume that the given figure is more symmetrical than may seem. The reason is that say, you want to find the value of x. Since in problem solving questions, you are required to find a single unique value of x, the value will stay the same even if you make the figure more symmetrical – provided it conforms to the given data.

Let us give an example from Official Guide 13th edition to show you what we mean:

Question: In the figure shown, what is the value of v+x+y+z+w?

(A) 45

(B) 90

(C) 180

(D) 270

(E) 360

We see that the leg with the angle w seems a bit narrower – i.e. the star does not look symmetrical. But the good news is that we can assume it to be symmetrical because we are not given that angle w is smaller than the other angles. We can do this because the value of v+x+y+z+w would be unique. So whether w is much smaller than the other angles or almost the same, it doesn’t matter to us. The total sum will remain the same. Whatever is the total sum when w is very close to the other angles, will also be the sum when w is much smaller. So for our convenience, we can assume that all the angles are the same.

Now it is very simple to solve. Imagine that the star is inscribed in a circle.

Now, arc MN subtends the angle w at the circumference of the circle; this angle w will be half of the central angle subtended by MN (by the central angle theorem discussed in your book).

Arc NP subtends angle v at the circumference of the circle; this angle v will be half of the central angle subtended by NP and so on for all the arcs which form the full circle i.e. PQ, QR and RM.

All the central angles combined measure 360 degrees so all the subtended angles w + v + x + y + z will add up to half of it i.e. 360/2 = 180.

Answer (C)

There are many other ways of solving this question including long winded algebraic methods but this is the best method, in my opinion.

This was possible because we assumed that the figure is symmetrical, which we can in problem solving questions!

But beware of question prompts which look like this:

– Which of the following cannot be the value of x?

– Which of the following must be true?

You cannot assume anything here since we are not looking for a unique value that exists. If a bunch of values are possible for x, then x will take different values in different circumstances.

If we know that the unknown has a unique value, then we are free to assume as long as we are working under the constraints of the question. Finally, we would like to mention here that this is a relatively advanced technique. Use it only if you understand fully when and what you can assume.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

In a previous post, we saw that

“Two consecutive integers can have only 1 common factor and that is 1.”

This implies that N and N+1 have no common factor other than 1. (N is an integer)

Similarly,

N + 5 and N + 6 have no common factor other than 1. (N is an integer)

N – 3 and N – 2 have no common factor other than 1. (N is an integer)

2N and 2N + 1 have no common factor other than 1. (N is an integer)

We are sure you have no problem up until now.

How about:

N and 2N+1 have no common factor other than 1. (N is an integer)

It is a simple application of the same concept but makes for a 700 level question!

2N and 2N+1 have no common factor other than 1 – we know

The factors of N will be a subset of the factors of 2N. It will not have any factors which are not there in the list of factors of 2N. So if 2N and another number have no common factors other than 1, N and the same other number can certainly not have any common factor other than 1.

Taking an example, say N = 6

Factors of 2N (which is 12) are 1, 2, 3, 4, 6, 12.

Factors of 2N + 1 (which is 13) are 1, 13.

2N and 2N + 1 can have no common factors.

Now think, what are the factors of N? They are 1, 2, 3, 6 (a subset of the factors of 2N)

They will obviously not have any factor in common with 2N+1 (except 1) since these are the same factors as those of 2N except that these are fewer.

So we can deduce the following (N and M are integers):

M and NM +1 will have no common factor other than 1.

8 and 8M + 1 will have no common factor other than 1.

M and NM – 1 will have no common factor other than 1.

and so on…

Here is the 700 level official question of this concept:

**Question**: If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?

Statement 1: x = 12u, where u is an integer.

Statement 2: y = 12z, where z is an integer.

**Solution**:

x = 8y + 12

We need to find the greatest common divisor of x and y. We have 8y in the equation. A couple of immediate deductions:

The factors of y will be a subset of the factors of 8y.

The difference between x and 8y is 12 so the greatest common divisor of x and 8y will be a factor of 12 (discussed in this post a few weeks back).

This implies that the greatest factor that x and y can have must be a factor of 12.

Looking at the statements now:

Statement 1: x = 12u, where u is an integer.

