Find the smallest no. Which when divided by 55 and 121 leaves the remainder 16 in each case
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Answered by
13
LCM of 55 and 121 = 605
required number = 605+16
= 621
required number = 605+16
= 621
Answered by
6
on thinking for a time u will find that that the required no.is the sum of l.c.m. of 55 and 121 and 16.
now u can solve it
now u can solve it
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