Powerful Digit Sum

A googol ($10^{100}$) is a massive number: one followed by one-hundred zeros; $100^{100}$ is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only $1$.
Considering natural numbers of the form, $a^b$, where $a, b \lt 100$, what is the maximum digital sum?

The maximum digital sum of an arbitrary number does not depend on the number of digits it has, but rather depends on how many of those digits are 9. More 9’s lead to a higher digital sum.

It can be proven that raising 99 to an arbitrary power will yield a number with a higher digital sum than any number with base less than 99. The intuition is that a higher base means more multiplication per digits’ contribution to the final number.

Therefore, we need to find a base (a) such that a is less than 100 and the digital sum for a to the power of anything (b) is maximized. Since 99 has the largest possible digital sum (18), it seems a logical choice to test.

Now we should consider a good exponent. We could consider using 1, but that wouldn’t do much. Likewise, 2 wouldn’t be very beneficial, although it would be better. So we start using higher powers until we reach exponents that are too high to give a beneficial increase in digital sum. At that point, additional multiplication by the base begins to lead to more “9’s” rolling over to “0’s” and thus lowering the digital sum.

After examining various cases, it happens that the maximum digital sum occurs with $a$ and $b$ both as 99.

Another way to explore this problem without much computational work would be to investigate how the digital sum grows as the exponent increases. Observing patterns in the growth may allow you to identify when it reaches its maximum value.

Now we need to calculate the digital sum of $99^{99}$.
The easiest way for a human to do this would be to use a scientific calculator, like as Mathematica or Python to perform this computation, as it is quite difficult (if not practically impossible) to do it by hand.

You will find that the digital sum of $99^{99}$ is 648.

Thus for the natural numbers of the form, $a^b$, where $a, b \lt 100$, the maximum digital sum is 648.

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