12 Days of Christmas GMAT Competition 2025 - Day 10: If a, b, and c

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Bunuel

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26 Dec 2025, 10:00

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Bunuel

GMAT Club Official Explanation:

We have three consecutive integers a, b, and c with 10 < a < b < c. So we can write a = b - 1 and c = b + 1.

We want the remainder when (b - 1)^(b + 1) is divided by b.

When you expand this expression, all terms except the last one will be multiples of b and thus will be divisible by b. The last one will be (-1)^(b + 1). Hence, the question essentially boils down to finding the remainder when (-1)^(b + 1) is divided by b.

If b is odd, then b + 1 is even, so (-1)^(b + 1) = 1. And 1 divided by b will give the remainder of 1.

If b is even, then b + 1 is odd, so (-1)^(b + 1) = -1. And -1 divided by b will give the remainder of b - 1. For example, -1 divided by 20 gives the remainder of 20 - 1 = 19.

Answer: D.

P.S. The process for finding the remainder when dividing a negative integer by a positive integer follows the same principles as when dividing a positive integer by a positive integer.

For example:

What is the remainder when dividing 21 by 6? We find the closest multiple of 6 that is less than 21, which is 18. Then, we calculate 21 - 18, yielding a remainder of 3.
What is the remainder when dividing -23 by 7? Here, we find the closest multiple of 7 that is less than -23, which is -28. Then, we calculate -23 - (-28), resulting in a remainder of 5.

What about dividing -1 by 26? The closest multiple of 26 less than -1 is -26. So, the remainder is -1 - (-26) = 25.

Alternatively, consider this method:

What is the remainder when dividing -23 by 7? Dividing 23 by 7 gives a remainder of 2. To find the remainder for -23 divided by 7, subtract this 2 from the divisor. Thus, the remainder when dividing -23 by 7 is 7 - 2 = 5.

Similarly, what is the remainder when dividing -1 by 26? Dividing 1 by 26 gives a remainder of 1. Therefore, the remainder when dividing -1 by 26 is 26 - 1 = 25.

Hope it helps.

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26 Dec 2025, 09:16

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Bunuel

a^n/(a-1) will always have reminder 1
since b = a-1
the remainder will be alwys 1

I is true

for II) when b =2, remainder is b-1 = 1 hence holds true
a= 3, b= 2, c =1 remainder is 1 which is b-1

hecne I and II can be true
answer is D

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26 Dec 2025, 09:27

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If a, b, and c are consecutive integers, where 10 < a < b < c, which of the following could be the remainder when a^c is divided by b?

I. 1
II. b - 1
III. b

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

given
10<a<b<c
a = b-1 ; c= b+1

a^c divided by b will give remainders
when c is even then b is odd and remainder will be 1
when c is odd then b is even and remainder will be b-1

OPTION D ; I & II only

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26 Dec 2025, 09:32

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10<a<b<c
remainder when a^c divided by b

assume a=11 b=12 c=13
11^13/12 remainder is 11 as per option it is b-1

assume a=12 b=13 c=14
12^14/2 remainder is 1
so answer is both 1 and 2

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26 Dec 2025, 09:36

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b=a+1
c=a+2

a^(a+2)/(a+1)

(a+1-1)^(a+2)/(a+1)
(-1)^(a+2)

This can be either 1 or -1 depending on a being even or odd
Remainder >=0, so 2 possibilities are 1, b-1
I and II

Bunuel

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26 Dec 2025, 09:49

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Bunuel

Since a, b, c are consecutive integers with a < b < c, we have a = b - 1 and c = b + 1

Mod b: a = b - 1 = -1
so a^c = (-1)^(b + 1) (mod b)

if b is odd => b + 1 is even => remainder = 1
for ex: (a,b,c) = (12,13,14) then 12^14 divided by 13 leaves remainder 1

if b is even => b + 1 is odd => remainder = -1 = b - 1
for ex: (a,b,c) = (11,12,13). Then 11^13 divided by 12 leaves remainder 11 = b - 1

Remainder cant be b ( that would mean remainder is 0), which never happens here

Option D

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a, b & c are consecutive integers, where 10 < a < b=a+1 < c= a+2
Which of the following could be the remainder when a^c is divided by b?
a>10
b = a+1 > 11
c = a+2 > 12

The remainder when a^c = (b-1)^(b+1) is divided by b
= The remainder when (-1)^c is divided by b
= 1 when c is even
= b - 1 when c is odd

I. 1: Possible
II. b-1: Possible
III. b: Not possible since the remainder < divisor (b)

IMO D

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26 Dec 2025, 10:31

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Bunuel

The remainder of (a/b) will be -1 because a and b are consecutive integers.

(-1)^c can be -1 when c is odd or 1 when c is even. As we don't know the even odd nature of c, we can't conclude.

If c is odd, the remainder = b -1 , else the remainder is 1.

Option D

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26 Dec 2025, 11:14

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Bunuel

The stem is asking for "could be true", which means for each answer choice we just need to prove once that it can happen for the answer choice to be correct. Let's take a=11, b=12, c=13. If we look at powers of 11 and divide them by 12, we get:
$\frac{11^1}{12}=0R11$, $\frac{11^2}{12}=1R1$, $\frac{11^3}{12}=11R11$, $\frac{11^4}{12}=122R1$

This indicates a pattern in the remainder: odd powers leave remainder 11, even powers leave remainder 1. We can extrapolate this out to c=13 and onwards.

