Aptitude - Problems on H.C.F and L.C.M - Discussion

Explanation of the answer is very helpful, Thank You all.

@Suchita.

You are truly right this really works!

Must say this is the best Sol for this question. Thanks.

Any number whose digits sum is nine (9) then only the number is divisible by nine.

taking k = 1..means 843 , sum will be 8+4+3 = 15 not divisible by nine.

k = 2 is 1643 & sum of digit is 1+6+4+3 = 18 means 1+8 = 9 hence least number is k=2.

Find the least number which when divided by 12, leaves a remainder of , when divided by 15, leaves a remainder of 10, when divided by 16, leaves a remainder of 11.

(a)115 (b)235 (c)247 (d)475

Solve and find the answer.

@Suchita.

18 is divisible by 6. There is no remainder. But as per the question, we should get 3 as the remainder when we divide the number by 6.

Why are we using variable 'K'? Please tell the method not shortcut.

Please tell me, how to find the value of k?

@Itsnoble.

111111111 is divisible by 9.

You will not get this much time in solving the question in competitive exams. There is a shortcut in solving this.

First check divisibility of each number-3 with 5,6,7 and 8 and the divisibility of the number with 9.

a) 1677 -3=1674, it will not be divisible by 5 so no need to check with any other.
b)1683-3= 1680, Last digit is 0 hence divisible by 5. Divisible by 2 (even number) and 3 (sum = 15 is divisible by 3) hence divisible by 6.

Checking divisibility by 7-> 16-8*2= 0. hence divisible by 7.
It is divisible by 8 -> (8*21=168).
Sum of original number 1683 = 18 which is divisible by 9 hence 1683 is divisible by 9.
There is no other number smaller than 1683 Hence the answer is 1683.

Another shortcut which will work in this case is checking thew divisibility by 9.
only 1683 is divisible by 9 hence the answer is 1683. In case there is more numbers divisible by you have to check it as my previous answer.