Quarter Wit, Quarter Wisdom: Magic or Math - Part II

Quarter Wit, Quarter WisdomToday we will continue from where we left our last post. In the last post, we discussed that of any two consecutive integers, one and only one of them will be even. Out of 20 and 21, only 20 is even. Since 2 is a factor of 20, it will not be a factor of 21. Does that make sense? Sure. Every second number will have 2 as a factor.

On the same lines, can both the consecutive numbers have 3 as a factor? Let’s take the same example – 20 and 21. 3 is a factor of 21. Can it be a factor of 20 as well? Do we even need to check? Since 21 is a multiple of 3, the previous multiple must be 3 places before (i.e., 18) and the next multiple of 3 must be 3 places ahead (i.e., 24).

What do you conclude then? Two consecutive integers can only have 1 common factor and that is 1. This means that if we pick any two consecutive integers, they will have no common factor other than 1. Say if 5 were their common factor, the numbers would be 5/10/15… apart e.g. 25 and 30. They cannot be consecutive. If 11 were their common factor, the numbers would be at least 11 apart e.g. 11 and 22. They cannot be consecutive.

Out of three consecutive integers, two could have 2 as a common factor e.g. 20, 21 and 22. Both 20 and 22 have 2 as a common factor. But can 3 be a common factor of any two numbers? No. One and only one number will be a multiple of 3.

Another way to look at this – Say we have the following consecutive integers:

(N – 4), (N – 3), (N – 2), (N – 1), N, (N + 1), (N + 2), (N + 3), (N + 4)

We are given that 2, 3, 5 and 7 are factors of N. What can we say about the factors of the rest of the numbers?

  1. (N + 1) and (N – 1) both will NOT have any of 2, 3, 5 and 7 as factors. They are consecutive with N. If N is a multiple of 2, 3, 5 and 7, the next multiples of these numbers will be farther away.
  2. (N + 2) and (N – 2) both will have 2 as a factor. They are 2 steps away from N. Since N is a multiple of 2, they will be multiples of 2 too. (N + 2) and (N – 2) both will NOT have 3, 5 and 7 as factors. (N + 2) and (N – 2) are only two steps away from N. The next multiples of 3, 5 and 7 will be farther away.
  3. (N + 3) and (N – 3 ) will have 3 as a factor. They both are 3 steps away from N. Since 3 is a factor of N, it will also be a factor of these two numbers. They will not have 2, 5 and 7 as factors.
  4. (N + 4) and (N – 4) will have 2 as a factor. They are 4 steps away from N. Since N is a multiple of 2, they will be multiples of 2 too. But, they will NOT have 3, 5 and 7 as factors.

The diagram given below will help you visualize this concept.



Now think: If you pick any two consecutive integers, can they both have 4 as a factor? or 7 as a factor? or 99 as a factor? No! Once you get one multiple of 99, you will not get another one in the next 98 numbers. The next multiple will appear when you add 99 to this multiple. For example, say you pick 99. Can 100, 101, 102… be multiples of 99? No. The next multiple of 99 will be 198. Therefore, numbers from 100 to 197 will not be multiples of 99.

We can say that consecutive numbers will not have any common factor other than 1. (1 is a factor of every number.)

Let’s look at how knowing this property can be useful.

Question: For every positive even integer n, the function f(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of f(100) +1, then p is

(A) between 2 and 20
(B) between 10 and 20
(C) between 20 and 30
(D) between 30 and 40
(E) greater than 40

Solution:

First of all, the question sounds much more complicated than it actually is. Just put some values for n and try and figure out what the function looks like.