Now we know that x has 12 as a factor. The problem is that we don’t know whether y has 12 as a factor.

y could be 3 —> x = 8*3 + 12 = 36 (a multiple of 12). Here greatest common divisor of x and y will be 3.

or y could be 12 —> x = 8*12 + 12 = 108 (a multiple 12). Here greatest common divisor of x and y will be 12.

So this statement alone is not sufficient.

Statement 2: y = 12z, where z is an integer.

This statement tells us that y also has 12 as a factor. So now do we just mark (C) as the answer and move on? Well no! It seems like an easy (C) now, doesn’t it? We must analyse this statement alone.

Substituting y = 12z in the given equation:

x = 8*12z + 12

x = 12*(8z + 1)

So this already gives us that x has 12 as a factor. We don’t really need statement 1.

Since both x and y have 12 as a factor and the highest common factor they can have is 12, greatest common divisor of x and y must be 12.

This statement alone is sufficient to find the greatest common divisor of x and y.

Answer (B)

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

Question: In a certain business, production index p is directly proportional to efficiency index e, which is in turn directly proportional to investment i. What is p if i = 70?

Statement I: e = 0.5 whenever i = 60

Statement II: p = 2.0 whenever i = 50

This was the issue that was raised:

If one were to follow the method given in the post on **joint variation, **one would arrive at this solution:

p/e = k (a constant)

e/i = m (another constant)

Hence, p*i/e = n is the joint variation expression

(where k, m and n are constants)

So we get that p is inversely proportional to i, that is, p*i = Constant

Statement II gives us the values of p and i which can help us get the value of the Constant.

2*50 = Constant

The question asks us the value of p given the value of i = 70. If Constant = 100,

p = 100/70.

But actually, this is wrong and the value that you get for p in this question is different.

The question is “why is it wrong?”

Valid question, right? It certainly seems like a joint variation scenario – relation between three variables. Then why does’t it work in this case?

The takeaway from this question is very important and before you proceed, we would like you to think about it on your own for a while and then proceed to the the rest of the discussion.

Here is how this question is actually done:

Taking one statement at a time:

“production index p is directly proportional to efficiency index e,”

implies p = ke (k is the constant of proportionality)

“e is in turn directly proportional to investment i”

implies e = mi (m is the constant of proportionality. Note here that we haven’t taken the constant of proportionality as k since the constant above and this constant could be different)

Then, p = kmi (km is the constant of proportionality here. It doesn’t matter that we depict it using two variables. It is still just a number)

Here, p seems to be directly proportional to i!

So if you have i and need p, you either need this constant directly (as you can find from statement II) or you need both k and m (statement I only gives you m).

So the issue now is that is p inversely proportional to i or is it directly proportional to i?

Review the joint variation post – In it we discussed that joint variation gives you the relation between 2 quantities keeping the third (or more) constant.

p will vary inversely with i if and only if e is kept constant.

Think of it this way: if p increases, e increases. But we need to keep e constant, we will have to decrease i to decrease e back to original value. So an increase in p leads to a decrease in i **to keep e constant**.

But if we don’t have to keep e constant, an increase in p will lead to an increase in e which will increase i.

It is all about the sequence of increases/decreases

Here, we are not given that e needs to be kept constant. So we will not use the joint variation approach.

Note how the independent question is framed in the joint variation post:

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical M present and inversely proportional to the concentration of chemical N present. If the concentration of chemical N is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical M required to k**eep the reaction rate unchanged**?

You need relation between N and M when reaction rate is constant.

You are given no such constraint here. So an increase in p leads to an increase in e which in turn, increases i.

So let’s complete the solution to our original question:

p = ke

e = mi

p = kmi

Statement I: e = 0.5 whenever i = 60

0.5 = m * 60

m = 0.5/60

We do not know k so we cannot find p given i and m.

This statement alone is not sufficient.

Statement II: p = 2.0 whenever i = 50

2 = km * 50

km = 1/25

If i = 70, p = (1/25)*70 = 14/5

This statement alone is sufficient.

Answer (B)

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

For example, look at this sentence:

A New York City ordinance of 1897 regulated the use of bicycles, mandated a maximum speed of eight miles an hour, required cyclists to keep feet on pedals and hands on handlebars at all times, and granted pedestrians right-of-way.

