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k is a positive integer and 225 and 216 are both divisors of
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Updated on: 02 Feb 2020, 13:29
Updated on: 02 Feb 2020, 13:29
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Question Stats:
66% (02:04) correct
33% (01:58) wrong based on 278 sessions
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k is a positive integer and 225 and 216 are both divisors of k. If \(k = 2^a \times 3^b \times 5^c\), where a, b and are positive integers, what is the least possible value of \(a + b + c\)?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Kudos for the right answer and explanation
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Kudos for the right answer and explanation
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Re: k is a positive integer and 225 and 216 are both divisors of
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02 Feb 2020, 11:19
02 Feb 2020, 11:19
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Carcass wrote:
k is a positive integer and 225 and 216 are both divisors of k. If \(k - 2^a \times 3^b \times 5^c\), where a, b and are positive integers, what is the least possible value of \(a + b + c\)?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Kudos for the right answer and explanation
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Kudos for the right answer and explanation
\(225 = 5^2 * 3^2\) and \(216 = 2^3 * 3^3\)
Since 225 and 216 are both divisors of \(k\), it must be that \(k\) is a multiple of the LCM of 225 and 216
=> \(k\) = Multiple of \(5^2 * 3^3 * 2^3\)
Since \(k = 2^a \times 3^b \times 5^c\), and we need the minimum values of \(a, b, c\), we take:
=> \(k\) = \(5^2 * 3^3 * 2^3\) = \(5^c * 3^b * 2^a\)
=> \(a = 3\), \(b = 3\), \(c = 2\)
=> \(a + b + c = 8\)
Answer E
General Discussion
Re: k is a positive integer and 225 and 216 are both divisors of
[#permalink]
26 Jul 2021, 06:00
26 Jul 2021, 06:00
2
Easiest way to solve this question is to do the prime factorization of both 225 and 216.
225: (3)^2 (5)^1
216:(3)^3 * (2)^3 * (5)^2
The least common multiple is therefore a = 3, b =3, c = 2.
Finally, 3 + 3 + 2 = 8.
B.O.L
225: (3)^2 (5)^1
216:(3)^3 * (2)^3 * (5)^2
The least common multiple is therefore a = 3, b =3, c = 2.
Finally, 3 + 3 + 2 = 8.
B.O.L
