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

1 2 3

sanjitscorps18

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27 Dec 2025, 06:35

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Let
a = x-1
b = x
c = x+1

Now (x-1)^(x+1) divided by x
Taking (x-1)mod(x) we get the remainder as -1

If x is odd, then x+1 is even
=> Remainder = (-1)^even = 1

If x is even, then x+1 is odd
=> Remainder = (-1)^odd = -1
Since this is a negative remainder we add the divisor => the remainder = -1+x = x-1
Since x = b we get remainder = b-1

The divisor b itself cannot be remainder as it would lead to remainder 0 and for that the other numbers have to multiples while it is given that these numbers are consecutive. Not possible.

Hence only I and II are possible

Option D

Bunuel

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

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firefox300

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27 Dec 2025, 06:42

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3 consecutive integers: b-1, b and b+1

(b-1)^(b+1) can be expanded and all the addends will be divisible by b except (-1)^(b+1)

remainder = (-1)^(b+1)

I. 1: true if b is odd
II. b-1: true if b is even (remainder=-1)
III. b: it's never true because remainder<b

The correct answer is D

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27 Dec 2025, 07:08

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Using remainder theorem:

Remainder ((b-1)^(b+1))/b = (-1)^(b+1)

That can be 1 or -1 depending if b is odd or even.
Remainder = -1 = divisor - 1 = b-1

III is not possible as remainder must be less than divisor.

The answer is D

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27 Dec 2025, 07:24

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a=b-1 < b < c=b+1

a^c = (b-1)^(b+1) = (b+1)C0 * b^(b+1)*(-1)^0 + (b+1)C1 * b^b*(-1)^1 + ... + (b+1)Cb * b^1*(-1)^b + (b+1)C(b+1) * b^0*(-1)^(b+1)

All the terms of a^c are divisible by b except the last one: (b+1)C(b+1) * b^0*(-1)^(b+1) = (-1)^(b+1), which is the remainder.

(-1)^(b+1) = 1 if b odd (as in I).
(-1)^(b+1) = -1 if b even, if remainder is -1, remainder = b-1 (as in II).

A remainder must be less than b so it can never equal b (as in III).

Answer D

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27 Dec 2025, 07:34

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Bunuel

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We need to understand that consecutive numbers are coprimes, so b can not divide any power of a so III is out.

Now a rem b is always b-1 or -1

So the reminder would depend on power raised if it's odd remainder will be -1 aka b-1
if it's even reminder is +1 1

So I and II possible.

Correct Answer D

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27 Dec 2025, 07:44

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All consecutive integers a, b and c are greater than 10

NB a<b<c

Let's take an even number for b, say, 14

a = 13 & c = 15

When $a^c /b$ in this case, we get $13^{15} / 14$. The remainder here would be $(13-14)^{15}$ which gives us -1. Therefore, the remainder would be 14-1 = 13 ---> which is a (or b-1)

So option II is plausible.

If b is odd, say, b = 27, a = 26 and c = 28

For $26^{28} /27$, the remainder $(26-27)^{28}$ which gives us a nice, round 1 (since the power is even). That checks out option I.

This pattern applies for all even and odd numbers in the given range.

And finally, it isn't possible to get a remainder equal to the divisor (b).

Therefore, the answer is D. I & II only

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27 Dec 2025, 07:54

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consecutive numbers: n-1, n, n+1 with n>=12

binomial theorem:
remainder of ((n-1)^(n+1))/n = (-1)^(n+1)

Two cases:
if n+1 is even (n is odd) -> remainder = 1
if n+1 is odd (n is even) -> remainder = -1, what means remainder = n(divisor) - 1

From the options, I and II are allowed (in fact, these are the only two possible remainders).
Remainder can never be equal to divisor as indicated in III.

IMO D

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27 Dec 2025, 08:01

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10<a<b<c => q=n, b=n+1, c=n+2
We are asked =>

n^(n=2) (mod n+1) (remainder when a^c is divided by b)

We can see that
n mod (n+1) = -1 (b = a+1)
Hence, n^(n+2) = (-1)^(n+2) mod (n+1)

Now, when n is odd => n+2 is odd=> remainder = -1 => b-1
When n is even => n+2 is even => remainder is 1 => 1

Hence the remainders can be 1 and b-1 => D

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27 Dec 2025, 08:13

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Logic : a, b ,c are consecutive integers (a=b-1,c=b+1)
Expression: (b-1)^b+1 divided by b (mod b)

using Binomial expansion or remainders : (-1)^b+1 (mod b)

Case 1 : If b is even : b+1 is odd . Remainder is (-1)^odd= -1 ≡b-1(mod b) ✅
Case 2 : (If b is odd ) : b+1 is even . Remainder is (-1) ^even = 1 (mod b)✅
Case 3 : A remainder must be strictly less than it's divisor . Therefore can't be b .❌

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27 Dec 2025, 08:41

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We know that,
10 < a < b < c

=> a = b - 1
=> c = b + 1

We need to know remainder when a^c / b

=> a^c = (b-1)^(b+1)
Let symbol for modulo be ~=
We use the concept of modulo here
=> (b-1) ~= -1 (mod b)
=> (b-1)^(b+1) ~= (-1) ^ (b+1) (mod b)

We check for different cases for when b is even or odd

b is even:
=> b+1 is odd and b-1 is odd
=> (-1)^b+1 = -1 ~= b-1 (mod b)

=> Remainder = b - 1

b is odd:
=> b+1 is even and b-1 is even
=> (-1)^(b+1) = 1

=> Remainder = 1

b as a remainder is not possible ever since the given expression is divided by b, hence remainder is always lesser than b

D. I and II only

-----------------------------------------------------------------------

If modulo concept is hard to grasp, then one could
try substituting
a=11,b =12, c=13 ----> Remainder = 11 = b-1
a = 12, b = 13 , c=14 -----> Remainder = 1
b is never possible since remainder always lesser than b

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29 Mar 2026, 02:45

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This is straightforward.
Consecutive integers thus the reminder if of the form (-1)^x

x can be even or odd.
Thus Remainder = 1 or -1. -1 = b-1
Hence 1,2 are true.

Answer: Option D

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Let me break this down step by step.

Since a, b, c are consecutive integers: b = a + 1 and c = a + 2.

Key Insight: Since b = a + 1, we know a = b - 1. This means when you divide a by b, the remainder is b - 1, which behaves exactly like -1. In math terms: a is equivalent to -1 when dividing by b.

So a^c divided by b gives the same remainder as (-1)^c divided by b.

Now there are only 2 cases:

Case 1 — c is even: (-1)^even = 1, so remainder = 1.
Example: a = 12, b = 13, c = 14. Then 12^14 divided by 13 gives remainder 1. ✓

Case 2 — c is odd: (-1)^odd = -1, which is the same as b - 1 when dividing by b.
Example: a = 11, b = 12, c = 13. Then 11^13 divided by 12 gives remainder 11 = 12 - 1 = b - 1. ✓

Since c = a + 2, when a is even c is even (Case 1 applies), and when a is odd c is odd (Case 2 applies). Both scenarios are possible.

So Statement I (remainder = 1) — YES, possible.
So Statement II (remainder = b - 1) — YES, possible.
Statement III (remainder = b) — IMPOSSIBLE. A remainder must always be less than the divisor. You can never get a remainder of b when dividing by b.

Answer: D — I and II only.