Find the greatest number which on dividing 391 and 318 leaves remainders 7 and 6 respectively.
Since on dividing 391 by the required no. the remainder is 7Â
then 391-7 ie. 384 will be exactly divisible by the required no.
Similarly,318-6 ie. 312 will be exactly divisible by the required no.
Now we will do prime factorisation of 384 and 312,
384 =Â $$2^2\times\ 3$$
312 =Â $$2^3\times\ 3\times\ 13$$
H.C.F =Â $$2^3\times\ 3=24$$
Required number is 24.
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