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Suppose a 6 digit number N is formed by rearranging the digits of the number 123456. If N is divisible by 5, then the set of all possible
remainders when N is divided by 45 is
(A) {30} (B) {15, 30} (C) {0,15,30} (D) {0, 5, 15, 30}
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Its an ISI question please someone add appropriate tag
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What's ISI ?
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Indian Statistical Institute.
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Its an amazing question total 120 combination are there have tried 10 and getting 30 as remainder, answer is either a or b but not which one
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2 Answers

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answer is option a) {30 }, bcoz no. is divisble by 5 ( i.e last digit is fixed ) , so we have to find remainder when no. is divive by 45 , for this first divide the no. by 5 and then divide the resultant by 9 and any no. is divisible by 9 if its sum is divisible by 9. So (1+2+3+4+6)%9 = 6 .  so remainder is 6*5 = 30
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The remainder is coming out to be 7.How did you get 6? And why did you not include 5 while adding the digits?
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N = 45q + r
q and r are quotient and remainders rsep.

now reduce the whole equation to modulus 9
N = r = 3 (mod9)
and the only possible value for such an "r" is 30 = 3 (mod9)
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