Puzzle of the week: Which two digits end 7 to the power of 9,999?

The number 7⁹⁹⁹⁹ has 8,451 digits. Printed in small type with narrow spacing, it would fill two A4 pages. But the puzzle does not require calculating or printing the number. It asks only for its final two digits.

The answer is 43. The powers of 7 follow a repeating pattern in their last two places: 01, 07, 49, 43. After that, the sequence starts again with 01.

The reason lies in remainders when numbers are divided by 100. The last two digits of a number are its remainder modulo 100. When a power is multiplied by 7, only that remainder matters for determining the next two digits. Any earlier digits represent multiples of 100 and cannot affect the final pair.

For example, multiplying 49 by 7 gives 343, whose last two digits are 43. Whenever the remainder 49 appears again in the sequence, multiplying by 7 produces 43 again, so the pattern continues periodically.

The cycle has length four. If the exponent is divisible by 4, the final two digits are 01. If it leaves a remainder of 1, they are 07; a remainder of 2 gives 49; and a remainder of 3 gives 43. Since 9,999 leaves a remainder of 3 when divided by 4, 7⁹⁹⁹⁹ ends in 43.