Question

The sum of two number is 36 and their HCF is 4. How many pairs of such numbers are possible?​

Answer 1

Step-by-step explanation:

${\red{\huge{\underline{\mathbb{Solution:-}}}}} \\ \\ Given, \\ HCF \: of \: two \: numbers \: = 4 \\ Hence, \: the \: numbers \: can \: be \: given \: by \: 4a \: and \: 4b \\ where \: a \: and \: b \: are \: co \: prime \: \\ \\ then \\ 4a + 4b = 36 \\ 4(a + b) = 36 \\ a + b = \frac{36}{4} \\ \\ a + b = 9 \\ \\ possible \: pairs \: satisfying \: this \: condition \: are \: \\ (1 ,\: 8) \:, \: (4 ,\: 5) \: ,\: (2 \: 7) \\ \\ Number \: of \: possible \: pair \: is \: 3.$

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