The number 70,843,029,071,054,204,781,573,025,995,798,769,907,739,696,815,288,329,699,328 is the twenty-third power of a positive integer.
Find the positive integer using just a simple calculator with pen and paper.
If N^23 ends in 8, then N must end in 2 because:
(a) 23 is 3 mod 4, and
(b) powers of 2 end in {6,2,4,8} when the exponent is {0,1,2,3} mod 4; whereas powers of 8 end in {6,8,4,2} when the exponent is {0,1,2,3} mod 4.
N^23 has 59 digits so N must have 3 digits, because:
100^23 has 47 digits
1000^23 has 70 digits
With your calculator, you find:
10^(58/23) = 332.45979322709417
10^(59/23) = 367.4661940736688
So N is between 333 and 367, and since it ends in a 2, it must be in:
{342, 352, 362}
The sod(N^23) is 302, therefore N must be 2 mod 3.
Of the three candidates, only 362 is 2 mod 3.
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Posted by Larry
on 2023-07-18 08:16:13 |
Solution