Is everything ok here? Well, it certainly seems so. We have four elements in parallel:

regulated …

mandated …

required …

granted …

But actually, there is a problem in this sentence:

‘regulated…’ will not be parallel to the rest of the three elements. The rest of the three elements will be in parallel.

Before we explain why, let’s take a simpler example:

The girl sitting next to me wears blue everyday, eats only waffles, and listens to music in office.

The sentence will not be ‘The girl sits next to me…’ because ‘sit’ is not parallel to other verbs. “sit” modifies the girl and is not used as a verb here. It is a present participle modifier modifying ‘girl’. It specifies the girl about whom we are talking.

Similarly, in the original sentence, ‘regulate’ is modifying ‘ordinance of 1897’. It is telling you which ordinance of 1897.

The other verbs ‘mandated’, ‘required’ and ‘granted’ are used as verbs and are parallel. They are assimilated under ‘regulate’. They tell you how the ordinance regulated.

How did it regulate?

mandated …

required …

granted …

Hence, you cannot use ‘regulated’ here. You must use ‘regulating’ – the present participle modifier to modify the ordinance. So you have to think logically – are the items in the given list actually parallel? Are they equal elements? If yes, then they need to be grammatically parallel too; else not.

Here is the complete official question:

Question: A New York City ordinance of 1897 regulated the use of bicycles, mandated a maximum speed of eight miles an hour, required of cyclists to keep feet on pedals and hands on handlebars at all times, and it granted pedestrians right-of-way.

(A) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required of cyclists to keep feet on pedals and hands on handlebars at all

times, and it granted

(B) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, granting

(C) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists that they keep feet on pedals and hands on handlebars

at all times, and it granted

(D) regulating the use of bicycles, mandating a maximum speed of eight miles an

hour, requiring of cyclists that they keep feet on pedals and hands on

handlebars at all times, and granted

(E) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, and granted

Solution:

From our above discussion, we know that we have choose one of (C), and (E).

(A), (B) and (D) put regulate parallel to the other verbs.

Still, let’s point out all the errors of these options:

(A) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required of cyclists to keep feet on pedals and hands on handlebars at all

times, and it granted

Parallelism problem – regulated cannot be parallel to mandated and other verbs. Also, ‘mandated’ is not parallel to ‘it granted’. Besides, ‘required of X to do Y’ is unidiomatic.

(B) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, granting

Parallelism problem – ‘regulated’ is parallel to ‘mandated’ though it should not be.

‘granting’ is not parallel to ‘mandated’ and ‘required’ though it needs to be parallel.

You also need an ‘and’ before the last element of the list ‘and granted …’

(D) regulating the use of bicycles, mandating a maximum speed of eight miles an

hour, requiring of cyclists that they keep feet on pedals and hands on

handlebars at all times, and granted

This is not a valid sentence because the main clause does not have a verb. ‘regulating…’, ‘mandating…’ and ‘requiring…’ are the present participle modifiers.

‘granted…’ is not parallel to the other elements. Besides, ‘requiring of X that they do Y’ is unidiomatic.

Now let’s look at the leftover options:

(C) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists that they keep feet on pedals and hands on handlebars

at all times, and it granted

‘it granted’ is not parallel to the other verbs. Besides, ‘required X that they do Y’ is unidiomatic.

(E) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, and granted

Perfect! All issues sorted out!

Answer (E)

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

That said, there is a set of questions in which we should think of these strategies. Number plugging is very useful when you have one or two variables in the options. Algebra can be time consuming in these cases because of equation manipulation required.

Similarly, some questions beg you to use the process of elimination. Their question stem goes something like ”which of the following options can be the value of x?”, “which of the following options cannot be the sum of a and b?” etc. These questions are framed like this because often they have multiple solutions. x could possibly take many different values but the options would have only one of them. So it makes sense to check which values x can take from the options. Let’s look at one such instance of a tricky question where process of elimination can be very useful.

Question:

A list of numbers has six positive integers. Three of those integers are known – 4, 5 and 24 and three of those are unknown – x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13

(B) 12

(C) 11

(D) 10

(E) 5

Solution: The question gives us concrete information about mean – it is 10 – but not about median – it is between 7 and 8 (exclusive). What can we say about median from this? That it cannot be 7 or 8 but anything in between. But we know that the list has all integers. When we have even number of integers, we know that the median is the average of the middle two numbers – when all are placed in increasing order. So can the average of the two middle numbers be, say, 7.1? Which two positive integers can average to give 7.1? None! Note that if the average of two integers is a decimal, the decimal must be (some number).5 such as 7.5 or 9.5 or 22.5 etc. This happens in case one number is odd and the other is even. In all other cases, the average would be an integer.

