Question

7. A number m is such that when it is divided by 30, 36, and 45 the remainder is always 7, find
the smallest possible value of m. (3 marks)​

Answer 1

Answer:

Step-by-step explanation:

First we have to find the lcm of 30,36 and 45

30=3*5*2

45=3*5*3

36=3*3*2*2

lcm =180

therefore m=180+7=187                since 7 is the remainder in all cases

Answer 2

The given question is A number m is such that when it is divided by 30, 36, and 45 the remainder is always 7, find

the smallest possible value of m.

let us find the lcm for these numbers

The factors for 30 is 2*3*5.

The factors for 36 is 3*3*2*2

The factors for 45 is 5*3*3

Therefore the LCM is 2*3*5*2*3=180

It is said the the remainder is always 7,then add 7 with the LCM, we get

180+7=187.

Therefore, the final answer is 187

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