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

Question

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

How can we solve this one without using Binomial theorem?

vtselvan · Answer

i think if we are solving by this method it will be correct

Solve for 3^5/4 ...
that will give 243/4, so reminder is 3 
i think 3 is the answer...

jallenmorris · Answer

You have to figure out the units digit of each.

3^5....^1 = 3, ^2 = 9 ^3 = 7, ^4=1 ^5 =3

5^(anything) = 5

7^1=7, ^2=9, ^3 = 3; ^4=1  . Divide the power of 9 by 4 because it repeats every 4 and that gives you 1 left over, so ^9 will have same units digit as ^1...7

11^(anything) = 1

Now, figure out the units digit of the product of all these numbers. 3*5*1*1 = 5 units digit, divided by 4 could be a remainder of 3 or a remainder of 1.

for example 15 / 4 = remainder of 3, but 25 = emainder of 1.

We have to determine how many times the product is divisible by 2. If it is divisible by 2 and even number of times, the remainder will be 1. If it is divisible by 2 an odd number of times, the remainder when divided by 4 will be 3. (this is not counting a remainder. Such as 9/2 =  4\frac{1}{2}  9/4 remainder is 1, but 15/2 =  7\frac{1}{2}  15/4 remainder is 3.

3^1 in this manner has only one 2 in it. 3^2 has four 2's in it with remainder of 1 3^3 has thirteen 2's in it with remainder of 1. I see a pattern here, when the power of 3 is odd, the remainder when divded by 4 will be 3. When the power of 3 is even, the remainder when divided by 4 will be 1. Does it work for all numbers?

The pattern is there, but it's in a slightly different form.

I'm too tired to see if this goes anywhere. Sorry.

IanStewart · Answer

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      when divided by 4.

Since      = -1, which has a remainder of 3 when divided by 4, the answer is 3.

GMAT TIGER · Answer

=  /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.

Nerdboy · Answer

No comprendo :oops:

GMAT TIGER · Answer

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?

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

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