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 Terminating in three 8s (Posted on 2011-05-11)
Without using any calculating aids, determine the minimum value of a positive base ten integer N, such that N3 ends in 888.

 No Solution Yet Submitted by K Sengupta No Rating

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 unaided solution | Comment 1 of 3
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 2011-05-11 15:29:26

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