# Two number, both greater than 29, have GCD = 29 and LCM = 4147. The sum of the numbers is: Option 1) Option 2) Option 3) Option 4) Option 5) -

$Product \: of\: numbers = 29\ast 4147.$

$Let \: the\: numbers \: be 29a \: and\: 29b.\:$

$Then, 29a \ast 29b = \left ( 24\ast 4147 \right )$

$=> ab = 143.$

$Now,\: co-primes \: with\: product \: 143 \:are \:\left ( 1, 143 \right ) \:and \:\left ( 11, 13 \right ).$

$So, \:the\: numbers\: are\: \left ( 29 \ast 1, 29\ast 143 \right )\:and\:\left ( 29 \ast 11, 29\ast 13 \right )$

$Since \: both \: numbers\: are \: greater\: than\: 29,$

$\: the\: suitable\: pair \: is \: \left ( 29 \ast 11, 29 \ast 13 \right )$

$i.e., \left ( 319, 377 \right ).$

$Required\: sum = \left ( 319 + 377 \right ) = 696.$

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