Re: What is the last two digits of 11^10 - 9 ?
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17 Nov 2021, 07:31
11^1 ends in 11, 11^2 ends in 21, 11^3 ends in 31, and so on, so 11^10 will end in '01', and after subtracting 9, we'll get a number ending in 92. This isn't something I'd expect the GMAT to ever test though.
There are a few ways to see why powers of 11 will exhibit this pattern. Using the binomial theorem makes it clear, but that's beyond the scope of the GMAT. If you know that the familiar units digit principle (when we multiply integers, the units digit of the answer comes only from the units digits of the numbers we're multiplying together) extends to tens digits (we only need the tens and units digits of our numbers to predict the tens and units digits of their product), you can then just imagine multiplying 11 by any number ending in "A1", where "A" is an arbitrary tens digit. We can then see what our last two digits must be:
11*A1 = (10 + 1)(A*10 + 1) = A*100 + A*10 + 10 + 1 = A*100 + (A + 1)*10 + 1
and we can see that our new tens digit, A+1, is one greater than our old one which was A (unless A = 9, in which case our new tens digit becomes 0). So when we multiply 11 by 11, we get something ending in 21, then when we multiply that by 11, we get something ending in 31, and so on.