Without using any calculating aids, determine the minimum value of a positive base ten integer N, such that N^{3} ends in 888.
If the solution exists it is a three digit number which can be written as (100a + 10b + c).
If we cube this expression, 17 of the 27 resulting terms are 10000 or greater. The sum of the other 10 terms is
300ac^2 + 300b^2c + 30bc^2 + c^3.
Since c^3 must end in 8, c=2 and the above simplifies to
1200a + 600b^2 + 120b+8
The 120b term must end in 80 so b is either 4 or 9, the above simplifies to either
b=4: 1200a + 10088
b=9: 1200a + 49688
With the b=4 case, 1200a must end in 800 so a=4 or a=9
With the b=9 case, 1200a must end in 200 so a=1 or a=6
The smallest possibility is then N=192
The full set of three digit numbers whose cubes end in 888 is:
442, 942, 192, 692

Posted by Jer
on 20110511 15:29:26 