Solution to Problem #65

• In order for the last digit of our cube to be 1, the last digit of our original number must be 1.
• 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.
• 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.
• Continuing in this manner, we find that 1871517841 is the smallest solution.