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If the least common multiple of m and n is 24, then what is
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30 May 2018, 01:22
30 May 2018, 01:22
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Question Stats:
85% (01:29) correct
15% (00:57) wrong based on 20 sessions
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If the least common multiple of m and n is 24, then what is the first integer larger than 3070 that is divisible by both m and n?
(A) 3072
(B) 3078
(C) 3084
(D) 3088
(E) 3094
(A) 3072
(B) 3078
(C) 3084
(D) 3088
(E) 3094
Re: If the least common multiple of m and n is 24, then what is
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05 Jun 2018, 13:10
05 Jun 2018, 13:10
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Expert Reply
Least common multiple of \(m\) and \(n\) is 24.
So for a number larger than 3070 to be divisible by both \(m\) and \(n\), it must be divisible by the least common multiple of m and n as well.
Using the gre calculator \(\frac{3070}{24}=127.9166\).
Clearly multiple of 24 larger than 3070 is \(128 \times 24=3072\).
So for a number larger than 3070 to be divisible by both \(m\) and \(n\), it must be divisible by the least common multiple of m and n as well.
Using the gre calculator \(\frac{3070}{24}=127.9166\).
Clearly multiple of 24 larger than 3070 is \(128 \times 24=3072\).
Re: If the least common multiple of m and n is 24, then what is
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08 Jun 2018, 09:26
08 Jun 2018, 09:26
1
24=23∗324=23∗3
As per divisibility rule -
1. For a number to be divisible by 8, the last 3 digits must be divisible by 8.
2. For a number to be divisible by 3, the sum of the digits must be divisible by 3.
Check the options only (A) follows
1. 072 - Divisible by 8
2. 3072 - Sum of the digits is 12 , which is divisible by 3
As per divisibility rule -
1. For a number to be divisible by 8, the last 3 digits must be divisible by 8.
2. For a number to be divisible by 3, the sum of the digits must be divisible by 3.
Check the options only (A) follows
1. 072 - Divisible by 8
2. 3072 - Sum of the digits is 12 , which is divisible by 3