Since the median is given to be between 7 and 8, the median of the list of the six positive integers must be 7.5 only.

Now we know that the mean = 10 and median = 7.5

**Method 1: Algebra/Logic**

Let’s try to solve the question algebraically/logically first.

There are 6 elements in the list. The average of the list is 10 which means the sum of all 6 elements = 6*10 = 60

4 + 5 + 24 + x + y + z = 60

x + y + z = 27

Median of the list = 7.5

So sum of third and fourth elements must be 7.5 * 2 = 15

There are two cases possible:

Case 1: Two of the three integers x, y and z could be the third and the fourth numbers. In that case, since already 4 and 5 are less than 7.5, one of the unknown number would be less than 7.5 (the third number) and the other two would be greater than 7.5.

The sum of the third and fourth elements of the list is 15 so

15 + z = 27

z = 12

So, two numbers whose sum is 15 such that one is less than 7.5 and the other greater than 7.5 could be

5 and 10

6 and 9

7 and 8

x, y and z could take values 5, 6, 7, 8, 9, 10 and 12.

Case 2: The known 5 could be the third number in which case one of the unknown numbers is less than 5 and two of the unknown numbers would be more than 7.5.

If the third number is 5, the fourth number has to be 10 to get a median of 7.5. Hence, 10 must be one of the unknown numbers.

The sum of the other two unknown numbers would be 27 – 10 = 17.

One of them must be less than 5 and the other greater than 10. So possible options are

4 and 13

3 and 14

2 and 15

1 and 16

x, y and z could take various values but none of them could be 11

Answer (C)

**Method 2: Process of Elimination**

Let’s now try to look at the process of elimination here and see if we can find an easier way.

The three unknowns need to add up to 10*6 – 4 – 5 – 24 = 27.

Two of the given options are 5 and 10. They have a median of 7.5 so lets assume that two of the unknown numbers are 5 and 10 (5 can be one of the unknowns since we are not given that all six integers need to be distinct). If two unknowns make up third and fourth numbers in the list and have a median of 7.5, their sum would be 15 and the third unknown will be 12 (to get the mean of 10). This case (5, 10, 12) satisfies all conditions so options (B), (D) and (E) are out of play.

Now we are left with two options 13 and 11. Check any one of them and you will know which one is not possible. Let’s check 13.

From the given options, any number greater than 7.5 must be either the fourth number or the fifth number. 13 cannot be the fourth number since the third number would need to be 2 in that case to get median 7.5. But we have 4 and 5 more than 2 so it cannot be the third number. So 13 must be the fifth number of the list. We saw in the case above that if two unknowns are third and fourth numbers then the fifth number HAS TO BE 12. So the already present 5 must be the third number and the fourth number must be 10. In that case, the leftover unknown would be 4 (to get a sum of 27). So the three unknowns would be 4, 10 and 13. This satisfies all conditions and is possible. Hence answer must be (C). 11 will not be possible.

Let’s see what would have happened had you picked 11 to try out. If 11 were the fourth number, to get a median of 7.5, we would need 4 as the third number. That is not possible since we already have a 5 given. So 11 must have been the fifth number. This would mean that the already present 5 and one unknown 10 would make the median of 7.5. So the third unknown in this case would be 6 (to get a sum of 27). But 6 would be the third number and the median in this case would be (6 + 10)/2 = 8. So one of the numbers cannot be 11.

Answer (C)

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

Today, let’s look at a puzzle where symmetry helps.

Question: A spider is sitting on one corner of a cube. It wants to get to the most distant corner but it can crawl only along the edges of the cube and cannot revisit a place where it has already been. In how many different ways can the spider go to the most distant corner?

(A) 6

(B) 12

(C) 18

(D) 24

(E) 30

Solution: The question is a puzzle type combinatorics questions. It seems like we will have to painstakingly calculate the various paths that the spider can take. But notice that the figure we have is a cube – a symmetrical figure. Let’s draw the figure to see what the question is asking.

