What is the sum of digits of the least number which when divided by 21, 28, 30 and 35 leaves the same remainder 10 in each case but is divisible by 17?
L.C.M. (21,28,30,35) = 420
Least number that will leave remainder 10 will be of the form = $$420n+10$$
Now, if $$n=1$$, number = 430, which is not divisible by 17
But, when $$n=2$$, number = 850, which is divisible by 17
Thus, sum of digits of number 850 = $$8+5+0=13$$
=> Ans - (B)
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