VeritasPrepKarishma wrote:
gmattokyo wrote:
If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?
A. 47
B. 48
C. 49
D. 50
E. 51
As suggested above, the quickest way to solve this question would be to ballpark it.
We know we are looking for 43 + 44 + 45...... = 372
Now 40*8 = 320.... 372 is more than 320
40*9 = 360.... but it is very close to 372. When we add the numbers, we add 48, 49 etc which have an excess of 8, 9 etc as compared to 40. Hence, we need the sum to be some difference away from 372.
Therefore, we need to add only 8 numbers, not 9.
Keep in mind, when going from 43 to n to get 8 numbers, n will be 50, not 51.
43 to 50 give us 8 numbers (50 - 43 + 1).
We can also use some divisibility properties, because we have a sequence of evenly spaced integers.
If the number of terms in the sequence is odd, the sum of the numbers is a multiple of the middle term.
If the number of terms is even, the total sum is a multiple of the sum of the two middle terms.
If we factorize 372 we get 2*2*3*31. By combining the factors we cannot get a divisor of 372 in the range of 40-50, but 3*31 = 93 = 46+47.
So, we have an even number of terms, with 46 and 47 the two middle terms, first term 43, then the last must be 50.
Not shorter than your solution, but it's fun to play with divisibility rules, sometimes just for the sake of practice...
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PhD in Applied Mathematics
Love GMAT Quant questions and running.