Let’s continue the discussion of last two digits we started last week. We discussed the concept of pattern recognition and how it can help us determine the last two digits in case of numbers raised to some powers. Today we look at what happens when there is no pattern to determine! What if we are asked to determine the last two digits of the product of a bunch of numbers. We know that getting the last digit in this case is very easy – just multiply the last digits of the numbers together. But last TWO digits would seem much more complicated.
Actually, we can find the last two digits quite easily in most such cases by using the concepts of remainders.
There are two concepts you need to understand before we go on to see how to solve such questions:
I. When you divide a number by 100, the remainder is formed by the last two digits of the number. Say, you divide 138 by 100, the remainder will be 38 (last two digits). Take another example – divide 1275 by 100, the remainder will be 75 and so on.
II. When you divide (px + a)(qx + b)*…*(tx + e) by x, the remainder will be the remainder obtained by dividing a*b*…*e by x. This should remind you of the binomial theorem we discussed many weeks ago. When we multiply all these terms together (px + a), (qx + b) etc, each term obtained will have at least one x except the last term which is obtained by multiplying the remainders together. To get a better idea, let’s take some numbers:
Let’s say we need to find the remainder when we divide 12*23*52*81 by 10.
K = 12*23*52*81 = (10 + 2)*(20 + 3)*(50 +2)*(80 + 1)
When you multiply these four terms together, you will get many terms such as 10*20*50*80, 10*20*50*1, 10*20*2*80 etc. All these will have a multiple of 10 except the last one. The last one will be 2*3*2*1 = 12. That doesn’t have a multiple of 10. Now divide 12 by 10 to get the remainder 2. So when you divide K by 10, the remainder will be 2.
Now, let’s look at a question:
Question 1: What are the last two digits of 63*35*37*82*71*41?
Solution: Using concept 1, we know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.
Remainder of (63*35*37*82*71*41)/ 100
Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:
Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.
So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.
We need the
Remainder of (63*7*37*41*71*41*5*2)/10*5*2
Remainder of (63*7*37*41*71*41)/10
Now using concept 2, let’s write the numbers in form of multiples of 10
Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10
Remainder of 3*7*7*1*1*1/10
Remainder of 147/10 = 7
Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.
When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.
Next week, we will see some more complicated questions using these and other fundamentals.
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!