Manhattan GMAT Challenge Problem of the Week - 19 Mar 10

by , Mar 19, 2010

Welcome back to this week's Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

Question

The tens digit of [pmath]6^17[/pmath] is

(A) 1

(B) 3

(C) 5

(D) 7

(E) 9

Solution

We know that there must be a pattern, since we cant be expected to expand [pmath]6^17[/pmath] out to all its digits. In other words, we must be able to spot a repeating cycle of digits.

The only way forward is to compute tens digits for powers of 6, starting with [pmath]6^1[/pmath], and see what we get. To go up, multiply the previous result by 6 and drop any higher digits than the tens, but we have to keep the units digit (which, as well see, will be 6 every time).

[pmath]6^1[/pmath] = 6 (no tens digit)

[pmath]6^2[/pmath] = 6 [pmath]6^1[/pmath] = 36 (tens digit = 3)

[pmath]6^3[/pmath] = 6 [pmath]6^2[/pmath] = ..16 (tens digit = 1)

[pmath]6^4[/pmath] = 6 [pmath]6^3[/pmath] = ..96 (tens digit = 9)

[pmath]6^5[/pmath] = 6 [pmath]6^4[/pmath] = ..76 (tens digit = 7)

[pmath]6^6[/pmath] = 6 [pmath]6^5[/pmath] = ..56 (tens digit = 5)

[pmath]6^7[/pmath] = 6 [pmath]6^6[/pmath] = ..36 (tens digit = 3)

Whew the numbers finally started repeating! The cycle is 3, 1, 9, 7, 5 which is 5 terms long. Every power will have the same tens digit as the 5th larger power, so [pmath]6^2[/pmath], [pmath]6^7[/pmath], [pmath]6^12[/pmath], and most importantly [pmath]6^17[/pmath] will all have 3 as their tens digit.

Notice that the pattern didnt start until [pmath]6^2[/pmath]. [pmath]6^1[/pmath] doesnt have a tens digit (or has a tens digit of 0, but this digit is never repeated later in the cycle).

The correct answer is (B) 3.

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