Integer properties

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If x and y are positive integers, the greatest value of x for which 57! / (7^x) is an integer is what percent greater than the greatest value of y for which
57! /(13^y) is an integer?

A. 71%

B. 100%

C.125%

D.225%

E. 250%

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Hi neha.njd,

This question involves some BIG numbers, but is ultimately about prime factorization.

We're told that X and Y are positive integers. We're then asked to find the GREATEST possible value of X and Y given two facts.

First, we're told that 57!/(7^X) is an integer.

57! = (57)(56)(55).....(3)(2)(1)

To figure out the greatest possible value of X, we have to find all of the 7s contained in that gigantic product....

7s can be 'found' in 7, 14, 21, 28, 35, 42, 49 (this has TWO 7s) and 56.....so X = 9

To find the greatest possible value of Y, we have to do the same thing (but we have to find all of the 13s contained in 57!....

13s can be 'found' in 13, 26, 39 and 52....so Y = 4

We're asked what PERCENT GREATER X is than Y, so we need the Percentage Change Formula:

(New - Old)/Old = (9-4)/4 = 5/4 = 1.25 = 125%

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

[Parth_23Nov]

Take X for calculation
x for which 57! / (7^x) is an integer
i.e X = 9 for which 57! / (7^x) is an integer

7x8 = 56 (Within 57, 56 highest number divisible by 7.)

Similarly
y for which 57! /(13^y) is an integer
i.e y = 4 for which 57! / (7^x) is an integer

13x4 = 52 (Withing 57, 52 highest number divisible by 13)


So if you check %change of X on Y = ((9-4)/(4))x100 = 125%

[GMATGuruNY]

If x and y are positive integers, the greatest value of x for which 57! / (7^x) is an integer is what percent greater than the greatest value of y for which
57! /(13^y) is an integer?

A. 71%

B. 100%

C.125%

D.225%

E. 250%

57! = 57*56*55*....*3*2*1.

To determine the greatest value of x for which 57!/(7^x) is an integer and the greatest value of y for which 57!/(13^y) is an integer, we need to count how many 7's and how many 13's can divide into 57!.
Put another way:
We need to count how many 7's and how many 13's are contained within 57!.
To count, simply divide increasing POWERS OF 7 AND 13 into 57.

Every multiple of 7 within 57! provides AT LEAST ONE 7:
57/7 = 8 --> eight 7's.
Every multiple of 7² within 57! provides a SECOND 7:
57/7² = 1 --> one more 7.
Thus, the total number of 7's contained within 57! = 8+1 = 9.
Result:
Since there are nine 7's contained within 57!, the greatest possible value of x for which 57!/(7^x) is an integer = 9.

Every multiple of 13 within 57! provides ONE 13:
57/13 = 4 --> four 13's.
Thus, the total number of 13's contained within 57! = 4.
Result:
Since there are four 13's contained within 57!, the greatest possible value of y for which 57!/(13^y) is an integer = 4.

Since 8 is 100% greater than 4, x=9 is A LITTLE MORE than 100% greater than y=4.

The correct answer is C.

For a similar problem, check here:
https://www.beatthegmat.com/how-to-find- ... 71343.html