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13 Oct 2025, 00:37
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
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Question Stats:
81% (00:45) correct
19%
(01:06)
wrong
based on 101
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If the sum of three consecutive integers is odd, then the first and the last integers must be
(A) odd, even
(B) odd, odd
(C) even, odd
(D) even, even
(E) none of the above
(A) odd, even
(B) odd, odd
(C) even, odd
(D) even, even
(E) none of the above
13 Oct 2025, 00:55
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Let the three consecutive integers be
n,n+1,n+2.
The sum is
n+(n+1)+(n+2)=3n+3=3(n+1), which is always a multiple of 3, and will be odd only if
n+1 is odd (i.e., n is even).
So
n (first integer) is even, and n+2 (last integer) is also even.
Thus, the correct answer is:
(D) even, even.
n,n+1,n+2.
The sum is
n+(n+1)+(n+2)=3n+3=3(n+1), which is always a multiple of 3, and will be odd only if
n+1 is odd (i.e., n is even).
So
n (first integer) is even, and n+2 (last integer) is also even.
Thus, the correct answer is:
(D) even, even.
13 Oct 2025, 08:42
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Bunuel
Let the three consecutive digits be a, b, c respectively. Where a is the first term, and c is the last term.
Given that : a+b+c = odd.
We know that:
Even + Even = Even
Even + Odd = Odd
Odd + Even = Odd
Odd + Odd = Even.
There are two cases ,
CASE 1:
Even + Odd + Even = odd + Even = Odd.
CASE 2:
Odd + Even + Odd = Odd + Odd = Even.
Case 1 satisfies. There fore the First term and Third term are both Even.
Option D
