How do you find out the second last digit of a large exponent?

Answer 1

Let us look at this with an example. Say you want to find the second last digit of 5,283,641267.

The first step is to recognise that the 5,283,600 part of the mantissa will contribute nothing to the last two digits. So you can ignore them completely. That simplifies the question to finding the second last digit of 41267.

The hard work starts now. You examine the last two digits of the powers of 41 to see if there is a pattern.

1) 41*1 = 41

2) 41*41 = 1681 => 81

3) 81*41 = 3321 => 21

4) 21*41 = 861 => 61

5) 61*41 = 2501 => 01

6) 01*41 = 41

The last two digits of 411are the same as the last two digits of 416. The difference in exponents is 5 and that is the key number for this example. What that says is that every 5th power you return to where you started. As a result, you can reduce the exponent by 5 (or multiples of 5) without changing the last two digits.

The exponent in the question is 267. Find the remainder when 267 is divided by 5. The remainder is 2 and so the last two digits are the same as those of 412, that is, 81. And so the second last digit is 8.

Warnings:

• Some numbers do not return to their starting point (odd multiples of 2 or 5: for example, 02). But 22and 222both end in 04. The difference in exponents is 20 and so you can reduce the exponent by multiples of 20. Except that, if at the end of these reductions, the remainder is 1, you need to look at 221and not 21.
• Many numbers have a cycle of length 20, so be patient.