JeffTargetTestPrep wrote:

Bunuel wrote:

When 999^500 is divided by 5, the remainder is

A. 0

B. 1

C. 2

D. 3

E. 4

We may recall that when dividing a number by 5, the remainder will be the same as when we divide the units digit of that number by 5. Thus, to determine the remainder when 999^500 is divided by 5, we need to get the units digit of 999^500, or simply 9^500.

Let’s start by evaluating the pattern of the units digits of 9^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 9. When writing out the pattern, notice that we are ONLY concerned with the units digit of 9 raised to each power.

9^1 = 9

9^2 = 1

9^3 = 9

9^4 = 1

The pattern of any base of 9 repeats every 2 exponents. The pattern is 9–1. In this pattern, all positive exponents that are even will produce a 1 as its units digit. Thus:

9^500 has a units digit of 1, and since 1/5 has a reminder of 1, the remainder when 999^500 is divided by 5 is 1.

Answer: B

nymfan14 wrote:

I saw the 1/9 pattern you outlined above but 9^5 has a unit digit of 9. 9/5 = r4 so I chose E. What am I missing?

@nymfan , glad you posted.

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You made a small mistake.

\(9^5\) has a units digit of 9, as does every odd power of 9. And true, a units digit of 9 will leave a remainder of 4.

But \(9^{500}\) is an

even power. Every even power of 9 has a units digit of 1 (and, if divided by 5, will leave a remainder of 1). Easy mistake.

I hope that helps.

For more on cyclicity, see Bunuel Last Digit of a Power (scroll down)
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