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 impossibly great 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 lowered

a little bit.

(The number to be changed is 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

ResultsThe original value was \(n = 196\)

Type of change => effect on \(n\)

Exponent 2, decreased by 1: \(n=14\)

Factor 2, decreased by 1: \(n=49\)

Factor 7, decreased by 1: \(n=144\)

(3) Conclusion: The least decrease in \(n\) comes from lowering the largest

factor, 7, by 1.

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

The largest factor is 29. Lowering it by 1 will result in the least decrease of \(n\).

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 as that above. Decreasing the greatest factor, 5, by 1 creates the least change (and decreasing the exponent creates the most change).
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In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"