Question 1131959: the 41st root of 2 199 023 255 552 is an integer. What integer, and how was the answer found?
Found 3 solutions by ikleyn, greenestamps, math_helper:
Answer by ikleyn(52866)   (Show Source):
Answer by greenestamps(13206)  (Show Source):
You can put this solution on YOUR website!
There are lots of things we can do using estimation and logical reasoning to determine that the integer is 2.
To begin with, the units digit is 2, so the integer has to be even. After that....
(1)10^41 is a number with 42 digits. If the 41st power of an integer is a number with only 13 digits, then the integer has to be VERY small. So just with that much reasoning, the integer almost has to be 2.
(2) We can gain confidence that the integer is 2 by noting that the units digit of the given number is 2. No power of either 4 or 6 has units digit 2, so the only possibilities for the integer are 2 and 8. And 8 is almost certainly too big.
(3) Finally, to confirm that 2 is the integer we are looking for, we can use the (very useful!) approximation 2^10 = 10^3 to see that 2^41 will be a number with about 13 digits:
2^41 = 2(2^40) = 2((2^10)^4) = 2((10^3)^4) = 2(10^12)
That is a 13-digit number with first digit 2 -- which is what the given number is.
Answer by math_helper(2461)  (Show Source):
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