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

Notice that 47 is 1 less than 48, and 49 is 1 greater than 48.

So we can write: (47)(49) = (48 - 1)(48 + 1)
= 48² - 1
= 48² - 8 + 7
= 8[(48)(6) - 1] + 7
= (some multiple of 8) + 7

This tells us that (47)(49) is 7 greater than some multiple of 8
So when we divide by 8, the remainder will be 7

Answer: E

Cheers,
Brent

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