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07 May 2019, 14:37
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Difficulty:
95%
(hard)
Question Stats:
35% (02:08) correct
65%
(02:22)
wrong
based on 143
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How many ordered pairs of positive integers (m,n) satisfy
GCD \((m^3, n^2) = 2^2 * 3^2\) and LCM \((m^2, n^3)= 2^4 * 3^4 * 5^6\)
where, GCD- Greatest common divisor and LCM- least common multiple
A. 1
B. 2
C. 4
D. 6
E. 8
GCD \((m^3, n^2) = 2^2 * 3^2\) and LCM \((m^2, n^3)= 2^4 * 3^4 * 5^6\)
where, GCD- Greatest common divisor and LCM- least common multiple
A. 1
B. 2
C. 4
D. 6
E. 8
prime factorization of m and n must consists of only the primes 2,3 and 5
m=\(2^a 3^b 5^c\)
n= \(2^d 3^e 5^f\)
where a,b,c,d,e and f are non-negative integers
Now GCD(\(m^3 n^2\))= \(2^2 3^2\) , we will get min(3a, 2d)=2, min(3b, 2e)=2 and min(3c, 2f)=0
we get d=1 e=1 and c=0 or f=0 { a,b,c,d,e and f are integers}
From LCM(\(m^2 n^3\)}= \(2^4 3^4 5^6\), we get max(2a,3d)=4, max(2b,3e)=4 and max(2c,3f)=6
We get a=2,b=2 and c=3 or f=2
Case1-
When c=0 then f=2
m=\(2^2 3^2 5^0\)
n= \(2^1 3^1 5^2\)
Case2-
When f=0 then c=3
m= \(2^2 3^2 5^3\)
n= \(2^1 3^1 5^0\)
Hence only two pairs of (m,n) are possible
m=\(2^a 3^b 5^c\)
n= \(2^d 3^e 5^f\)
where a,b,c,d,e and f are non-negative integers
Now GCD(\(m^3 n^2\))= \(2^2 3^2\) , we will get min(3a, 2d)=2, min(3b, 2e)=2 and min(3c, 2f)=0
we get d=1 e=1 and c=0 or f=0 { a,b,c,d,e and f are integers}
From LCM(\(m^2 n^3\)}= \(2^4 3^4 5^6\), we get max(2a,3d)=4, max(2b,3e)=4 and max(2c,3f)=6
We get a=2,b=2 and c=3 or f=2
Case1-
When c=0 then f=2
m=\(2^2 3^2 5^0\)
n= \(2^1 3^1 5^2\)
Case2-
When f=0 then c=3
m= \(2^2 3^2 5^3\)
n= \(2^1 3^1 5^0\)
Hence only two pairs of (m,n) are possible
General Discussion
elifceylan
Joined: 18 Aug 2018
Last visit: 24 Jul 2019
Posts: 9
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Given Kudos: 97
Location: Turkey
Concentration: Strategy, Technology
Schools: Other Schools (S)
GPA: 3.31
16 May 2019, 10:59
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nick1816 not sure if I understood your approach. Would you please share the official explanation?