f(2) = 2

f(4) = 2*4

f(6) = 2*4*6

and so on…

f(100) = 2*4*6*…*98*99*100 = (2*1)*(2*2)*(2*3)*…*(2*48)*(2*49)*(2*50)

f(100) = (2^50) * 1*2*3*…*48*49*50

Can I say that all numbers from 2 to 50 are definitely factors of f(100)? Sure. We can see above that they are. Then what can I say about the factors of f(100) + 1? Since f(100) and f(100) + 1 are consecutive integers, can I say that they share only one common factor and that is 1? Yes, I can. We just saw this concept above. This means that f(100) + 1 will not have any factor lying between 2 and 50, inclusive because each of these numbers is a factor of f(100). So if p is a prime factor of f(100) + 1, can we say that it must be greater than 50? Yes, we can. We know that p cannot be 1 since p is a prime factor. The next factor of f(100) + 1 must be greater than 50. Since it is greater than 50, it will definitely be greater than 40 too.

Answer (E)

I think you will agree that the solution is much simpler than what you would have first expected. This is an important number property and could be useful in a range of situations. Make sure you understand it well, and as always, keep practicing!

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!

10 Responses

  1. Himanshu says:

    Two take away :
    1) Two consecutive integers are always co-prime
    2) if n is a factor of p, then n will be always be a factor of either p+n or p-n

  2. Karishma says:

    Yes.
    If n is a factor of p, then n will be a factor of both (p+n) and (p-n).
    (p is not 0.)

  3. Moni says:

    Thanks for the article, Karishma!

    I am having a confusion here.

    If we take two consecutive integers,e.g 7 & 8, then 8 has a prime factor of 2 & 7 has a prime factor of 7. So, here , the prime factor of 8 is not larger than the prime factor of 7 , right?

    Similarly, if we take 80 & 81 ,then 80 has a prime factor of 2 & 5 , whereas 81 has 3.
    So, 3 is less than 5 .so, prime factor of 81 is smaller than that of 80.

    Please let me know if i am missing something.

    • Karishma says:

      Yes, you are right. It is not necessary that the greater number must have the greater prime factor. Why did you think it must be so?
      The only thing is that when you have two consecutive integers, they will have no common factor except 1.

      • Moni says:

        I am getting confused because the answer choice says “greater than 40″ . That means the smallest prime factor of f(100)+1, which is a consecutive integer of f(100), is greater than 40 , which is actually greater than 50 here.

        Please clarify.

  4. Moni says:

    I didn’t mean to say i don’t understand “greater than 50″.That was a modifier i used unnecessarily.

    I meant to ask that here prime factor of f(100) +1 is greater than 50.

    Anyhow, i understood it by testing for smaller factorials. For 2 consecutive integers e.g. for 20 & 21 , 20=2^2*5 and 21=3*7. Here prime factor 3 of 21 is less than the prime factor 5 of 20. However, in case of factorials, the case would be different.It will be a multiplication of many factors starting from 1.So,all these cannot be a factor of its next consecutive integer.

    • Karishma says:

      Yes, in case of factorials, say, if your number is 50!, the number next to it i.e. 50! + 1 cannot have any of the first 50 numbers as its factor (except 1). This is so because each one of the first 50 numbers is a factor of 50!.

      In case of 20 and 21
      20 = 2^2 * 5
      21 = 3*7
      the point is that 21 cannot have 2 and 5 as factors. We cannot derive anything about the magnitude of the prime factors of two consecutive integers.

  5. Moni says:

    Thanks for confirming my understanding !

  6. Sachin says:

    Hi Karishma,
    How is the following

    f(100) = 2*4*6*…*98*99*100 = (2*1)*(2*2)*(2*3)*…*(2*48)*(2*49)*(2*50)

    equal to

    f(100) = (2^50) * 1*2*3*…*48*49*50

    ?
    Regards,
    Sach

    • Karishma says:

      When you have (2*1)*(2*2)*(2*3)*…*(2*48)*(2*49)*(2*50), you can separate the 2s out and the other numbers.

      (2*2*2*…*2*2*2) * (1*2*3*…*48*49*50)

      2^{50} * (1*2*3*…*48*49*50)

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