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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
Show more
\(51*51 = 2601\)

\(\frac{2601}{13} =\) 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
Show more
­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.­
Show more


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
Show more
­
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

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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.
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­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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Gmatguy007

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?­
Show more
­
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.  
 
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­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

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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.
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­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: 19 Aug 2025, 07:23
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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, when we expand this expression, all terms except the last one will contain \(52=13*4\), so they leave no remainder upon division by 13. The last term will be \((-1)^{25}=-1\). Therefore, the question becomes: what is the remainder when -1 is divided by 13? The answer is 12, as \(-1=13*(-1)+12\).

Answer: A.
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not able to understand the concept. Can you elaborate?
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When 51^25 is divided by 13, the remainder obtained is: 19 Aug 2025, 07:37
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ayuj2401
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, when we expand this expression, all terms except the last one will contain \(52=13*4\), so they leave no remainder upon division by 13. The last term will be \((-1)^{25}=-1\). Therefore, the question becomes: what is the remainder when -1 is divided by 13? The answer is 12, as \(-1=13*(-1)+12\).

Answer: A.
not able to understand the concept. Can you elaborate?
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Step 1: Rewrite 51
Since 52 is a multiple of 13, it’s convenient to write 51 as 52 - 1. This makes expansion easier because every term with a factor of 52 will automatically be divisible by 13.

Step 2: Expand (52 - 1)^25
When you expand (52 - 1)^25, you get a sum of terms. Every term except the very last one has 52 in it, multiplied by some coefficient.

Step 3: Remove divisible terms
Because 52 is divisible by 13, all those terms with a factor of 52 are also divisible by 13 and leave remainder 0.

Step 4: The last term
The only term left after ignoring all multiples of 13 is the final one, which comes from (-1)^25. That value is -1.

Step 5: Find the remainder
Now we want the remainder when -1 is divided by 13. We can write -1 as 13 * (-1) + 12, so the remainder is 12.

Thus, remainder when 51^25 is divided by 13 is 12.

For more details, check the previous discussion.
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