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­Find the remainder when 17^17 is divided by 64:

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playthegame
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­Find the remainder when 17^17 is divided by 64: [#permalink] New post   27 Mar 2024, 10:45
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­Find the remainder when 17^17 is divided by 64:

A 0
B 1
C 17
D 34
E 64­
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Re: ­Find the remainder when 17^17 is divided by 64: [#permalink] New post   27 Mar 2024, 21:09
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An easier way here would be using the options.

Given an even number is dividing an odd number the remainder has to be odd

That leaves us with only 2 options - 1 and 17

Let’s consider 17 first:

If 17 is the remainder, (17^17-17)/64 must leave a remainder of 0

= 17(17^16-1)/64

Now, 17^16 - 1 can be broken down as (17^8 -1)(17^8+1) = (17^4-1)(17^4+1)(17^8+1) and so on. Hence it is divisible by 17-1 = 16 and every other factor of it will be divisible by 2.

Hence, 17^17-17 must be divisible by 64

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­Find the remainder when 17^17 is divided by 64: [#permalink] New post   27 Mar 2024, 18:42
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playthegame wrote:
­Find the remainder when 17^17 is divided by 64:

A 0
B 1
C 17
D 34
E 64­

­
Remainder(\(\frac{17^{17}}{64}\)) = ­\(\frac{17^{16}*17}{64}\)

= Remainder(­\(\frac{17^{16}}{64}\)) * Remainder(­\(\frac{17}{64}\))

= Remainder(­\(\frac{((16+1)^{2})^8}{64}\)) * Remainder(­\(\frac{17}{64}\))­

= Remainder(­\(\frac{(256 + 1 + 32)^8}{64}\)) * Remainder(­\(\frac{17}{64}\))­

⇒ Remainder(­\(\frac{(256 + 1 + 32)}{64}\)) = 33

A remainder of 33, can be represented as 64 - 31, hence the remainder of 33 can be expressed as -31. 

Remainder(­\(\frac{(-31)^8}{64}\)) * Remainder(­\(\frac{17}{64}\))­

= Remainder(­\(\frac{((-31)^2)^4}{64}\)) * Remainder(­\(\frac{17}{64}\))­

⇒ Remainder(­\(\frac{(-31)^2)}{64}\))

⇒ Remainder(­\(\frac{961}{64}\)) = 1

= Remainder(­\(\frac{(1)^4}{64}\)) * Remainder(­\(\frac{17}{64}\))­

\(1 * 17 = 17\)

Option C

Another method to solve this question is using theorems. However, such concepts are not required to solve any questions in GMAT.­
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Re: ­Find the remainder when 17^17 is divided by 64: [#permalink] New post   27 Mar 2024, 19:39
gmatophobia wrote:
playthegame wrote:
­Find the remainder when 17^17 is divided by 64:

A 0
B 1
C 17
D 34
E 64­

­
Remainder(\(\frac{17^{17}}{64}\)) = ­\(\frac{17^{16}*17}{64}\)

= Remainder(­\(\frac{17^{16}}{64}\)) * Remainder(­\(\frac{17}{64}\))

= Remainder(­\(\frac{((16-1)^{2})^8}{64}\)) * Remainder(­\(\frac{17}{64}\))­

= Remainder(­\(\frac{(256 + 1 - 32)^8}{64}\)) * Remainder(­\(\frac{17}{64}\))­

⇒ Remainder(­\(\frac{(256 + 1 - 32)}{64}\)) = -31

Remainder(­\(\frac{(-31)^8}{64}\)) * Remainder(­\(\frac{17}{64}\))­

= Remainder(­\(\frac{((-31)^2)^4}{64}\)) * Remainder(­\(\frac{17}{64}\))­

⇒ Remainder(­\(\frac{(-31)^2)}{64}\))

⇒ Remainder(­\(\frac{961}{64}\)) = 1

= Remainder(­\(\frac{(1)^4}{64}\)) * Remainder(­\(\frac{17}{64}\))­

\(1 * 17 = 17\)

Option C

Another method to solve this question is using theorems. However, such concepts are not required to solve any questions in GMAT.­

­Great thanks. So re: your comment about theorems which ones are you suggesting? Binomial theorem? Would you be able to describe the theorem you think is applicable in this case?

I think Fermat's little theorem and the remainder theorem I was not able to really apply.

I ended up taking 17 square out as 289 then that gave me a remainder of 33 when divided by 64 (289-256) and I ended up reducing from there.

Thanks!
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Re: ­Find the remainder when 17^17 is divided by 64: [#permalink] New post   29 Mar 2024, 02:05
gmatophobia wrote:
playthegame wrote:
­Find the remainder when 17^17 is divided by 64:

A 0
B 1
C 17
D 34
E 64­

­
Remainder(\(\frac{17^{17}}{64}\)) = ­\(\frac{17^{16}*17}{64}\)

= Remainder(­\(\frac{17^{16}}{64}\)) * Remainder(­\(\frac{17}{64}\))

= Remainder(­\(\frac{((16-1)^{2})^8}{64}\)) * Remainder(­\(\frac{17}{64}\))­



 

Is this a typo or am I lacking some concept?

How \(17 = 16-1 \)

Thank you.­
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Re: ­Find the remainder when 17^17 is divided by 64: [#permalink] New post   29 Mar 2024, 02:17
stne wrote:
Is this a typo or am I lacking some concept?

How \(17 = 16-1 \)

Thank you.­

­Corrected typo. 

Hope it helped.
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Re: ­Find the remainder when 17^17 is divided by 64: [#permalink] New post   30 Mar 2024, 06:09
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­This is a problem of Binomial expansion

17 ^ 17  divided by 64  ( we should note 64 can be written as 16 * 4 )

hence we can write 17 ^ 17 as  ( 16 + 1 ) ^ 17

(x + y) ^ 17 Binomial expansion is :

C0 * x^17 * y^0 + C1 * x^16 * y^1 + ..... + C16 * x^1 * y^16 + C17 * x^0 * y^17

C0 * 16^17 * 1^0 + C1 * 17^16 * 1^1 + ..... + C16 * 16^1 * 1^16 + C17 * 16^0 * 1^17

The portion marked in green is the only portion not divisible by 64 rest all terms have atleast 16^2 as power hence divisible by 64

C16 * 16^1 * 1^16 + C17 * 16^0 * 1^17 = 17 * 16 * 1 + 1  =  16(16 + 1) + 1

Again 16*16 is divisible by 64 .. remiander is 16*1 + 1 = 17
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Re: ­Find the remainder when 17^17 is divided by 64: [#permalink] New post   30 Mar 2024, 09:14
(16+1)^17=17C1 (16)^17+17C1 (16)^17+… 17C1(16)+17C0
=16^2 z+ 17x16+1=16^2a+16+1=16^2b+17

Remainder is 17

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Re: ­Find the remainder when 17^17 is divided by 64: [#permalink]
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