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akshayuberoi
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If m is a positive integer and m^3 is divisible by 48 then what is Updated on: 05 Nov 2020, 05:51
Originally posted by akshayuberoi on 05 Nov 2020, 05:42.
Last edited by Bunuel on 05 Nov 2020, 05:51, edited 1 time in total.
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C
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74% (01:33) correct 26% (01:33) wrong based on 293 sessions

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If m is a positive integer and m^3 is divisible by 48 then what is minimum possible value of m ?

A) 6
B) 9
C) 12
D) 18
E) 48
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C
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If m is a positive integer and m^3 is divisible by 48 then what is 05 Nov 2020, 05:48
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48 = (2^4) *(3^1)

As, (m^3)/48
m should contain : 2^2 * 3 = 12


If we select m=6 , then m^3 does not have enough 2's to cancel it out.
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If m is a positive integer and m^3 is divisible by 48 then what is 05 Nov 2020, 07:14
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akshayuberoi
If m is a positive integer and m^3 is divisible by 48 then what is minimum possible value of m ?

A) 6
B) 9
C) 12
D) 18
E) 48

m^3 is divisible by 48= 2^4*3. That means m^3 have at least 2^4*3 in it. But, to get a positive integer cube root of m, m^3 must contain at least (2^6)*(3^3). Thus, m must have at least, (2^2)*(3^1) as its factor. The minimum possible value of m= 2^2*3=12 (C) Ans.
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If m is a positive integer and m^3 is divisible by 48 then what is 05 Nov 2020, 11:33
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If m is a positive integer and m^3 is divisible by 48 then what is minimum possible value of m ?

A) 6
B) 9
C) 12
D) 18
E) 48

Solution:

m^3 = 48K
m^3 = 2^ 4 * 3^1

Since m is an integer.

The min value of m = 2^2 * 3^ 1 = 12 ( Hence C)
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If m is a positive integer and m^3 is divisible by 48 then what is 09 Nov 2020, 13:34
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akshayuberoi
If m is a positive integer and m^3 is divisible by 48 then what is minimum possible value of m ?

A) 6
B) 9
C) 12
D) 18
E) 48
Solution:

Recall that a perfect cube has prime factors whose exponents are each multiples of 3. For example, we know that 216 = 2^3 x 3^3, so we are assured that 216 is a perfect cube (6^3 = 216).
Since 48 = 16 x 3 = 2^4 x 3, we need two more 2’s and two more 3’s to make a perfect cube. So we see that m^3 is at least 2^6 x 3^3 = (2^2 x 3)^3. Thus, m is at least 2^2 x 3 = 12.

Answer: C
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If m is a positive integer and m^3 is divisible by 48 then what is 09 Nov 2020, 13:48
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\(48 = 2^4.3\)
Let's go from lowest to highest choices
(A) 6=2x3, so this won't cut it as we need 4 as a power of 2
(B) 9, no point looking here as there is no factor of 2
(C) \(12=2^2.3\) Awesome so this will work.

Answer: C
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If m is a positive integer and m^3 is divisible by 48 then what is 26 Mar 2025, 04:20
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Is there any standard method to solve such type of questions, I'm often puzzled on questions of this type. KarishmaB can you pls explain
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If m is a positive integer and m^3 is divisible by 48 then what is 27 Mar 2025, 05:21
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napolean92728
Is there any standard method to solve such type of questions, I'm often puzzled on questions of this type. KarishmaB can you pls explain

Yes. Split 48 into its prime factors. \(48 = 2^4 * 3\)

Given that \(m^3 = 48a = 2^4*3*a\)

In m^3, exponents of all prime factors must be a multiple of 3 (just like exponents of primes must be a multiple of 2 in perfect squares)
So at the minimum, we should have m^3 = 2^6 * 3^3 and hence m must be at least 2^2 * 3 = 12

I have a couple of posts that explain the basics of this:

https://anaprep.com/number-properties-f ... -a-number/
https://anaprep.com/number-properties-f ... ct-square/

You can also check out the entire explanation from my content using the free 3 day trial. Check Factors and Factorization under Number Properties. It will all start to make sense.
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If m is a positive integer and m^3 is divisible by 48 then what is 17 May 2025, 01:45
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Accordingly, if in the given question m would have been m^2 or m^5, should the exponents of the prime factors have been in multiples of 2 and 5 respectively?
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napolean92728
Is there any standard method to solve such type of questions, I'm often puzzled on questions of this type. KarishmaB can you pls explain

Yes. Split 48 into its prime factors. \(48 = 2^4 * 3\)

Given that \(m^3 = 48a = 2^4*3*a\)

In m^3, exponents of all prime factors must be a multiple of 3 (just like exponents of primes must be a multiple of 2 in perfect squares)
So at the minimum, we should have m^3 = 2^6 * 3^3 and hence m must be at least 2^2 * 3 = 12
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If m is a positive integer and m^3 is divisible by 48 then what is 17 May 2025, 02:04
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shori345
Accordingly, if in the given question m would have been m^2 or m^5, should the exponents of the prime factors have been in multiples of 2 and 5 respectively?
KarishmaB
napolean92728
If m is a positive integer and m^3 is divisible by 48 then what is minimum possible value of m ?

A) 6
B) 9
C) 12
D) 18
E) 48

Is there any standard method to solve such type of questions, I'm often puzzled on questions of this type. KarishmaB can you pls explain

Yes. Split 48 into its prime factors. \(48 = 2^4 * 3\)

Given that \(m^3 = 48a = 2^4*3*a\)

In m^3, exponents of all prime factors must be a multiple of 3 (just like exponents of primes must be a multiple of 2 in perfect squares)
So at the minimum, we should have m^3 = 2^6 * 3^3 and hence m must be at least 2^2 * 3 = 12

Yes, if we had m^2 or m^5 instead of m^3, then the powers of the primes in m^2 and m^5 would have to be multiples of 2 and 5 respectively.
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