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

Question

(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

generis · Accepted Answer

The answer is D.

When faced with huge numbers and/or exponents, use smaller numbers and/or exponents that follow the prompt's pattern.

If we are looking for the least change (decrease), we need the value that has the most weight to be decreased a little.

To find the smallest decrease in  n , we need the value that has the  most weight  to be  decreased a little . 
• Changing a small number such as 2, to 1, cuts the product in half. 
• Changing a larger number such as 4, to 3, decreases the product by 1/4.

(And that which gets decreased must be a factor, not the exponent. Decreasing the exponent by 1, see below, is not decreasing by a little.)

Think of n = (2*7) ^2 . That equals 196.

By definition, an exponent yields exponential increase in value. 
Decreasing the exponent will yield an exponential decrease - not likely to be the least change.

(2*7) ^2  = 196

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

The smallest decrease comes from decreasing the largest factor by 1.

You can use (2*3*4*5) ^3  (increasing positive factors to an odd power) and come up with the same result. The essential "ingredients" in the prompt's  n  are the same. 
When I see insane values, I think: go with small but similar values.

Answer D

chetan2u · Answer

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 ..

Zhenek · Answer

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

generis · Answer

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

Results 
The 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 D

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 as that above. Decreasing the greatest factor, 5, by 1 creates the least change (and decreasing the exponent creates the most change).

ScottTargetTestPrep · Answer

If the numbers are “too large”, we can use much smaller numbers to see a similar effect. Let’s say n = (3 x 4)^2 = 12^2 = 144.

If 3 becomes 2, we have (2 x 4)^2 = 8^2 = 64. So the value of n decreases by 80.

If 4 becomes 3, we have (3 x 3)^2 = 9^2 = 81. So the value of n decreases by 63.

If 2 becomes 1, we have (4 x 3)^1 = 12^1 = 12. So the value of n decreases by 132.

We see that decreasing the exponent by 1 will decrease the value of n the most; however, decreasing the largest factor in the base will decrease the value of n the least. Therefore, in the given expression of n, decreasing the largest factor, 29, by 1 will decrease the value of n the least.

Answer: D

EgmatQuantExpert · Answer

Given

• A number n =  (17 × 19 × 23 × 29)^{31}

To Find

• The number that must be decreased by 1 to get the least decrease in n.

Approach and Working Out

Although it's better to assume a smaller value in this case, I'll try to give a more proper approach here.

• n =  (17 × 19 × 23 × 29)^{31} 
 o If we decrease 17 by 1 then,
  The term will be n × ( \frac{16}{17} ) ^{31} 
 Similarly for 19, 23, and 29, the terms will be, n × ( \frac{18}{19} ) ^{31} , n × ( \frac{22}{23} ) ^{31} , n × ( \frac{28}{29} ) ^{31} 
 We can see that the highest among these fractions is 28/29, hence this will have the least decrease among these 4. 
o If we decrease 31 by 1, then the term becomes n/(17 × 19 × 23 × 29)
o This term is way less than  \frac{28}{29} .
o However, one can approximately calculate the term.

•  (28/29)^{31}  will be  (0.96)^{31} .
 o We can try to get an approximate value by squaring 0.95 for 5 times. The logic here is ( x^2 ) ^5 =  x^{32} .
o  0.95^2  -> 0.90, 
o  0.90^2  -> 0.81, 
o  0.81^2  -> 0.65
o  0.65^2  ->0.42
o  0.42^2  -> .17
o So the decrease will be only about 83%
o Where as the decrease for the term 1/(17 × 19 × 23 × 29) is way less than that as 1/17 only gives a decrease of 93%. 
• We can now conclude that the least decrease will be when 29 is decreased by 1.

Correct Answer: Option D

Kinshook · Answer

(17 * 19 * 23 * 29)^{31} = n 
Asked: Lowering which of the following numbers by one will result in the least decrease of n?

Out of 17, 19, 23 & 29, decreasing 29 to 28 will result in least decrease of n since 28/29 ratio is highest.
Decreasing power of 31 to 30 will decrease the number by 17*19*23*29 times.

IMO D

StacyArko · Answer

Kindly assist with options D and E. I’m getting:
(17*19*23)^31 for option D and 17*19*23*29 for option E and interpreting that the decrease in option E is smaller. Kindly assist. Thank you

Posted from my mobile device

chetan2u · Answer

No, by decreasing power by 1, the difference is way too much.... (17*19*23*29)^{31}  - (17*19*23*29)^{30}  .
 (17*19*23*29)^{30}(17*19*23*29-1)  ~ (17*19*23*29)^{31}  , as (17*19*23*29-1) ~ 17*19*23*29

While in D, it is  (17*19*23*29)^{31}  - (17*19*23*28)^{31} = (17*19*23)^{31}(29^{31}-28^{31}) , which is way smaller than  (17*19*23*29)^{30}(17*19*23*29-1)

StacyArko · Answer

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 ..

Posted from my mobile device    
No, by decreasing power by 1, the difference is way too much.... (17*19*23*29)^{31}  - (17*19*23*29)^{30}  .
 (17*19*23*29)^{30}(17*19*23*29-1)  ~ (17*19*23*29)^{31}  , as (17*19*23*29-1) ~ 17*19*23*29

While in D, it is  (17*19*23*29)^{31}  - (17*19*23*28)^{31} = (17*19*23)^{31}(29^{31}-28^{31}) , which is way smaller than  (17*19*23*29)^{30}(17*19*23*29-1)

Thanks so much.

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(171923*29)^{31} = n. Lowering which of the following numbers by one

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