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# For any positive integer n greater than 1, n! denotes the product of a

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Director
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For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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Updated on: 09 Feb 2020, 06:09
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95% (hard)

Question Stats:

30% (01:46) correct 70% (03:01) wrong based on 33 sessions

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For any positive integer n greater than 1, n! denotes the product of all the integers from 1 to n, inclusive.
If A is a positive integer such that the greatest number that divides both $$A^3$$ and 13! is 448, which of the following can be the value of A?

A. 14
B. 56
C. 140
D. 196
E. 448

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Originally posted by lnm87 on 09 Feb 2020, 05:47.
Last edited by lnm87 on 09 Feb 2020, 06:09, edited 1 time in total.
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Re: For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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09 Feb 2020, 06:05
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Is that A3 = $$A^3$$ or $$A*3$$??
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Re: For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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09 Feb 2020, 06:10
You are right.
Thanks for pointing the error. Corrected it.
shameekv1989 wrote:
Is that A3 = $$A^3$$ or $$A*3$$??

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Re: For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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09 Feb 2020, 06:42
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If $$A^3$$ is divisible by 448 (which is $$2^6$$*7) then A must have atleast one 7 -> $$A^3$$ must have 7^3 = 343

The only value that is greater than 343 is 448 - Answer - E
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Re: For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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09 Feb 2020, 11:00
shameekv1989 wrote:
If $$A^3$$ is divisible by 448 (which is $$2^6$$*7) then A must have atleast one 7 -> $$A^3$$ must have 7^3 = 343

The only value that is greater than 343 is 448 - Answer - E

My friend you were right till halfway. Revisit, i am sure you would find why.
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For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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Updated on: 09 Feb 2020, 11:43
Answer would be I think 196 since 448= 2^6 *7
And 196^3 = 7^6*2^6
And it would be divisible by 448

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Originally posted by Apt0810 on 09 Feb 2020, 11:07.
Last edited by Apt0810 on 09 Feb 2020, 11:43, edited 1 time in total.
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Re: For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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09 Feb 2020, 11:13
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lnm87 wrote:
shameekv1989 wrote:
If $$A^3$$ is divisible by 448 (which is $$2^6$$*7) then A must have atleast one 7 -> $$A^3$$ must have 7^3 = 343

The only value that is greater than 343 is 448 - Answer - E

My friend you were right till halfway. Revisit, i am sure you would find why.

You are right. That's A^3 not A; Anyways :-

Minimum that A has is one 7 and two 2's. Two 2's is fixed but 7 is flexible. Thus it can be 7^2 * 4 = 196

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For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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09 Feb 2020, 11:39
1
lnm87 wrote:
For any positive integer n greater than 1, n! denotes the product of all the integers from 1 to n, inclusive.
If A is a positive integer such that the greatest number that divides both $$A^3$$ and 13! is 448, which of the following can be the value of A?

A. 14
B. 56
C. 140
D. 196
E. 448

The greatest common divisor (GCD) of $$A^3$$ and $$13!$$ is 448

$$448 = 4 * 112 = 4 * 4 * 28 = 4 * 4 * 4 * 7 = 2^6 * 7$$

Observe that $$13!$$ has only $$7^1$$ as factor
Also, highest power of $$2$$ in $$13! = [13/2] + [13/4] + [13/8] = 6 + 3 + 1 = 10$$ (where $$[n]$$ denotes the greatest integer less than or equal to $$n$$)

Since $$2^6 * 7$$ is the GCD of $$A^3$$ and $$13!$$, we can conclude that $$2^6 * 7$$ must be a factor of $$A^3$$

Thus, we have:

1. $$A^3$$ must NOT have the power of 2 greater than 6 (since 13! has power of 2 as 10, if $$A^3$$ had a power of 2 greater than 6, the GCD would also have a power of 2 greater than 6) => Highest power of 2 in A must be 2 i.e. A is a multiple of $$2^2$$ (not more than that) ... (i)

2. $$A^3$$ must have $$7^3$$ as a factor as well (it is the cube of a number), implying $$A$$ is a multiple of $$4 * 7$$ i.e. $$28$$ ... (ii)

Working with options:

A. $$A = 14$$ ---- not possible since A should be a multiple of 28 ---> from (ii)

B. $$A = 56 = 2^3 * 7$$ ---- Violates (i)

C. $$A = 140 = 2^2 * 7 * 5$$ ---- If A is a multiple of 5, and since 13! is also a multiple of 5, the GCD would have 5 as a factor ---- Violates given info

D. $$A = 196 = 2^2 * 7^2$$ ---- Satisfies both (i) and (ii). Note: even though A has a factor $$7^2$$, $$13!$$ has a factor only $$7^1$$, hence GCD would still have $$7^1$$ as factor

E. $$A = 448 = 2^6 * 7$$ ---- Violates (i)

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Re: For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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09 Feb 2020, 13:10
1
Both A and 13! must have common factors of minimum one 7 and exactly two 2s. No other factors is allowed.

Eliminate choices A, B, E since the number doesn't have exact factors of 2^2.
A. 14=2*7
B. 56=2^3*7
E. 448=2^6*7

Eliminate choice C because it has additional factor 5. If so, A^3 and 13! must have been divisible by 448*5^2
C. 140=2^2*7*5

Both A and 13! must have common factors of minimum one 7 and exactly two 2s.
D. 196=2^2*7^2

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For any positive integer n greater than 1, n! denotes the product of a  [#permalink]

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09 Feb 2020, 19:20
Apt0810 wrote:
Answer would be I think 196 since 448= 2^6 *7
And 196^3 = 7^6*2^6
And it would be divisible by 448

Posted from my mobile device

Good that you corrected the mistake made earlier, a mistake that i also made under timed condition. Anyway great learning
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For any positive integer n greater than 1, n! denotes the product of a   [#permalink] 09 Feb 2020, 19:20
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