(17*19*23*29)^{31} = n. Lowering which of the following numbers by one

The answer is D.

When faced with huge numbers and/or exponents, use smaller numbers and/or exponents that follow the prompt's pattern.

If we are looking for the least change (decrease), we need the value that has the most weight to be decreased a little.

To find the smallest decrease in \(n\), we need the value that has the most weight to be decreased a little.
• Changing a small number such as 2, to 1, cuts the product in half.
• Changing a larger number such as 4, to 3, decreases the product by 1/4.

(And that which gets decreased must be a factor, not the exponent. Decreasing the exponent by 1, see below, is not decreasing by a little.)

Think of n = (2*7)\(^2\). That equals 196.

By definition, an exponent yields exponential increase in value.
Decreasing the exponent will yield an exponential decrease - not likely to be the least change.

(2*7)\(^2\) = 196

Decrease exponent 2 by 1. (2*7)\(^1\) = 14

Now decrease factor 2 by 1. (1*7)\(^2\) = 49

Now decrease factor 7 by 1. (2*6)\(^2\) = 144

Original value => 196
Exponent: 2 - 1 => 14
Factor 2 - 1 => 49
Factor 7 - 1 => 144

The smallest decrease comes from decreasing the largest factor by 1.

You can use (2*3*4*5)\(^3\) (increasing positive factors to an odd power) and come up with the same result. The essential "ingredients" in the prompt's \(n\) are the same.
When I see insane values, I think: go with small but similar values.

Answer D