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13 May 2019, 23:33
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Difficulty:
85%
(hard)
Question Stats:
40% (02:06) correct
60%
(02:14)
wrong
based on 45
sessions
History
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How many ordered pairs (a, b) of positive integers satisfy the equation a*b + 63 = 20*lcm}(a, b) + 12*gcd(a,b), where gcd(a,b) denotes the greatest common divisor of a and b, and lcm(a,b) denotes their least common multiple?
A. 0
B. 2
C. 4
D. 6
E. 8
A. 0
B. 2
C. 4
D. 6
E. 8
17 May 2019, 02:47
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a*b= lcm(a,b) * gcd(a,b)
Let lcm(a,b)= l
and gcd(a.b)=g
Hence we can re-write the equation as
lg+63=20l+12g
or l=(12g-63)/g-20
or l= 12+(177/g-20)
l must be a positive integer, hence 177/g-20 will also an integer
g-20 can take 8 values -177, -59, -3, -1,1,3,59 or 177
check for each value, it must satisfy 2 conditions
1. l and g must be positive
2. l is a multiple of g
Only case satisfied both conditions when g-20=1, g=21
l= 12+177/1=189
gcd(a,b)=21
a=21x and b=21y
lcm(a,b)=189
only possible when either x=1 and y=9, or when x=9 and y=1
Hence 2 solutions are possible
Let lcm(a,b)= l
and gcd(a.b)=g
Hence we can re-write the equation as
lg+63=20l+12g
or l=(12g-63)/g-20
or l= 12+(177/g-20)
l must be a positive integer, hence 177/g-20 will also an integer
g-20 can take 8 values -177, -59, -3, -1,1,3,59 or 177
check for each value, it must satisfy 2 conditions
1. l and g must be positive
2. l is a multiple of g
Only case satisfied both conditions when g-20=1, g=21
l= 12+177/1=189
gcd(a,b)=21
a=21x and b=21y
lcm(a,b)=189
only possible when either x=1 and y=9, or when x=9 and y=1
Hence 2 solutions are possible
