Question

The greatest number that will divide 148, 246, and 623 leaving remainders 4, 6, and 11, respectively, is

• A. $12$
• B. $18$
• C. $24$
• D. $30$

Answer 1

The correct option is A $12$

The required greatest number=H.C.F of $(148-4),(246-6),(623-11)$
= H.C.F of $144,240,612$
Here, the HCF of $144,240,and612$ is

$2$	$144$	$240$	$612$
$2$	$72$	$120$	$306$
$3$	$36$	$60$	$153$
	$12$	$20$	$51$

$H.C.F=2\times 2\times 3=12$
$\therefore$ The required number is $12$.