## Blog

### 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 are told that n = kw + r, with a number of conditions on the possible values of the variables (k and n are positive integers, w is an integer, and r is non-negative). An example is given to us:

13 = 32 + 4. The question is this: for a given k and w, what is the largest possible value of r?

Let’s keep with the example. The given k and w are 3 and 2, respectively, in the expression 32. The question becomes “how big can r get?” At first, it might seem that there’s no cap on the size of r, but if you consider n = 30, for instance, you can write it using a larger power of 3:

30 = 32 + 21
30 = 33 + 3

So the first equation doesn’t fit the conditions (w has to be the largest integer such that r is not negative).

The tipping point is the next power of 3, namely 33 = 27. 27 itself would be written as 33 + 0, with r = 0, so the number that gives the largest r for k = 3 and w = 2 must be 26:

26 = 32 + 17, giving r = 17.

At this point, we could take a number-testing approach: which answer choice equals 17 when k = 3 and w = 2? After a little computation, we’d find that the answer is (A).

We can also take a more algebraic approach. The maximum r is going to come when n is the integer just below the next power of k above kw, in other words when n equals kw+1 – 1.

Plug this expression for n into the equation:

kw+1 – 1 = kw + r

Now split kw+1 into k(kw):

k(kw) – 1 = kw + r

Finally, subtract kw from both sides:

(k – 1)(kw) – 1 = r

The left side matches the expression in choice (A).

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