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# (17*19*23*29)^{31} = n. Lowering which of the following numbers by one

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Math Expert
Joined: 02 Sep 2009
Posts: 43917
(17*19*23*29)^{31} = n. Lowering which of the following numbers by one [#permalink]

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12 Nov 2017, 00:54
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$$(17*19*23*29)^{31} = n$$. 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
[Reveal] Spoiler: OA

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Joined: 17 Mar 2015
Posts: 121
(17*19*23*29)^{31} = n. Lowering which of the following numbers by one [#permalink]

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12 Nov 2017, 02:10
The idea here is that the resulting value of N will be equal to the product of n and a fraction $$\frac{a-1}{a} = 1 - \frac{1}{a}$$.
The question can be rephrased as "what is the value of a, taken from the numbers in the product, so that the fraction is the largest". The obvious answer is the largest value available, since in this case $$1 - \frac{1}{a}$$ is maxed out for the available numbers.

Another approach is to do a direct test with simplier numbers:
Take 2 3 and 4, for example
2 * 3 * 4 = 24

1*3*4 = 12 (reduce the lowest by 1)
2*2*4 = 16 (reduce the middle by 1)
2*3*3 = 18(reduce the highest by 1)

The least decrease is shown from the last equation (24 - 18 = 6)

D
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Joined: 02 Aug 2009
Posts: 5667
Re: (17*19*23*29)^{31} = n. Lowering which of the following numbers by one [#permalink]

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12 Nov 2017, 03:03
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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

Hi..
When you decrease one of the four, the overall decrease in the product would be the product of other three, so decreasing the largest will result in decrease in the product by an amount EQUAL to the product of the smallest 3 number.
So answer will be decrease the largest by 1, meaning answer is D.

And this is why..
17*19*23*29...
17*19*23*(29-1)=17*1 9*23*29-17*19*23
So the decrease is 17*19*23, ofcourse with power 31 ..
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(17*19*23*29)^{31} = n. Lowering which of the following numbers by one [#permalink]

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12 Nov 2017, 20:06
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.

If 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).
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(17*19*23*29)^{31} = n. Lowering which of the following numbers by one   [#permalink] 12 Nov 2017, 20:06
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