What is the least number which when divided by 8, 12 and 16 leaves 3 as the remainder in each case, but when divided by 7 leaves no remainder ?

To solve this problem, we need to find the least common multiple (LCM) of 8, 12, and 16, and then add 3 to it.

Step 1: Find the LCM of 8, 12, and 16. The prime factors of 8 are 2, 2, and 2. The prime factors of 12 are 2, 2, and 3. The prime factors of 16 are 2, 2, 2, and 2.

The LCM is found by multiplying the highest power of all prime numbers that appear in the factorization of each number. So, the LCM of 8, 12, and 16 is 2^4 * 3 = 48.

Step 2: Add 3 to the LCM. 48 + 3 = 51.

So, 51 is the least number which when divided by 8, 12, and 16 leaves 3 as the remainder in each case.

Step 3: Check if 51 is divisible by 7. 51 divided by 7 gives a quotient of 7 and a remainder of 2.

So, 51 is not the number we are looking for.

Step 4: Find the next multiple of 48 (the LCM of 8, 12, and 16) that is divisible by 7. This is done by adding 48 to 51 repeatedly until we find a number that is divisible by 7.

Doing this, we find that 96 + 3 = 99 is the next number that leaves a remainder of 3 when divided by 8, 12, and 16, and leaves no remainder when divided by 7.

So, 99 is the least number which when divided by 8, 12, and 16 leaves 3 as the remainder in each case, but when divided by 7 leaves no remainder.