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# 10! is divisible by 3x5y, where x and y are positive integer

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10! is divisible by 3x5y, where x and y are positive integer [#permalink]  12 Aug 2018, 16:00
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Question Stats:

64% (01:00) correct 36% (00:55) wrong based on 50 sessions
10! is divisible by $$3^x5^y$$, where x and y are positive integers.

 Quantity A Quantity B The greatest possible value for x Twice the greatest possible value for y

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Sandy
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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]  17 Aug 2018, 16:09
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Expert's post
Explanation

First, expand 10! as 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.

(Do not multiply all of those numbers together to get 3,628,800—it’s true that 3,628,800 is the value of 10!, but analysis of the prime factors of 10! is easier in the current form.)

Note that 10! is divisible by 3x5y, and the question asks for the greatest possible values of x and y, which is equivalent to asking, “What is the maximum number of times you can divide 3 and 5, respectively, out of 10! while still getting an integer answer?”

In the product 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, only the multiples of 3 have 3 in their prime factors, and only the multiples of 5 have 5 in their prime factors. Here are all the primes contained in 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 and therefore in 10!:

10 = 5 × 2
9 = 3 × 3
8 = 2 × 2 × 2
7 = 7
6 = 2 × 3
5 = 5
4 = 2 × 2
3 = 3
2 = 2
1 = no primes

There are four 3’s and two 5’s total. The maximum values are x = 4 and y = 2. Therefore, the two quantities are equal.
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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]  09 Mar 2019, 03:24
As the maximum values are x = 4 and y = 2, should not be A greater?
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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]  09 Mar 2019, 04:05
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$$10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1$$

$$(2*5) * (3^2) * (2^3) * 7 * (2*3) * 5 * 3 * (2^2) * 1$$

$$\frac{7 * 5^2 * 3^4 * 2^7 * 1}{3^x 5^y}$$

As you clearly see the quantity must be divided by $$3^x$$ and $$5^y$$.

In the numerator $$3^4$$ and $$5^2$$ , which means that the exponent of 3 is $$4 = x$$ and the exponent of 5 is $$2=y$$

A > B

I think also the answer should be A

Hope this helps.

Regards
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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]  09 Mar 2019, 07:06
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Carcass wrote:
$$10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1$$

$$(2*5) * (3^2) * (2^3) * 7 * (2*3) * 5 * 3 * (2^2) * 1$$

$$\frac{7 * 5^2 * 3^4 * 2^7 * 1}{3^x 5^y}$$

As you clearly see the quantity must be divided by $$3^x$$ and $$5^y$$.

In the numerator $$3^4$$ and $$5^2$$ , which means that the exponent of 3 is $$4 = x$$ and the exponent of 5 is $$2=y$$

A > B

I think also the answer should be A

Hope this helps.

Regards

Be careful.
Quantity B = TWICE the greatest possible value for y

Cheers,
Brent
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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]  09 Mar 2019, 09:55
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Thanks to both of you, it is clear now.

I often make this kind of silly mistakes, I need to read more carefully.

Thanks again!
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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]  09 Mar 2019, 09:56
Expert's post

Trueeeeeee

x = 4 and y= 2 but B is twice. So, y = 4

Thank you Sir
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Re: 10! is divisible by 3x5y, where x and y are positive integer   [#permalink] 09 Mar 2019, 09:56
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