I. From the power of 11 example above, we know that 1 remainder is possible. This is correct.
II. From the example above, we know that b-1 is possible. This is correct.
III. This can never be true. If we were dividing by b and there was b left over, we could divide that by b and increase the quotient by 1. Remainder always has to be less than the divisor. This is incorrect.

Therefore D is our answer.

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26 Dec 2025, 11:47

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Remainder of a^c/b = Remainder of (a/b) to the power c
b=a+1
Remainder of a/b =-1
Ans= (-1)^c
if c is even, Remainder is 1. if c is odd, Remainder = -1 = b-1

Ans D

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the remainder can never be equal or bigger than the divisor, so option III is not possible

example for options I and II that work
a = 11, b = 12, c = 13
11^13 / 12 -> remainder of b-1

a = 12, b = 13, c = 14
12^14 / 13 -> remainder of 1

Answer D

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Firstly, division by b remainder will be between 0 and b-1
So statement 3 is not possible

Note: % means remainder (like in C language or any programming languages)

if a=11, b=12, c=13

$a^{c}$ % b

$11^{13}$ % 12

$(11 \text{ % }12) ^ {13}$

$(-1)^{13}$

-1

remainder = -1 or 11 or in other words b-1

if a=12, b=13, c=14

$a^{c} $ % b

$12^{14}$ % 13

$(12 \text{ % }13)^{14}$

$(-1)^{14}$

1

reminder = 1

so statement 1 and 2 are possible

ans: Option D

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26 Dec 2025, 12:33

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given a, b & c are consecutive integers such that 10<a<b<c.
we need to find the remainder when a^c is divided by b?
lets discuss the only two possible cases regarding the scenario;
1. a=11 b=12 c=13 when a is odd
remainder of (11^13)/12=> (-1)^13 = -1 = 11= a which is equal to (b-1) so II is correct.
(using concept of negative remainders)
2. a=12 b=13 c=14 when a is even
remainder of (12^14)/13=>(-1)^14 = 1 so I is correct
also III can never be true as its the divisor itself and a divisor can never be a remainder.
so D I & II correct

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Bunuel

I dont know how i will explain this , but let me try
eg- If 2 is divide by 3 , remainder is 2, but in remainder concept we can say remainder modulo = -1,
and if 3/2 ... remainder modulo = rem = 1 (remains same)
since now if we add 3 to -1 we get the remain = 2 now lets move to the question
let a= k, b= k+1, c= k+2
concept--- remainder (a X b)= remainder(a) X remainder (b)
Rem(a^c)/b= rem (a/b) x rem(a/b)..........till c terms
now we know that a=k and b= k+1
so modulo rem (a/b)= -1
so rem [(a^c)/b]= (-1)^ c

Now if c= even , rem= 1
if c= odd , rem modulo = -1,
rem= b-1
Ans=D

or

if u don't want to go this way
use smaller nos like 2,3,4... and u will get the answer

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26 Dec 2025, 14:27

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Bunuel

A,B,& C are consecutive integers

B is not possible so we have 1 & 2 left

a =11 b = 12 and c = 13

11^13 / 12 = reminder is 12 so b - 1 is possible

a= 12 b = 13 and c =14

12^14 / 13 = reminder is 1 so 1 is possible

So Ans - D

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26 Dec 2025, 17:00

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Say a, b, c are 12, 13, and 14 respectively

The multiples of 12 ends up with the unit digits 2,4,8 and 6 and the cycle repeats every 4 multiples. So, 12^14 ends up with a unit digits of 4. Now every multiple of 12 that has a unit digit of 4, when divided by 13 leaves a remainder of 1. For eg, 12^2=144 and 144÷13 has a remainder of 1.Same occurs with (12^14)÷13.

Another example, if a,b,c were 15,16 and 17 respectively all multiples of 15 ends up with a 5.
(15^1)÷16 leaves a remainder of 15 ie.b-1. Same happens with (15^17)÷16.

So,answer D, remainder 1 or b-1

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26 Dec 2025, 20:41

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What would be the remainder when a^c is divided by b, given a,b,c are consecutive.

2 cases are possible - Even, Odd, Even & Odd, Even, Odd

Case 1 : 11,12,13 ->
11^13/12 can be written with negative remainders as (-1)^13 = -1, or (b-1)

Case 2 : 12,13,14 ->
12^14/13 = (-1)^14 = 1

As for the third option, a division by b can never yield a remainder of b, it has to be less than b.

So the answer is Option 1&2

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26 Dec 2025, 22:39

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Condition is given such that 10<a<b<c , where a ,b ,c are consecutive integers. Let's take few examples to solve this question.

Let a = 11 ,b = 12 , c= 13.

11^13 when divided by 12 will give remainder 11. It means the value is similar to b-1. So , II will be correct.

Let's a =12 ,b =13 , c=14 .

12^14 when divided by 13 will give remainder 1. So I is also correct.

And if we take any further example then remainder will either be II or I. It cannot be divisor itself.

So answer is I,II which is D

Bunuel