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

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.

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

This will satisfy all the conditions.

I understand you trick, Thank you @Jignesh Rajput.

How Lcm of 90 come? Please explain.

@Priya it is so easy.

6 = 2 * 3
9 = 3 * 3
15 = 3 * 5
18 = 2 * 3 * 3.

Take one number from common multiple and multiply rest digits.
2*3*3*5.

90k + 4/7 find k.
90k/7 + 4/7.

We have to distribute 90 k as multiple of 7 and divisible by 7.
12 * 7k + 6k + 4/7.
12 * 7k/7 + 6k + 4/7.

Now we can put the value of k in 6k+4/7.

Hey guys, what will we take in the right-hand side? Please help me out.

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.

Here is my simple explanation.

I think everyone is understanding how did they get ab = 12,
now if first no. is 13a,
and second no. is 13b,
factors of 12 can be (1,12) (3,4) (6,2),

Now, see if you include any common factor in a and b both then that will increase the H.C.F as HCF if highest common factor like if we take factor 6 and 2 then there is common factor between 6 and 2 which is 2 so HCF will change now to 26 that's why we can't take 6 and 2.
So only 2 pairs (1,12) (3,4).