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.
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