A) Zero
B) Only 1
C) Only 2
D) Many
Correct Answer: A
Solution :
HCF of two numbers\[=16\] and LCM of two numbers\[=146\] Let the first number, \[x=16a\] and the second number, \[y=16b\] We know that, HCF \[\times \] LCM = Product of two numbers \[\Rightarrow \] \[16\times 146=x\times y\] \[\Rightarrow \] \[16\times 146=16a\times 16b\] \[\Rightarrow \] \[a\times b=\frac{146}{6}=9.125\] The number of such pairs is zero, since, \[a\times b\] is not a integer.You need to login to perform this action.
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