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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]
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Expert Reply
\(10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1\)

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

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

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.

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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]
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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]
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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]
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:evil:

Trueeeeeee

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

C is the answer.

Thank you Sir 8-)
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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]
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Simply write down the factorial value of 10! i.e.
10.9.8.7.6.5.4.3.2.1
Then write down prime factors of above and add power values of 2 and 3, from which x=4 and y=2.
Therefore A and B are equal.
Answer is C

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Re: 10! is divisible by 3x5y, where x and y are positive integer [#permalink]
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If the question was 100! we cannot do from traditional method (100 X 99 X 98... X 1)
Better solution:
100/3 -> 33
100/9 -> 11
100/27 -> 3
100/81 -> 1
Add all -> 33+11+3+1 = 48 (x=48)

100/5 -> 20
100/25 -> 4
Add all -> 24 (y=24)

X=2Y

Similarly for 10!
10/3 -> 3
10/9 -> 1
Add all -> 4 (x = 4)

10/5 -> 2 (y = 2)

X=2Y
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10! is divisible by 3x5y, where x and y are positive integer [#permalink]
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We have to basically find out the number of times the factors 3 and 5 exist in 10!

10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

3 occurs twice in 9, once in six and once in 3, a total of 4 times. Therefore x=4

5 occurs once in 10 and once in 5. Therefore y=2

Quantity A = The greatest possible value for x = 4
Quantity B = Twice the greatest possible value for y = 2 x 2 = 4

They are BOTH EQUAL

The answer is C.
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