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How many ordered pairs of positive integers (m,n) satisfy

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nick1816
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How many ordered pairs of positive integers (m,n) satisfy [#permalink] New post   07 May 2019, 14:37
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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
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B
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nick1816
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How many ordered pairs of positive integers (m,n) satisfy [#permalink] New post   Updated on: 22 Aug 2019, 12:44
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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

Originally posted by nick1816 on 08 May 2019, 12:03.
Last edited by nick1816 on 22 Aug 2019, 12:44, edited 1 time in total.
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Re: How many ordered pairs of positive integers (m,n) satisfy [#permalink] New post   16 May 2019, 10:59
nick1816 not sure if I understood your approach. Would you please share the official explanation?
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Re: How many ordered pairs of positive integers (m,n) satisfy [#permalink] New post   17 May 2019, 01:33
I saw this question on a website and believe me other methods are way more difficult than mine.
I can give you an example, that might help you to understand my approach.
Consider two numbers 6 and 9
6= 2*3 and 9=2^0*3^2
Now GCD(6,9)= 3=3^1*2^0
exponent of 3 is minimum value between 1 and 2
exponent of 2 is minimum value between 1 and 0

LCM(6,9)=18=2^1*3*2
exponent of 3 is maximum value between 1 and 2
exponent of 2 is maximum value between 1 and 0

elifceylan wrote:
nick1816 not sure if I understood your approach. Would you please share the official explanation?
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How many ordered pairs of positive integers (m,n) satisfy [#permalink] New post   17 May 2019, 03:18
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nick1816 wrote:
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


Since GCD of m^3 and n^2 is 2^2 * 3^2, let

\(m^3 = 2^2 * 3^2a\) (a must be at least 2*3 to get 2^3*3^3. So m must be at least 2*3)
\(n^2 = 2^2 * 3^2b\) (b could be 1 or a perfect square. So n must be at least 2*3)
where a and b are co-primes.

Since LCM of m^2 and n^3 is 2^4 * 3^4 * 5^6,

- m^2 must have 2^4 * 3^4 (since n^3 cannot have 4 as an exponent). So m is at least 2^2 * 3^2
- Either m^2 or n^3 could have 5^6 but not both m and n because 5 does not appear in the GCD.
- n must have 2*3 and not higher powers of 2 and 3.

Various cases of m, n

Case 1:
m = 2^2*3^2*5^3
n = 2*3

Case 2:
m = 2^2*3^2
n = 2*3*5^2

Answer (B)
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Re: How many ordered pairs of positive integers (m,n) satisfy [#permalink] New post   28 Mar 2020, 07:22
nick1816 wrote:
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


\(Let m = 2^a * 3^b * 5^c & n = 2^d *3*e*5^f\)

Since GCD \((m^3, n^2) = 2^2 * 3^2\)
min (3a,2d) = 2; min(3b, 2e) = 2; min(3c,2f) = 0
d = 1; e = 1; Either c or f = 0

Since LCM \((m^2, n^3)= 2^4 * 3^4 * 5^6\)
max(2a,3d) = 4 ; max(2b,3e) = 4 ; max(2c,3f) = 6
a = 2; b = 2; c = 3 & f =0 or c = 0; f=3

m = 2^2*3^2 & n = 2*3*5^3

or

m = 2^2*3^2*5^3 & n = 2*3

Only 2 solutions are possible

IMO B
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Re: How many ordered pairs of positive integers (m,n) satisfy [#permalink] New post   27 Jun 2023, 21:38
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Re: How many ordered pairs of positive integers (m,n) satisfy [#permalink]
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