If m equals the sum of the even integers from 2 to 16

Question

If m equals the sum of the even integers from 2 to 16, inclusive, and n equals the sum of the odd integers from 1 to 15, inclusive, what is the value of m - n ?

(A) 1
(B) 7
(C) 8
(D) 15
(E) 16

Bunuel · Accepted Answer

Even integer from 2 to 16, inclusive (2, 4, 6, ..., 16) as well as odd integers from 1 to 15, inclusive (1, 3, 5, ..., 15) represent evenly spaced set (aka arithmetic progression). Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms, where the mean of the set is (first+last)/2. (Check Number Theory chapter of Math Book for more: math-number-theory-88376.html) There are 8 even integers from 2 to 16, inclusive, their sum equals to (first+last)/2*(number of terms)=(2+16)/2*8=72; There are 8 odd integers from 1 to 15, inclusive, their sum equals to (first+last)/2*(number of terms)=(1+15)/2*8=64; Difference: 72-64=8, Answer: C. Or: 2+4+6+8+10+12+14+16-(1+3+5+7+9+11+13+15)= (2-1)+(4-3)+(6-5)+(8-7)+(10-9)+(12-11)+(14-13)+(16-15)=1+1+1+1+1+1+1+1=8. Answer: C.

PareshGmat · Answer

2 * 1 = 2

2 * 8 = 16

There are 8 even integers from 2 to 16

For every even integer, there is a 1 less odd integer whose subtraction will give result 1

(16-15) + (14-13) ................. (2 - 1)

This is repeated 8 times = 8 * 1 = 8

Answer = C

babu4478 · Answer

Sum of 1st n odd integers is n^2 & Sum of 1st n even integers is n^2+n

-> (sum of n even integers) - (sum of n odd integers) = (n^2+n) - (n^2) = n

Here, There are 8 even integers between 2 and 16 (inclusive) and 8 odd integers between 1 and 15 (inclusive).

-> n= 8

Ans:C

bumpbot · Answer

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If m equals the sum of the even integers from 2 to 16

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If m equals the sum of the even integers from 2 to 16

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