## Blog

### The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a 700 level GMAT question on our blog. Today, they have followed up with the answer:

We cannot compute 2200 in anywhere near the time allotted, so we should look for a pattern in much simpler problems that we can scale up to 2200.

The simpler problems we should solve are these:

What is the remainder when you divide 2 by 7?
What is the remainder when you divide 22 by 7?
What is the remainder when you divide 23 by 7?
What is the remainder when you divide 24 by 7?
… and so on with the powers of 2.

2 divided by 7 leaves remainder 2.
22 (which equals 4 ) divided by 7 leaves remainder 4.
23 (which equals 8 ) divided by 7 leaves remainder 1.
24 (which equals 16 ) divided by 7 leaves remainder 2.
25 (which equals 32 ) divided by 7 leaves remainder 4.
26 (which equals 64 ) divided by 7 leaves remainder 1

We should stop as soon as we notice that the cycle will repeat itself forever in this pattern: [2, 4, 1]. Every third remainder is the same. (From here on out, “remainder” always means “remainder after we divide by 7.”) Since every third remainder is the same, we should look at the remainder when the power is a multiple of 3. The remainders of 23 and 26 are 1. Thus, the remainder of 2 raised to a power that is any multiple of 3 is 1.

Now, 200 is not a multiple of 3, but we can look for a multiple of 3 near 200. 201 is a multiple of 3 (its digits add to 3), so 2201 has a remainder of 1. Finally, we notice that the remainder of 2200 must be one position earlier in the cycle than the remainder of 2201. Since the cycle is [2, 4, 1], the remainder of 2200 is 4.

The correct answer is D.

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