Mathletes Problem

[magicdanw]

Hi. I just got back from a Mathletes competition. 5/6 correct! :biggrin: But, the one I didn't get right, I can't figure out how to do. Here's the problem:

[DarkerLine]

The usual method of finding units digits of high powers is to recognize, first, that only the units digit of the base of the exponent matters (the units digit of 6847¹⁰² is the same as the units digit of 7¹⁰²), and second, that the units digits repeat (with a cycle of at most 4, actually), so since the units digits of 7^x repeat as 7,9,3,1,7,.. , the units digit of 7¹⁰² is the same as the units digit of 7²=49, so it's 9.

In this case, you're lucky that the top is 72 and not 77, because then the result is actually an integer. You can factor 72 as 2³*3² and 96 as 2⁵*3, so the fraction expands as:

2^3*153*3^2*153/(2^5*49*3⁴⁹).

This simplifies to be 2²³⁴*3²⁵⁷. Using the method I outlined in the first paragraph, you know that the units digit is the same as that of 2²*3¹, which is 12, so the units digit is 2.

[magicdanw]

Thanks!

[luby]

[DarkerLine]

You take the remainder of the power when divided by 4 (except you'd take 4 instead of 0 in the situations where it matters). The 4 is because units digits repeat every 4 steps, which is either a funny observation or a consequence of a theorem about prime numbers.