Manhattan GMAT Challenge Problem of the Week – 26 Oct 2010

by Manhattan Prep, Oct 26, 2010

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Question

If x is a positive integer greater than 1, what is the sum of the multiples of x from x to [pmath]x^2[/pmath], inclusive?

(A) [pmath]x(x + 1)(x-1)[/pmath]

(B) [pmath]x^2(x + 1)/2[/pmath]

(C) [pmath]x^2(x-1)[/pmath]

(D) [pmath](x^3 + 2x)/2[/pmath]

(E) [pmath]x(x-1)^2[/pmath]

Answer

We can approach this problem with algebra or by plugging numbers. Even though the latters probably faster and easier, its 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 courseits just x times x, or the xth multiple of x.

So the list we care about is x, 2x, 3x up to [pmath]x^2[/pmath].

The sum of these numbers can be written as x + 2x + 3x + + [pmath]x^2[/pmath].

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.

Its 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 were 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 + + [pmath]x^2[/pmath], we just multiply by x again to get [pmath]x^2(x + 1)/2.[/pmath]

Plugging numbers:

Pick x = 2. The sum of multiples of 2 from 2 to [pmath]2^2[/pmath] is just 2 + 4 = 6. Check the answers:

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

(B) [pmath]2^2(3 + 1)/2 = 6 [/pmath]

(C) [pmath]2^2(2-1)=4[/pmath]

(D) [pmath](2^3 + 2*2)/2 = 6[/pmath]

(E) [pmath]2(2-1)^2=2[/pmath]

Okay, all we can eliminate is C and E. Try x = 3. The sum of multiples of 3 from 3 to [pmath]3^2[/pmath] is just 3 + 6 + 9 = 18. Check the remaining answers:

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

(B) [pmath]3^2(3 + 1)/2 = 18[/pmath]

(D) [pmath](3^3 + 2*3)/2 = 16.5[/pmath]

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