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

Simply we can solve this by,
5*6*7*8 = 1680.

And then after 1680+3 = 1683
The answer is 1683.

Firstly, find the L.C.M of 5,6,7,8. That will give us 840.
That means 840 is divisible by 5,6,7 and 8.

According to the question, the number must leave remainder 3 when divided by 5,6,7,8. So 840+3 is the number.

Now number must be divisible by 9 and not leave any remainder. But 843 doesn't divide by 9 properly so we look for another number.

840*1 + 3 no divisible by 9.
840*2 +3 = 1683. Divisible by 9 => Hence it's the answer.

840k+3 can anybody explain this step?

Take=1,2,3,4 till reminder is zero.

Let the required number be (840k+3)/9.
If k = 1 take 840(*1)+3/9=93.667=reminder is not 00
If k = 2 840(*2)+3/9=187= reminder is 00.
Take you k=2 because reminder is 0.
The solution is 840k+3=1683.
The least value of k for which (840k+3) is divisible by 9 is k=2.

Let's check the options from the divisibility rule of 9. i.e; the sum of all no should be divisible by 9.

1+6+7+7=21/9 =2.222 - NOT DIVISIBLE.
1+6+8+3=18/9=2 - DIVISIBLE.
2+5+2+3=12/9=1.3 - NOT DIVISIBLE.
3+3+6+3=15/9=1.6 -NOT DIVISIBLE.

So, Answer is Option b)1683.

Lcm of 5, 6, 7, 8 = 840.

So, divide all the numbers in the option by 840.

i) 1677/840= 840 ×1+remainder( of course gonna be more than 3
Since in question given that 5,6,7,8
when divided a number leaves the remainder 3)

Moving on to the second option:
1683/840 = 2×840(i.e 1600)+3.

(Leaves remainder 3 ---> condition satisfied)

Checking second condition ( leaves no remainder when divided by 9)
1683/9 = 187.
So, yeah the option is B) 1683.

( Also the third and fourth option leaves the remainder when divided by 9).