What a great question!

We could test the principle ourselves with smaller digits. Say we have (2 * 3 * 4)�, or 24�.

The four possible reductions are

(1 * 3 * 4)�, or 12�
(2 * 2 * 4)�, or 16�
(2 * 3 * 3)�, or 18�
(2 * 3 * 4)�, or 24�

We want the LEAST possible reduction, so we want the number CLOSEST to 24�. Conceptually speaking, that means we want the largest result. It's clearly either 18� or 24�, so let's compare them:

18� vs 24�

(3*6)� vs (4*6)�

3� * 6 vs 4�

The left side is bigger, so 18� is the biggest number, and we want to reduce the largest number in the ( )s by one.

From here, we could generalize that this will probably work in most other cases, making D the likeliest answer.

AAPL wrote:$$\left(17\cdot19\cdot23\cdot29\right)^{31}=n$$
Lowering which of the following numbers by one result in the least decrease of n?

A. 17
B. 19
C. 23
D. 29
E. 31

n = (17*19*23*29)³¹.
Option E implies that the exponent decreases by 1.
If we decrease the exponent by 1, we get:
(17*19*23*29)³�.
Calculating the difference between the two expressions, we get:
(17*19*23*29)³¹ - (17*19*23*29)³� = (17*19*23*29)³�(17*19*23*29 - 1).
The difference in red is quite large and thus not the least possible decrease.
Eliminate E.

The correct answer must be one of the 4 factors inside the parentheses -- 17, 19, 23, or 29 -- rendering the exponent irrelevant.
To decrease (17*19*23*29)³¹ as little as possible, we must decrease 17*19*23*29 as little as possible.

Consider an EASY CASE of 4 factors:
3abc.
If we decrease the blue factor by 1 and leave the 3 red factors unchanged, we get:
2abc.
Calculating the difference between the two products, we get:
3abc - 2abc = abc.
Implication:
The decrease in the product is equal to the product of the 3 factors that are UNCHANGED (abc).

Thus, to decrease 17*19*23*29 as little as possible, the 3 UNCHANGED FACTORS must be as small as possible:
17*19*23.
Since 17, 19 and 23 are unchanged, the factor that must decrease by 1 is the remaining factor:
29.

The correct answer is D.