MGMAT Adv Quant - WS#5, Question 42

47/8 is not 7 remainder 7. It's 5 remainder 7. What if the 49 had been first?

(49)(47) divided by 8. What is the remainder?

(49/8)(47) = (6 R1)(47) = some number remainder 1?

Try a simple example. Is 4 * 5 / 3 = some number remainder 1? 4/3 is 1 R 1. 20/3 = 6 remainder 2.

Why make this difficult by trying to find a short cut? It's a simple multiplication problem and then division problem that takes a few seconds.

47
x49
----
2303

2303 / 8 = 287 R 7

Or.

To find the remainder when dividing the product of two numbers by a divisor, divide each number by the divisor, then multiply the remainders to find the final remainder.

49/8 = 6 r1 47/8 = 5 r7 1*7 = 7

Try it with any pair or set of numbers.

13*14*15 divided by 3. What is the remainder?

13/3 = 4 r1 14/3 = 4 r2 15/3 = 5 r0 1*2*0=0

13*14*15=2730 and 2730/3 = 910 (no remainder).

A couple of shortcuts:

1. 47/8 has a remainder of 7. 49/8 has a remainder of 1. Multiply the remainders: 7*1=7

2. (47)(49)=(48-1)(48+1)=48^2-1

This is clearly one less than a multiple of 8, so it must have a remainder of 7.

Reaminders follow a constant pattern, here 2 Nos - 1 less than multiple of 8(47) and the other greater than a multiple of 8 (49).
Take the smallest such pair you can think of. (Say 7 and 9)

7*9 = 63 -> Remainder 7 when divided by 8.

Every such pair pretty much follows the same pattern.

There are a number of shortcuts or ways to remember the pattern. But the bottom line is if you spend a few seconds on this and can't think of a shortcut, quickly perform the multiplication and division.

Someone else might chime in with how many seconds to consider a shortcut before performing simple math.

GmatMathPro wrote: 1. 47/8 has a remainder of 7. 49/8 has a remainder of 1. Multiply the remainders: 7*1=7

This is what I originally did when solving the problem the first time... even thought it looked weird, I stuck by it and chose the correct answer. But when reviewing the question later, I thought that it looked weird, since there is only one 8 in the denominator and this required dividing each by 8.

But I remembered seeing this in the MGMAT books. The solution I posted above was a "modified" version of this which I threw together after the fact when reviewing. All in all, when doing the problem, I got confused, but still ended up picking the right answer, and then moved on.

Thanks everyone for the clarifications!
--Rishi