Bunuel wrote:

\((17*19*23*29)^{31} = n\). Lowering which of the following numbers by one will result in the least decrease of n?

A. 17

B. 19

C. 23

D. 29

E. 31

When I see insane values, I think: go with small but similar values.

Keep the essential "ingredients" the same, that is, use increasing positive factors to an integral power greater than 1.

The theory: To achieve the least change in n, we need the value that has the most weight to be decreased a little.

(Not likely to be the exponent, which carries all of n's weight more than once.)

1) Try something such as n = (2*7)\(^2\) = 196

2) Decrease EACH number - 2, 7, and the exponent 2 -- by 1

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

3) Conclusion: The smallest decrease comes from lowering the largest

factor, 7, by 1.

The same will hold for this prompt's set of numbers.

The largest factor is 29. Lowering it by 1 will have the least effect.

Answer DIf you don't trust the simpler math, try (2*3*4*5)\(^3\), which is closer to the prompt (four increasing positive factors to an odd power) - and arithmetic heavy. You will get the same result with the complicated simulation. Decreasing the factor 5 by 1 creates the least change (and decreasing the exponent creates the most change).