There are two positive integers - a and b. And it is given that - sum of their hcf and lcm is 91. Find out the number of possible pairs which satisfy the above condition.
Answers
Answer:
16
Step-by-step explanation:
Hi,
Given that a and b are 2 positive integers
Given that Sum of H.C.F and L.C.M is 91.
But, we know that HCF is factor of LCM,
hence HCF should be also a factor of 91.
Factors of 91 are 1, 7, 13 and 91
But H.C.F cannot be equal to 91, because if H.C.F = 91, then
L.C.M will be at least 91 and it violates the given sum of 91
Hence , H.C.F could be either 1 or 7 or 13
Case 1: If H.C.F = 1
then L.C.M = 90
But we know that product of numbers ab = L.C.M*H.C.F
Hence, a b = 90 = 2*3²*5
Since H.C.F is 1 , entire 9 should be factor of one integer, else
hcf would become 3
ab = 2*9*5
The possible number of writing as product of 2 factors are
= (1+1)*(1+1)*(1 + 1)
Total number of ways of pairing are (1+1)*(1+1)*(1 + 1)
= 8
Hence 8 possible pairs could be formed whose H.C.F would
be 1
Case 2: If H.C.F = 7
then L.C.M = 84
But we know that product of numbers ab = L.C.M*H.C.F
Hence, a b = 7*7*14 = 7*7*2*7
But we know a is of form 7X and b is of form 7Y where
H.C.F(X, Y) = 1
Hence, ab = 49XY = 7*7*2*7
XY = 2*7
The possible number of writing as product of 2 factors are =
(1+1)*(1+1)
Possible number of ways of choosing pairs are (1 + 1)*(1 + 1)
= 4
Hence 4 possible pairs could be formed whose H.C.F would
be 7.
Case 3: If H.C.F = 13
then L.C.M = 78
But we know that product of numbers ab = L.C.M*H.C.F
Hence, a b = 13*13*6 = 13*13*2*3
But we know a is of form 13X and b is of form 13Y where
H.C.F(X, Y) = 1
Hence, ab = 169XY = 13*13*2*3
XY = 2*3
The possible number of writing as product of 2 factors are =
(1+1)*(1+1)
Possible number of ways of choosing pairs are (1 + 1)*(1 + 1)
= 4
Hence 4 possible pairs could be formed whose H.C.F would
be 13.
Hence, in total (8 + 4 + 4) = 16 pairs are possible.
Hope, it helps !
Answer:
Given that a and b are 2 positive integers
Given that Sum of H.C.F and L.C.M is 91.
But, we know that HCF is factor of LCM,
hence HCF should be also a factor of 91.
Factors of 91 are 1, 7, 13 and 91
But H.C.F cannot be equal to 91, because if H.C.F = 91, then
L.C.M will be at least 91 and it violates the given sum of 91
Hence , H.C.F could be either 1 or 7 or 13
Case 1: If H.C.F = 1
then L.C.M = 90
But we know that product of numbers ab = L.C.M*H.C.F
Hence, a b = 90 = 2*3²*5
Since H.C.F is 1 , entire 9 should be factor of one integer, else
hcf would become 3
ab = 2*9*5
The possible number of writing as product of 2 factors are
= (1+1)*(1+1)*(1 + 1)
Total number of ways of pairing are (1+1)*(1+1)*(1 + 1)
= 8
Hence 8 possible pairs could be formed whose H.C.F would
be 1
Case 2: If H.C.F = 7
then L.C.M = 84
But we know that product of numbers ab = L.C.M*H.C.F
Hence, a b = 7*7*14 = 7*7*2*7
But we know a is of form 7X and b is of form 7Y where
H.C.F(X, Y) = 1
Hence, ab = 49XY = 7*7*2*7
XY = 2*7
The possible number of writing as product of 2 factors are =
(1+1)*(1+1)
Possible number of ways of choosing pairs are (1 + 1)*(1 + 1)
= 4
Hence 4 possible pairs could be formed whose H.C.F would
be 7.
Case 3: If H.C.F = 13
then L.C.M = 78
But we know that product of numbers ab = L.C.M*H.C.F
Hence, a b = 13*13*6 = 13*13*2*3
But we know a is of form 13X and b is of form 13Y where
H.C.F(X, Y) = 1
Hence, ab = 169XY = 13*13*2*3
XY = 2*3
The possible number of writing as product of 2 factors are =
(1+1)*(1+1)
Possible number of ways of choosing pairs are (1 + 1)*(1 + 1)
= 4
Hence 4 possible pairs could be formed whose H.C.F would
be 13.
Hence, in total (8 + 4 + 4) = 16 pairs are possible.