If S is the sum of x consecutive integers, S must be even if x is a mu

Question

If S is the sum of x consecutive integers, S must be even if x is a multiple of

(A) 6
(B) 5
(C) 4
(D) 3
(E) 2

stne · Answer

Let us try to test the options under various conditions and see if that option can survive.

IF S is a multiple of  2  then let there be  2  consecutive digits:

1+2 = 3   Not even,reject

IF S is a multiple of  3  then let there be  3  consecutive digits:

1+2+3= 6   Even,pass .
 2+3+4= 9  , Not even,reject.

IF S is a multiple of  4   then let there be  4  consecutive digits:

1+2+3+4= 10   EVEN, pass.
 2+3+4+5= 14   EVEN, pass.

It seems no matter how we choose  4  consecutive digits the summation of them always comes to EVEN.

Can there be any more test conditons that we can try to make this  4  digits summation odd? Does not look like so.

Since there can be only one answer among the options we can mark the answer as C and move on.

Alternatively one could further test summation of consecutive  5  and  6  digits and nullify option B and A respectively to arrive at option C.

Ans C

MagnusSamIBA · Answer

If the sum of x integers is even, the sum must have an even number of odd integers because the sum of two odd integers is even.

With x = 2, we get an odd sum (e.g., 1 + 2 = 3), so this doesn't work.

With x = 3, the sum alternates between even and odd (e.g., 1 + 2 + 3 = 6, but 2 + 3 + 4 = 9), so it doesn’t always work.

With x = 4, no matter the starting integer, we always get an even sum (e.g., 1 + 2 + 3 + 4 = 10, 2 + 3 + 4 + 5 = 14). This always works.

So, x = 4 ensures the sum is always even.

Answer: C (4)

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