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If x is a product of all integers from 2 to 24, inclusive, w

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If x is a product of all integers from 2 to 24, inclusive, w  [#permalink]

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New post Updated on: 06 Apr 2019, 08:21
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If x is a product of all integers from 2 to 24, inclusive, what is the greatest integer y for which 35^y is a factor of x?

(A) 2
(B) 3
(C) 5
(D) 7
(E) 9



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Originally posted by nm97 on 06 Apr 2019, 07:26.
Last edited by nm97 on 06 Apr 2019, 08:21, edited 1 time in total.
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Re: If x is a product of all integers from 2 to 24, inclusive, w  [#permalink]

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New post 06 Apr 2019, 07:42
Yes. How did you solve? Could you explain in detail? I'm having confusion since 35 involves 2 prime numbers, 5&7.

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Re: If x is a product of all integers from 2 to 24, inclusive, w  [#permalink]

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New post 06 Apr 2019, 07:46
I check the repetitions of 35 in 24!, which happens like 3 times. But this approach is better only the numbers are less if they ask the same thing like in 100! I would be clueless. Any suggestions for a quicker approach ?


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Re: If x is a product of all integers from 2 to 24, inclusive, w  [#permalink]

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New post 06 Apr 2019, 07:53
karthikkaushik91 wrote:
I check the repetitions of 35 in 24!, which happens like 3 times. But this approach is better only the numbers are less if they ask the same thing like in 100! I would be clueless. Any suggestions for a quicker approach ?


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What do you mean by repetitions?
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Re: If x is a product of all integers from 2 to 24, inclusive, w  [#permalink]

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New post Updated on: 06 Apr 2019, 08:05
karthikkaushik91 wrote:
I check the repetitions of 35 in 24!, which happens like 3 times. But this approach is better only the numbers are less if they ask the same thing like in 100! I would be clueless. Any suggestions for a quicker approach ?


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Exactly. I did the same way. 24! was a small number and removing 5&7 was easy.

I also thought about the same thing in different way. I found out during the first solving that there are fewer 7s than 5s. So if we take 7s in 24!, 7 appears 3 times in this range- 7,7(14=7*2) & 7(21=7*3). So total 3 7s within factors of x.
But honestly, it took a while to understand and shoot. Is there any simpler way to solve such questions?

If the question had any single prime number instead of 2 prime numbers(as in for 35= 7&5), simply use the formula of trailing zeros.


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Originally posted by nm97 on 06 Apr 2019, 07:57.
Last edited by nm97 on 06 Apr 2019, 08:05, edited 1 time in total.
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Re: If x is a product of all integers from 2 to 24, inclusive, w  [#permalink]

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New post 06 Apr 2019, 07:59
drakell wrote:
karthikkaushik91 wrote:
I check the repetitions of 35 in 24!, which happens like 3 times. But this approach is better only the numbers are less if they ask the same thing like in 100! I would be clueless. Any suggestions for a quicker approach ?


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What do you mean by repetitions?



drakell As we need to find how many 35s are factors of 24!, divide 35s from 24! till it is possible (repetition). If you try, only 3 times 35 is possible.
Hope it helps.

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Re: If x is a product of all integers from 2 to 24, inclusive, w  [#permalink]

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New post 06 Apr 2019, 08:05
Can u elaborate pls?


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Re: If x is a product of all integers from 2 to 24, inclusive, w  [#permalink]

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New post 06 Apr 2019, 08:11
35 is 7 multiplied by 5. 2 to 24 product will have three integers 7,14,21 , which can be divided by 7. So power of 7 is 3 in denominator limiting the overall power of 35 to 3.

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Re: If x is a product of all integers from 2 to 24, inclusive, w  [#permalink]

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New post 07 Apr 2019, 00:08
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karthikkaushik91 wrote:
Can u elaborate pls?


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The questions imply asks what is the highest power of 35 in 24 factorial.

We can write 35 as product of prime factor 5 and 7.

Now simply see that we have to find highest power of 7 to find the highest power of 35. Remember we always have to find the highest power of the highest prime number , which in this case is 7.


Highest power of 7 in 24! is 3.

I hope you get it now
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Re: If x is a product of all integers from 2 to 24, inclusive, w   [#permalink] 07 Apr 2019, 00:08
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