Question

What is the least number which when divisible by 4, 6, 8 and 9 leaves zero remainder in each case but when divided by 13 leaves a remainder of 7?

Answer 1

Hint: First of all, divide this question into two parts to solve easily. In part 1 find the LCM of the numbers 4, 6, 8 and 9 to get the least number which is divisible by them and leaves 0 remainder. Next, in part 2 divide the obtained number with 13 and if it leaves a remainder of 7 then the obtained number also satisfies our second condition. So, we can say that that number is our required answer.

Complete step-by-step answer:
Since we are looking for the number that leaves 0 as remainder with 4, 6, 8 and 9, we are actually looking for the LCM (least common multiple) as a part 1.
So, consider
Multiples of 4 \[ = 2 \times 2\]
Multiples of 6 \[ = 2 \times 3\]
Multiples of 8 \[ = 2 \times 2 \times 2\]
Multiples of 9 \[ = 3 \times 3\]
Hence LCM of 4, 6, 8 and 9 \[ = 2 \times 2 \times 2 \times 3 \times 3 = 72\]
Now, if we check we have
\[
  72 \div 4 = 18{\text{ (0 remainder)}} \\
  72 \div 6 = 12{\text{ (0 remainder)}} \\
  72 \div 8 = 9{\text{ (0 remainder)}} \\
  72 \div 9 = 8{\text{ (0 remainder)}} \\
\]
The next part is to ensure that there's a remainder of 7 (with that number) when we divide it with 13.
Incidentally 72 leaves remainder of 7, when divided by 13, which is given by
\[72 \div 13 = 13 \times 5 + 7{\text{ (7 remainder)}}\]
Therefore, the least number which when divided by 4, 6, 8 and 9 leaves zero remainder, and when divided by 13 leaves a remainder of 7 is 72.

Note: The LCM (least common multiple) of two or more numbers is the smallest number that is evenly divisible by all numbers in the set. Here we have used a prime factorization method to find the LCM of the numbers 4, 6, 8 and 9.