Solution to Problem #65
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In order for the last digit of our cube to be 1, the last digit of our
original number must be 1.
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The original number must therefore be of the form 10x + 1. Cubing
this we obtain 1000x3 + 300x2 +
30x + 1. The first two terms do not contribute to the last two
digits, so we just need to find an x such that 30x + 1 has
21 as its last two digits and x = 4 is the only solution.
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We now know that our original number is of the form 100y + 41.
Cubing this we obtain 1000000y3 +
1230000y2 + 504300y + 68921. Again, the first two
terms do not contribute to the last three digits, so we need to find a
y such that 300y + 921 has 321 as its last digits and
y = 8 is the only solution.
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Continuing in this manner, we find that 1871517841 is the smallest
solution.