raleigh wrote:
So your choices are 1, 11, and 21. The second term being 0 will force the tens digit to be 0 which will rule out 11 and 21 since the remainder has to be less than the divisor(25).
Doesn't 31/25 have a remainder of 6? When you divide a number ending in 1 by 25, the possible remainders are actually:
1, 6, 11, 16 and 25
While it's going to take forever and a day to actually calcluate the value of 101^101, if we do the first couple we can quickly see a pattern:
101*101 = 10101
10101*101 = 1010101
As the series progresses, we're just adding a "10" to the front of the expression.
When we divide by 25, all that matters is the last two digits (since 100 is divisible by 25). Once we recognize that 101^101 ends in "01", we know with certainty that the remainder is 1.
We also could have come to this conclusion by thinking in terms of quadratics:
101^2 = (100 + 1)^2 = 100^2 + 2(100*1) + 1 = 10000 + 100 + 1 = 10101
10101*101 = (10101)(100) + (10101)(1) = 1010100 + 1 = 1010101
and again, we can see the pattern that's developing.
