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13 Apr 2020, 08:55
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Difficulty:
95%
(hard)
Question Stats:
45% (02:09) correct
55%
(01:56)
wrong
based on 136
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How many pairs of positive integers x, y exist such that HCF (x, y) + LCM (x, y) = 91?
A. 10
B. 9
C. 8
D. 6
E. 7
Are You Up For the Challenge: 700 Level Questions
A. 10
B. 9
C. 8
D. 6
E. 7
Are You Up For the Challenge: 700 Level Questions
13 Apr 2020, 09:28
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x=ka
y=kb
GCD(x,y) = k
LCM(x,y) = k*a*b
k(1+ab) = 91
k must be a factor of 91. As, a and b are non-zero, k can't be 91.
k= 1, 7 or 13
Case 1 -
k= 1; LCM=x*y = 90= 2*3^2*5
Number of possible solutions = 2^2 = 4
Case 2- k =7; a*b = 12= 2^2*3
Number of possible solutions = 2^1 =2
Case 2- k =13; a*b = 6= 2*3
Number of possible solutions = 2^1 =2
Total unordered solutions = 4+2+2=8
y=kb
GCD(x,y) = k
LCM(x,y) = k*a*b
k(1+ab) = 91
k must be a factor of 91. As, a and b are non-zero, k can't be 91.
k= 1, 7 or 13
Case 1 -
k= 1; LCM=x*y = 90= 2*3^2*5
Number of possible solutions = 2^2 = 4
Case 2- k =7; a*b = 12= 2^2*3
Number of possible solutions = 2^1 =2
Case 2- k =13; a*b = 6= 2*3
Number of possible solutions = 2^1 =2
Total unordered solutions = 4+2+2=8
Bunuel
13 Apr 2020, 22:51
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firas92 AnirudhaS AdiBatman jackfr2
\(a*b = 90= 2*3^2*5 \)
a and b are co-primes. Hence, these 2 numbers have no shared factors. So either a or b is a multiple of \(3^2\). (You can't split it between a and b)
So we can write a as \(2^x*(3^2)^y*(5)^z\)
x,y and z can be 0 or 1
Total possible values of a = 2*2*2=8
This will give you number of ordered pairs of (a,b).
For example-
i) when x=y=z=0; a=1 and b= 90
ii) when x=y=z=1; a=90 and b= 1
Number of unordered pairs =\( \frac{2^3}{2}=2^2 =4\)
If number of prime factors of a*b is n, and a and b are coprimes, then number of possible ordered pair of (a,b) = \(2^n\) and number of unordered pair of (a,b) is \(2^{n-1}\)
\(a*b = 90= 2*3^2*5 \)
a and b are co-primes. Hence, these 2 numbers have no shared factors. So either a or b is a multiple of \(3^2\). (You can't split it between a and b)
So we can write a as \(2^x*(3^2)^y*(5)^z\)
x,y and z can be 0 or 1
Total possible values of a = 2*2*2=8
This will give you number of ordered pairs of (a,b).
For example-
i) when x=y=z=0; a=1 and b= 90
ii) when x=y=z=1; a=90 and b= 1
Number of unordered pairs =\( \frac{2^3}{2}=2^2 =4\)
If number of prime factors of a*b is n, and a and b are coprimes, then number of possible ordered pair of (a,b) = \(2^n\) and number of unordered pair of (a,b) is \(2^{n-1}\)
firas92
