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

You don't need to take anything on right-hand side. Just see whether the values of k when multiplied and added are a multiple of 7.

By formula;.

Given reminder is 4, and divisor is 6, 9, 15, 18 and lcm of 6, 9, 15, 18 is 90. To find dividend we have to multiply 90 with remainder 4 and then add remainder ie 4.

I understand you trick, Thank you @Jignesh Rajput.

Guys , here it is asked the least multiple so the answer is 94.

This will satisfy all the conditions.

Hi, friends it's very simple trick.

LCM of 6, 9, 15, 18 is 90.

We can simply write the answer 90 + 4 but in question, it should be multiple of 7.

So (90k + 4) / 7 now we have to find the value of 'k' to complete the answer.

The first method is to substitute the value of k = 1, 2, 3. And divide it by 7 and sees the result. It actually time taking.

Now the trick:

(90k + 4) / 7 = 90k / 7 + 4 / 7,

The remainder of 4 / 7 is 4,

Remainder 90k / 7 is 6k.

Now add both remainders 6k+4 it should be divided by 7.

Now substitute the value of k it comes out be 4.

Put 90 * 4 + 4 = 364.

In the equation,

AX + B,

-> A is the LCM of all the given values.
-> B is the remainder.
-> The first number of the form: 90A+4 is 364. Hence, the ans.

No need to be messed up. Here we are given a clue (multiple of 7) it means the no is to be divided by 7. We got only 364 divisible with 7.

Now just make a test by dividing 364 by 6. As it leaves a remainder of 4 so 364 is the correct answer.

Thanks for the answer @Jignesh.

Sir, is there any shortcut method present for this type of question other then options verification?

How find the value of k==4?