Now, assume that the spider is at A. In that case, he has to go to F – the farthest vertex from A. Every vertex has only one vertex farthest to it. C, E and G are equidistant from A but they are in the same plane. F is further off than C, E and G. So it needs to go from A to F:

Step 1:

It can crawl only along the edges so from A, it can take three different paths – AB or AH or AD. As far as F is concerned, all the points D, H and B are similar.

Step 2:

Now, from each of these 3 points, the spider has two path options. If it is at D, it can crawl on DE or DC. The third path from D leads back to A but the spider is not allowed to revisit a place. So there are only two forward options for it – DE or DC.

Similarly, if it is at H, it can crawl on HE or on HG. If it is at B, it can crawl on BG or BC. So the total number of paths that we have found till now are 3*2.

Till now, we hope you did not face any problems.

Step 3:

Now comes the tricky part. The spider is at one of three vertices – E, C or G. Assume it came the AD – DE route and is now at E. There are multiple ways in which it can reach F. The obvious one is directly from E to F. But it can also go to F via H because it has not visited a number of other vertices (H, G, B, C)

There are three ways in which it can reach F now:

- directly E to F
- a three path EH – HG – GF
- a five path EH – HG – GB – BC – CF

This takes care of all the ways in which it can reach F from E.

Since we found 3 different paths from E, it is obvious that we will find 3 different paths from C and from G too. It is a symmetrical figure and hence we don’t need to calculate the number of paths from each point. In any case, we have 3 ways to reach F now.

So total different paths to reach the farthest vertex = 3*2*3 = 18

Answer (C)

Hope you see how symmetry helps us reduce our work substantially.

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

But before I tell you what that question was, let’s solve a couple of questions which are similar to some questions you might have seen before (for the sake of brevity, let’s ignore the options):

**Question 1**: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if Dave should get either three paintings or five paintings? (All paintings should be given away).

Solution 1:

There are two ways of distributing the paintings in this case:

Dave gets 3 paintings and Mona gets the rest: You select 3 of the 10 paintings and give them to Dave. This can be done in 10C3 = 120 ways

Dave gets 5 paintings and Mona gets the rest: You select 5 of the 10 paintings and give them to Dave. This can be done in 10C5 = 252 ways

Total number of ways in which you can distribute the paintings = 120 + 252 = 372 ways

Simple enough, right? Let’s take a look at another simple similar question.

**Question 2**: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if Dave should get at least two paintings? (All paintings should be given away.)

Solution 2:

Dave should get at least two paintings so it means he can get 2 or 3 or 4 or more up to 10 paintings. Calculating all those cases would be tedious so this is a perfect opportunity to use ‘Total – Opposite’ method.

Total ways in which you can distribute 10 paintings between two people without any constraints: Each painting can be given away in two ways – either to Dave or to Mona. So the paintings can be distributed in 2*2*2*…*2 = 2^10 = 1024 ways

Number of ways in which Dave gets 0 paintings or 1 painting: 1 + 10C1 = 11 ways

So number of ways in which Dave gets 2 or 3 or 4 … upto 10 (i.e. at least 2 paintings) = 1024 – 11 = 1013 ways

Another ‘seen before, know how to solve it’ kind of question. Now let’s come to the question of the day which doesn’t look much different but actually is.

**Question 3**: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if both collectors should get an even number of paintings? (All paintings should be given away.)

Solution 3:

Paintings can be distributed in the following ways:

0, 10 – One person gets 0 paintings and the other gets 10

2, 8 – One person gets 2 paintings and the other gets 8

4, 6 – One person gets 4 paintings and the other gets 6

You will need to calculate each one of these ways and then add them. Note that the ‘Total – Opposite’ method does not work here because finding the number of ways in which each person gets odd number of paintings is equally daunting.

Case 1: 0, 10

One person gets 0 paintings and the other gets 10. This can be done in 2 ways – either Dave gets all the paintings or Mona gets them.

Case 2: 2, 8

One person gets 2 paintings and the other gets 8. Select 2 paintings out of 10 for Dave in 10C2 = 45 ways. Mona could also get the 2 selected paintings so total number of ways = 45*2 = 90 ways

Case 3: 4, 6

One person gets 4 paintings and the other gets 6. Select 4 paintings out of 10 for Dave in 10C4 = 210 ways. Mona could also get the 4 selected paintings so total number of ways = 210*2 = 420

Total number of ways such that each person gets even number of paintings = 2 + 90 + 420 = 512 ways

But 512 is 2^9 – in form, suspiciously close to 2^10 we used in question 2 above. Is there some logic which leads to the answer 2^(n-1)? There is!

