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555-605 Level|   Exponents|   Remainders|      
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Abhishek009
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When 51^25 is divided by 13, the remainder obtained is: 26 Jan 2024, 08:08
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LM
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0
\(51*51 = 2601\)

\(\frac{2601}{13} =\) Rem \(1\)

Now, \(51^{25} = 51^{12*2+1}\)

Thus, \(\frac{51^{25}}{13} = \frac{51^{12*2+1}}{13} = \frac{51^{24}}{13}*\frac{51}{13} \)

\(\frac{51^{24}}{13} =\) Rem \(1\), thus the question boils down to \(\frac{51}{13}\)


Calculate the rem of \(\frac{51}{13} = 13*3 + 12\), Rem is \(12\), Answer must be (A)
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When 51^25 is divided by 13, the remainder obtained is: 11 Mar 2024, 11:24
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Quote:
 When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0
­Why don't we solve it with the method of cyclicity? As in other similar questions (for instance, remainder of 43^86 divided by 13), I took the unit digit of 51 which has infinite cyclicity (remains 1) and I divided it with the 13 and the remainder is 1.­
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When 51^25 is divided by 13, the remainder obtained is: 11 Mar 2024, 11:41
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Gmatguy007
Quote:
 When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0
­Why don't we solve it with the method of cyclicity? As in other similar questions (for instance, remainder of 43^86 divided by 13), I took the unit digit of 51 which has infinite cyclicity (remains 1) and I divided it with the 13 and the remainder is 1.­


I believe your doubt is addressed here:

https://gmatclub.com/forum/when-51-25-i ... l#p1427592­
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When 51^25 is divided by 13, the remainder obtained is: 11 Mar 2024, 12:06
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Bunuel
Gmatguy007
Quote:
  When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0
­Why don't we solve it with the method of cyclicity? As in other similar questions (for instance, remainder of 43^86 divided by 5), I took the unit digit of 51 which has infinite cyclicity (remains 1) and I divided it with the 13 and the remainder is 1.­
 

I believe your doubt is addressed here:

https://gmatclub.com/forum/when-51-25-i ... l#p1427592­
­
I got it, so whenever we are interested only in the last digit we use cyclicity, whereas when we want to find the remainder of x^y when divided by z we use binomial. The only exception where we can use cyclicity for the remainder is when the divisor is either 2,5,10 like in the example I mentioned.

Thanks!
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When 51^25 is divided by 13, the remainder obtained is: 10 Apr 2024, 13:00
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KarishmaB
LM
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0
The method used by Bunuel above is the best way to get to the answer.
­KarishmaB Bunuel GMATCoachBen & others, I also chose (A) but by thinking that the remainder will be the same as if we divided 51 by 13. I tested with an example: remainder of 5/2 is 1 and remainder of 5^4/2 is 1.

Is my reasoning correct?­
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When 51^25 is divided by 13, the remainder obtained is: 11 Apr 2024, 00:24
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KarishmaB
LM
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0
The method used by Bunuel above is the best way to get to the answer.
­KarishmaB Bunuel GMATCoachBen & others, I also chose (A) but by thinking that the remainder will be the same as if we divided 51 by 13. I tested with an example: remainder of 5/2 is 1 and remainder of 5^4/2 is 1.

Is my reasoning correct?­
­
This is not correct. Division by 2, 5 or 10 is different and exponents of 5 always end with 5.  
The explanation of that is rather long and you can check it out on Sunday (details in my signature below) but this logic doesn't work with all dividends and divisors.

Consider this:
Divide 8 by 3. Remainder 2.
Divide 8^2 by 3. Remainder 1.
Divide 8^3 by 3. Remainder 2. 
Divide 8^4 by 3. Remainder 1.

There is a cyclicity to it but the remainders are not always the same.  
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When 51^25 is divided by 13, the remainder obtained is: 11 Apr 2024, 04:02
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KarishmaB
Gmatguy007
­KarishmaB Bunuel GMATCoachBen & others, I also chose (A) but by thinking that the remainder will be the same as if we divided 51 by 13. I tested with an example: remainder of 5/2 is 1 and remainder of 5^4/2 is 1.

Is my reasoning correct?­
­
This is not correct. Division by 2, 5 or 10 is different and exponents of 5 always end with 5.  
The explanation of that is rather long and you can check it out on Sunday (details in my signature below) but this logic doesn't work with all dividends and divisors.

Consider this:
Divide 8 by 3. Remainder 2.
Divide 8^2 by 3. Remainder 1.
Divide 8^3 by 3. Remainder 2. 
Divide 8^4 by 3. Remainder 1.

There is a cyclicity to it but the remainders are not always the same.  
 
­Okay, so it was just a coincidence. thank you for the assistance KarishmaB
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When 51^25 is divided by 13, the remainder obtained is: 29 May 2024, 17:22
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Bunuel
LM
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0
\(51^{25}=(52-1)^{25}\), now if we expand this expression all terms but the last one will have \(52=13*4\) in them, thus will leave no remainder upon division by 13, the last term will be \((-1)^{25}=-1\). Thus the question becomes: what is the remainder upon division -1 by 13? The answer to this question is 12: \(-1=13*(-1)+12\).

Answer: A.
­Hello Bunuel - Quick question as this appears several times in this style of question. How is "-1 [divided] by 13" 12? Wouldn't it be 0.769?
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When 51^25 is divided by 13, the remainder obtained is: 03 Jun 2025, 05:21
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