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(17*19*23*29)^31=n

This topic has 3 expert replies and 0 member replies

(17*19*23*29)^31=n

Post Sat Nov 04, 2017 8:08 am
$$\left(17\cdot19\cdot23\cdot29\right)^{31}=n$$
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

The OA is D.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

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Post Fri Nov 10, 2017 3:26 am
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)³⁰.
Calculating the difference between the two expressions, we get:
(17*19*23*29)³¹ - (17*19*23*29)³⁰ = (17*19*23*29)³⁰(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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Post Thu Nov 09, 2017 7:25 pm
What a great question!

We could test the principle ourselves with smaller digits. Say we have (2 * 3 * 4)⁵, or 24⁵.

The four possible reductions are

(1 * 3 * 4)⁵, or 12⁵
(2 * 2 * 4)⁵, or 16⁵
(2 * 3 * 3)⁵, or 18⁵
(2 * 3 * 4)⁴, or 24⁴

We want the LEAST possible reduction, so we want the number CLOSEST to 24⁵. Conceptually speaking, that means we want the largest result. It's clearly either 18⁵ or 24⁴, so let's compare them:

18⁵ vs 24⁴

(3*6)⁵ vs (4*6)⁴

3⁵ * 6 vs 4⁴

The left side is bigger, so 18⁵ 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.

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Post Mon Nov 06, 2017 10:31 pm
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

The OA is D.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
The correct answer would be one among 17, 29, and 31 since 19 and 23 lie between 17 and 29.

Let's discuss options A, D and E.

A. 17: After decreasing 17 by 1, we get 16. Thus, the new value of n = n*(16/17)^31;
D. 29: After decreasing 29 by 1, we get 28. Thus, the new value of n = n*(28/29)^31;
E. 17: After decreasing 31 by 1, we get 30. Thus, the new value of n = n/(17.19.23.29)

We see that among n*(16/17)^31, n*(28/29)^31, and n/(17.19.23.29), the value of n*(28/29)^31 would be highest. Thus, decreasing 29 by 1 will result in the least decrease of n.

The correct answer: D

Hope this helps!

-Jay
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