You have 10 different paintings. Each painting can be given to one of the 2 people in 2 ways. You do that with 9 paintings in 2*2*2… = 2^9 ways. When you distribute 9 paintings, one person will have odd number of paintings and one will have even number of paintings (0 + 9 or 1 + 8 or 2 + 7 or 3 + 6 or 4 + 5).

The tenth painting needs to be given to the person who has the odd number of paintings so you give the tenth painting in only one way. This accounts for all cases in which both get even number of paintings.

Total ways = 2^9 * 1 = 512

On the same lines, now think about this:

Question 4: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if both collectors should get an odd number of paintings? (All paintings should be given away.)

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

Without actually finding all the factors of two numbers, how do we know whether they have any common factors (ignoring 1)?

Let’s take some examples:

- If the integers are even, we know that they must have at least one common factor – 2. Let’s say we have two numbers 476 and 478. How many common factors can they have? We know that 2 is a factor common to them. Can they have any other common factor? Note that the difference between them is 2. So if 4 were a factor of 476, could it be a factor of 478? No. If 4 were a factor of 476, it would be a factor of 480 next (4 away from 476). Similarly, if 7 were a factor of 476, it would not be a factor of 478, definitely. It would be a factor of 483 (7 away from 476). In fact, since the difference between the two numbers is 2, the only factor they can have in common is 2.

- Now consider that the two numbers are 476 and 484. They have a difference of 8 between them. The common factors they can have are 2, 4 and 8 (the factors of 8). If any of these factors is a factor of 476, it will be a factor of 484 too. Obviously, 476 and 484 will have many other factors but they will not have any other common factor. 7 is a factor of 476. The next multiple of 7 will be 483 and the next will be 490. 7 cannot be a factor of 484.

- What happens when both integers are odd? Say 523 and 529. The difference between them is 6. The factors of 6 are 2 and 3. Both 523 and 529 are odd numbers so 2 cannot be their factor. If 3 is a factor of 523, it will be a factor of 529 too else it will not be a factor of both the numbers. Any other number can be a factor of one of them, but not both.

This is what we can deduce:

The only factors that CAN be common (it’s not necessary that they will be common) between two numbers are the factors of the difference between them.

If any factor of the difference between them is a factor of one of the numbers, it will be a factor of the other number too. If it is not a factor of one number, it will not be a factor of the other number.

Take a look at a question based on these concepts:

**Question**: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (x-m)?

Statement 1: m^2 – 10m + 16 = 0

Statement 2: x + 26 is a prime number.

**Solution**:

The two given positive integers are (x + m) and (x – m). x is a positive integer so m must be an integer too. Whether m is positive or negative, we don’t know.

To know the GCD of two numbers, we need to know their common divisors. As of now, we have no idea about their common divisors, but we know that the difference between the two numbers is 2m. Their common factors must be factors of 2m.

Let’s look at the two statements:

Statement 1: m^2 – 10m + 16 = 0

We know that the quadratic will give us two values for m so we will not be able to find a unique value for m. But let’s solve it in case we get some other clues from it.

m^2 – 10m +16 = 0

m^2 – 2m – 8m + 16 = 0

m (m – 2) – 8 (m – 2) = 0

(m – 2)*(m – 8) = 0

m is either 2 or 8. So 2m is either 4 or 16.

The factors of 2m will be 1, 2 and 4 and additionally, 8 and 16 (if 2m is 16). We have no idea whether x+m and x-m will have these factors so this statement alone is not sufficient.

Statement 2: x + 26 is a prime number.

What does it tell us about x? Other than 2, all prime numbers are odd numbers. Since x is a positive integer, x+26 cannot be 2. It must be a prime number greater than 2 and hence, must be odd. But 26 is even. So x must be an odd integer (Odd + Even = Odd). But we have no information about m so this statement alone is not sufficient.

Using both statements together, since x is an odd integer and m is definitely even (either 2 or 8), both the numbers (x + m) and (x – m) are odd integers. Odd integers will not have any of these factors: 2, 4, 8, 16.

