The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:

We can approach this problem with algebra or by plugging numbers. Even though the latter’s probably faster and easier, it’s good to have both tools up your sleeve.

Algebraic solution (harder):

The multiples of x are x, 2x, 3x, etc. The square of x is also a multiple of x, of course—it’s just x times x, or the xth multiple of x.

So the list we care about is x, 2x, 3x… up to x².

The sum of these numbers can be written as x + 2x + 3x + … + x².

We can now factor out an x to get x(1 + 2 + 3 + … + x).

Now, the sum that remains (that is, 1 + 2 + 3 + … + x) is a sum of consecutive integers, which is evenly spaced.

It’s easy to calculate the average (arithmetic mean) of an evenly spaced set: just add up the outermost numbers and divide by 2:

Average of {1, 2, 3, …, x} = (x + 1)/2

The number of numbers in that set is just x, since there are x consecutive integers between 1 and x, inclusive. (That sounds harder than it is! If x is 3, then all we’re saying is that in the set {1, 2, 3}, there are 3 numbers.)

Now, back to {1, 2, 3, …, x}. Since the average equals the sum divided by the number of numbers, the sum equals the average times that number of numbers.

So 1 + 2 + 3 + … + x = Average × Number = [(x + 1)/2]x

Finally, to get the original sum, x + 2x + 3x + … + x², we just multiply by x again to get x²(x + 1)/2.

Plugging numbers:

Pick x = 2. The sum of multiples of 2 from 2 to 2² is just 2 + 4 = 6. Check the answers:

(A) 2(3)(1) = 6

(B) 2²(3 + 1)/2 = 6

(C) 2²(2 – 1) = 4

(D) (2³ + 2×2)/2 = 6

(E) 2(2 – 1)² = 2

Okay, all we can eliminate is C and E. Try x = 3. The sum of multiples of 3 from 3 to 3² is just 3 + 6 + 9 = 18. Check the remaining answers:

(A) 3(4)(2) = 24

(B) 3²(3 + 1)/2 = 18

(D) (3³ + 2×3)/2 = 16.5