Question

Last two digit of the number (129)^129 will be

Answer 1

Answer: 89

Step-by-step explanation:

The last two digits of the powers of 9 follow a pattern.

9^1=09

9^2=81

9^3=29...

The pattern goes like:

09, 81, 29, 61, 49, 41, 69, 21, 89, 01 and again goes on like 09,81 and so on...

Notice that the consecutive odd terms in the pattern increase by 20 and consecutive even terms decrease by 20.

The pattern also repeats every 10 terms. So as the power is 129, divide it by 10. The remainder is 9.

So the ninth term in the pattern is the answer i.e., 89.

Thanks for asking. Hope it helps.

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