So (x + m) and (x – m) must have 1 as the only common factor. Hence their greatest common divisor must be 1.

Together, the two statements are sufficient to answer the question.

Answer (C)

To recap: Any common factor of two numbers has to be a factor of the difference between them. This also implies that the GCD of two numbers has to be a factor of the difference between them.

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In fact, what may look like an alphametic problem, might actually be a number properties problem only.

We will look at an example below:

Question:

In the correctly-worked multiplication problem above, each symbol represents a different nonzero digit. What is the value of C?

Statement 1: D is prime.

Statement 2: B is not prime.

Solution: We multiply two two-digit integers and get 1995. The good thing is that we know the result of the multiplication will be 1995. Usually, multiplication alphametics are harder since they involve multiple levels, but here the multiplication is actually a blessing. There are many many ways in which you can ADD two integers to give 1995 but there are only a few ways in which you can multiply two integers to give you 1995.

Let’s prime factorize 1995:

1995 = 3*5*7*19

We can probably count on our fingers the number of ways in which we can select AB and CD.

19 needs to be multiplied with one other factor to give us a two digit number since 5*3*7 = 105 (a three digit number) so AB and CD cannot be 19 and 105.

19*3 = 57, 5*7 = 35 – This is not possible since two of the four digits are same here – 5.

19*5 = 95, 3*7 = 21 – This is one option for AB and CD.

19*7 = 133 – Three digit number not possible.

Hence AB and CD can only take values out of 21 and 95.

As of now, C can be 2 or 9. We need to find whether the given statements give us a unique value of C.

Statement 1: D is prime

D is the units digit of CD. So D can be 1 or 5.

1 is not prime so CD cannot be 21. Hence, CD must be 95 and AB must be 21.

Hence, C must be 9.

This statement alone is sufficient.

Statement 2: B is not prime

If B is not prime then AB cannot be 95. Hence AB must be 21.

This means CD will be 95 and C will be 9.

This statement alone is sufficient.

Answer (D)

Note that the entire question was just about number properties – prime factors, prime numbers etc. Actually it required no iterative steps and no hit and trial. Rest assured that if it is a GMAT question, it will be reasoning based and will not require painful calculations.

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Question: The following alphametic shows multiplication of two numbers, IF and DR. The product you obtain is DORF.

What is the value of D + O + R + F?

(A) 17

(B) 20

(C) 22

(D) 23

(E) 30

Solution: The good thing is that we know D + O + R + F has a single value. This means there will be a logic to obtain the value of each of D, O, R and F.

As discussed last week, we first focus on the big picture, but we will have to go one level at a time.

(i) IF * R = OFF

(ii) IF * D = IF

(iii) OF + IF = DOR

A few interesting points to note from the above:

– From (ii), IF * D = IF

When you multiply IF by D, you get IF itself. This means that D must be 1. D can take no other value.

D = 1

– From (iii), F + F has unit’s digit of R.

Also O + I gives O as unit’s digit and 1 as tens digit (D of DORF obtained from above). How can this happen? Say, if O = 4, 4 + I = 14. This is possible only when I = 9 and there is a 1 carry over from the previous addition of F + F. This means that F must be 5 or greater to have a carryover of 1. It cannot be 5 because 5+5 will give you 10 making R = 0. This would mean that F*R would end in R (0). But in (i), F * R has unit’s digit of F, not R. So F cannot be 5.

D = 1, I = 9

– Another interesting point: From (i), F * R has unit’s digit of F. This is possible only when F = 0 or F = 5 or R = 6 (Think of multiplication tables of numbers to convince yourself why this is so)

Since F has to be greater than 5 (as seen above), R must be 6.

If R = 6, then from (iii), F + F has unit’s digit of 6 and a carryover of 1 so F = 8. When you add 8 + 8, you will get 16 (units digit of 6 and a carryover)

D = 1, I = 9, R = 6, F = 8

– From (i), when we multiply IF by R, we get OFF. That is, when we multiply 98 by 6, we get 588. So O must be 5.

This gives us: D + O + R + F = 1 + 5 + 6 + 8 = 20

Answer (B)

This question uses your understanding of numbers and how they are added and multiplied. It certainly takes time to get to the answer. Such questions can help you get a feel for numbers and their behavior.

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