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

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Intern
Joined: 19 Jul 2008
Posts: 23
What is remainder of: ((3^5)(5^7)(7^9)(11^13)) / 4 How can [#permalink]

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25 Aug 2008, 18:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

How can we solve this one without using Binomial theorem?
Intern
Joined: 25 Aug 2008
Posts: 3

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25 Aug 2008, 19:07
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...
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Joined: 30 Apr 2008
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25 Aug 2008, 19:12
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 [repeats after here]. 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.

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

How can we solve this one without using Binomial theorem?

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25 Aug 2008, 19:45
gmatcraze wrote:
What is remainder of:
((3^5)(5^7)(7^9)(11^13)) / 4

How can we solve this one without using Binomial theorem?

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.
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25 Aug 2008, 22:15
gmatcraze wrote:
What is remainder of:
((3^5)(5^7)(7^9)(11^13)) / 4

How can we solve this one without using Binomial theorem?

= [(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.
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Joined: 16 Jul 2008
Posts: 285

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26 Aug 2008, 00:56
GMAT TIGER wrote:
gmatcraze wrote:
What is remainder of:
((3^5)(5^7)(7^9)(11^13)) / 4

How can we solve this one without using Binomial theorem?

= [(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
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26 Aug 2008, 11:11
Nerdboy wrote:
GMAT TIGER wrote:
gmatcraze wrote:
What is remainder of:
((3^5)(5^7)(7^9)(11^13)) / 4

How can we solve this one without using Binomial theorem?

= [(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 multiply all the terms on the above expression, the final term will be a product of 3, 1, 3, and 3 and 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.

NoW comprendo?
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Re: remainder problem   [#permalink] 26 Aug 2008, 11:11
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# What is remainder of: ((3^5)(5^7)(7^9)(11^13)) / 4 How can

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