Question

the least multiple of 7 which leaves a remainder of 4 when divided by 6, 9, 15 and 18 is ?
a) 74
b)94
c)184
d)364

Answer 1

Answer:

Answer: Option D

Step-by-step explanation:

Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4 = 364.

Answer 2

Answer:

D) 364

Step-by-step explanation:

So,

For solving this sum, we need to find the LCM of 6,9,15 and 18.

After we solve it, we get it as 90.

Lcm(6,9,15,18) = 90

Now,

In the question they have given it should be a multiple of 7 and it leaves a remainder as 4.

For this, we can keep K as a constant. (If we multiple 90 by any number of K we get is as a multiple of 7)

So the equation will be,

N= 90K +4

Let's break 90 in terms of the multiple of 7.

So that will be = > 84K + 6K +4 =N

Now 84 K is a multiple of 7.

But, we need to prove if 6K + 4 is a multiple of 7.

So by trial and error method, we get K as 4.

Now we can substitute it in the equation we get is as,

90*4 + 4 = N

=> N= 364

Therefore,

The least multiple of 7 which leaves a remainder of 4 when divided by 6, 9, 15 and 18 is 364

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