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 Question from Raginee: How to find last two or three digits of a large exponent like 17^256?

Hi,

I am going to look at the last two digits, the tens and units digits.

First of all any integer $z$ can be written $z = 100 \times x + t$ where x is an an integer and t is a two digit integer. For example $73142 = 73100 + 42 = 100 \times 731 + 42.$ Thus using the binomial theorem

$z^2 = (100 \times x + t)^2 = 10000 \times x^2 + 200 \times x + t^{2}.$

Hence $z^2$ is the sum of three terms and the tens and units digits of the first two terms are zero. Thus the number formed by the tens and units digit of $z^2$ is the same as the number formed by the tens and units digit of $t^{2}.$ If you use the binomial theorem to expand $z^n$ where $n$ is a positive integer you will obtain a similar result. The number formed by the tens and units digit of $z^n$ is the same as the number formed by the tens and units digit of $t^{n}.$ I want to use this to determine the tens and units digits of $73142^{93}.$

I start by calculating powers of 42 keeping in mind that I only need to retain the tens and units digits.

Power Calculation Tens and units digits
1   42
2 42 x 42 = 1764 64
3 1764 x 42 88
4 88 x 42 96

Continuing this table I get

Power Calculation Tens and units digits
15   68
16 68 x 42 56
17 56 x 42 52
18 52 x 42 84
19 84 x 42 28
20 28 x 42 56

But 56 appeared earlier at $42^{16}$ so from this point on the rows repeat. From the 16th power onwards the numbers formed by the tens and units digit are 56, 52, 84, 28 and then repeat.

I want the tens and units digits of $73142^{93}.$ I now see that after the first 15 powers I have $93 - 15 = 78$ rows remaining in my table. But these rows repeat in multiples of 4. 78 divided by 4 is 19 with a remainder of 2 so 4 row pattern repeats 19 times and then there are two more rows. The tens and units digits of these two remaining rows are 56 and 52. Thus the tens and units digits of $73142^{93}$ are 5 and 2.

I hope this helps,
Penny

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