Question

# Find the smallest number which when divided by 12, 20, 30 or 60; leaves a remainder 5 each time.

Hint: To find such a number, first find the smallest number which is completely divisible by the numbers 12, 20, 30 and 60. This smallest number is nothing but Lowest Common Multiple (LCM) of these 4 numbers. If we add 5 to this number then dividing by any of these 4 numbers will always give remainder 5 and that would be the required answer.

To find such a number, first we need to find the smallest number which is completely divisible by the numbers 12, 20, 30 and 60. This smallest number is nothing but Lowest Common Multiple (LCM) of these 4 numbers. If we add 5 to this number then on dividing by any of these 4 numbers will always give remainder 5.
For example, suppose we have two numbers 10 and 15. LCM of 10 and 15 is 30. Now, suppose we add 4 to the LCM. Now the new number is 34. If we divide 34 by 10 or 15, we will always get 4 as remainder. We have already ensured that this number is the smallest by taking the LCM.
So, we have to find the Lowest Common Multiple (LCM) of 10, 12, 20 and 30.
To find LCM we have to write the given numbers as the product of the smallest factors.
\begin{align} & 10=1\times 2\times 5 \\ & 12=1\times 3\times 4 \\ & 20=1\times 2\times 2\times 5 \\ & 30=1\times 2\times 3\times 5 \\ \end{align}
Now we have 4 distinct factors- 1,2, 3, 4, 5. Since 2 has the highest power of 2 in 20 and all other factors 1,3,4,5 has the highest power of 1. Therefore,
$LCM=1\times 2\times 2\times 3\times 5=60$
Now to get the required number, we add 5 to this number
$60+5=65$