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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:

Since there are five labels, given in order to all the integers, the label alpha is given to 0, 5, 10, etc. – that is, the alpha’s are the multiples of 5 and end in 0 or 5. All the other labels correspond to non-multiples of 5 – in fact, they each correspond to particular remainders and particular units digits. For instance, the beta’s (1, 6, 11, 16, etc.), which all end in 1 or 6, also all leave a remainder of 1 after division by 5. The gamma’s correspond to a remainder of 2 (units digits = 2 or 7). Delta’s correspond to a remainder of 3 (units digits = 3 or 8), and epsilon’s correspond to a remainder of 4 (units digits = 4 or 9).

Now, a gamma raised to the seventh power will be large, even if we pick the smallest gamma (2 itself). But all we need is the units digit of the result. So compute the units digit in stages:

First power: units digit = 2
Second power: units digit = 2×2 = 4
Third power: units digit = 2×4 = 8 (remainder = 3)
Fourth power: units digit = 2×8 = 16 = …6 (units digit only) (remainder = 1)
Fifth power: units digit = 2×6 = 12 = …2 (units digit only) (remainder = 2)
Sixth power: units digit = 2×2 = 4 (remainder = 4)
Seventh power: units digit = 2×4 = 8 (remainder = 3)

Do the same for the delta.

First power: units digit = 3
Second power: units digit = 3×3 = 9 (remainder = 4)
Third power: units digit = 3×9 = 27 = …7 (remainder = 2)
Fourth power: units digit = 3×7 = 21 = …1 (remainder = 1)
Fifth power: units digit = 3×1 = 3 (remainder = 3)
Sixth power: units digit = 3×3 = 9 (remainder = 4)
Seventh power: units digit = 3×9 = 27 = …7 (remainder = 2)

Gamma7 gives us a remainder of 3. Delta7 gives us a remainder of 2. Adding the remainders, we get a remainder of 5, which is the same as a remainder of 0 (remember, we’re talking about division by 5).

So the sum gets a label of alpha.

The correct answer is A.

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