Question

Find the smallest number which when
divided by 42 and 48 leaves remainder
12 and 18 respectively.​

Answer 1

Answer:

We got LCM = 9576

Difference between divisor and reminder is

18–15=3

24–21=3

38–35=3

42–39=3

So the difference is common for all which is 3

LCM - Difference

= 9576 - 3

= 9573

9573 is the smallest possible number.

Answer 2

Concept introduction:

The quantity that is "left over" after conducting a calculation is referred to as the remnant in mathematics. The integer that remains after multiplying one integer by another to get an integer quotient is known as the remnant in mathematics.

Given:

Here it is given that diving by $42$ and $48$ leaves remainder $12$ and $18$respectively.

To find:

We have to find the acquired number.

Solution:

According to the question,

Given,

diving by $42$ and $48$ leaves remainder $12$ and $18$ respectively.

and $42-12=30$ and $48-18=30$ are divisible by the required numbers.

Therefore, the required number will be $30$ less than the LCM

LCM  is $336$

So, the acquired number is $336-30=306$.

Final answer:

The final answer of the question is $306$.

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