What is remainder of: ((3^5)(5^7)(7^9)(11^13)) / 4 How can

First, this wouldn't be a GMAT problem. It's not the type of question you see on the test, and what could the five answer choices be? When you divide by 4, there are only four possible remainders: 0, 1, 2 and 3. You might also notice that ((3^5)(5^7)(7^9)(11^13)) is odd, so the remainder could only be 1 or 3. It's very quick to get down to a 50-50 guess.

It's easiest to solve this problem with modular arithmetic: if you only care about remainders when dividing by a fixed number, you can replace any number in a product (or sum or difference) with another number that gives the same remainder. So in the above, I could, for example, replace the 11 with a 3, because 11 and 3 have the same remainder when divided by 4. So we could replace 11 with 3, 7 with 3, and 5 with 1:

(3^5)(5^7)(7^9)(11^13) will have the same remainder as (3^5)(1^7)(3^9)(3^13) when divided by 4. Since

(3^5)(1^7)(3^9)(3^13) = 3^27

we have a much simpler problem. Indeed, since 3^2 has a remainder of 1 when divided by 4, we can replace 3^2 with 1:

3^27 = 3*3^26 = 3*(3^2)^13 which has the same remainder as 3*(1^13) = 3. So the answer is 3.

It's actually even easier to replace the numbers as follows:

-replace 3 with -1
-replace 5 with 1
-replace 7 with -1
-replace 11 with -1

Thus, ((3^5)(5^7)(7^9)(11^13)) has the same remainder as [(-1)^5][(1)^7][(-1)^9][(-1)^13] when divided by 4.

Since [(-1)^5][(1)^7][(-1)^9][(-1)^13] = -1, which has a remainder of 3 when divided by 4, the answer is 3.

= [(3^5)(5^7)(7^9)(11^13)]/4
= (0+3)^5 (4+1)^7 (4+3)^9 (8+3)^13)/4
= (0+3)^5 (4+1)^7 (4+3)^9 (8+3)^13)/4

if we multiple all the terms on the abovce expression, the final term will be a product of 3, 1, 3, and 3. so their product is 27.
all other terms are divisible by 4 except this last one. therefore, reminder should be 3.

No comprendo

only consider reminders

1. (3x5)/4 has 3 reminder:
(0+3) (4+1) = 0x4 + 0x1 + 4x3 + 3x1
the reminder 3 is the product of two reminders 3x1.

2. (3x5x7)/4 has 1 reminder:
(0+3) (4+1) (4+3)
= (0x4 + 0x1 + 4x3 + 3x1) (4+3)
= 0x4x4 + 0x1x4 + 4x3x4 + 3x1x4 + 0x4x3 + 0x1x3 + 4x3x3 + 3x1x3

the reminder 1 is the product of two reminders 3x1x3.

FOLLOW THE PATTREN................



NoW comprendo?