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

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New post Updated on: 09 Feb 2020, 06:09
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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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New post 09 Feb 2020, 06:05
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Is that A3 = \(A^3\) or \(A*3\)??
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New post 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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New post 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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New post 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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New post 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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New post 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

Answer - D
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New post 09 Feb 2020, 11:39
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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)


Answer D
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New post 09 Feb 2020, 13:10
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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

FINAL ANSWER IS (D)
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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New post 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

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Good that you corrected the mistake made earlier, a mistake that i also made under timed condition. Anyway great learning :thumbup:
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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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