swerve wrote: ↑Tue Feb 25, 2020 4:19 pm
What is the remainder when \(47\cdot 49\) is divided by \(8\)?
A. 1
B. 3
C. 4
D. 5
E. 7
The OA is
E
Source: Manhattan Prep
Solution:
Let’s write 47 = 40 + 7 and 49 = 48 + 1. Then:
47*49 = (40 + 7)(48 + 1) = 40*48 + 40 + 7*48 + 7
Notice that the terms 40*48, 40, and 7*48 are all divisible by 8; thus, the remainder when these terms are divided by 8 is 0. So, the remainder when 47*49 is divided by 8 is equal to the remainder when 7 is divided by 8, which is 7.
Alternate Solution:
Let’s write 47 = 48 - 1 and 49 = 48 + 1. Then:
47*49 = (48 -1)(48 + 1) = 48^2 - 1 = 48^2 - 8 + 7
Since 48^2 - 8 is divisible by 8 (notice that both terms are divisible by 8), we see that the remainder must be the last term 7.
Yet Another Solution:
Let a\n denote the remainder when a is divided by n. Then (a * b) \ n = [(a\n) * (b\n)] \ n.
So instead of finding the product of 47 and 49, we can divide each number by 8 first and find the product of the remainders and divide that by 8.
Since 47/8 = 5 R 7 (so 47\8 = 7) and 49/8 = 6 R 1 (so 49\8 = 1), we have:
(47 * 49)\8 = [(47\8) * (49\8)]\8 = [7 * 1]\8 = 7\8 = 7
Answer: E
