Find the least number which when divided by 16, 24, 32 and 40 leaves the remainder 7, 15, 23 and 31 respectively.

Sure, let's solve this step by step.

Step 1: We need to find a number which is divisible by 16, 24, 32 and 40. The easiest way to do this is to find the Least Common Multiple (LCM) of these numbers.

Step 2: The prime factors of 16 are 2222, of 24 are 2223, of 32 are 22222 and of 40 are 222*5.

Step 3: The LCM is found by multiplying the highest power of all the prime numbers in these numbers together. So, the LCM of 16, 24, 32 and 40 is 2^5 * 3 * 5 = 960.

Step 4: Now, we need to find a number which when divided by 960 leaves a remainder of 7, 15, 23 and 31. This means we are looking for a number of the form 960k + r, where k is an integer and r is the remainder.

Step 5: If we subtract the remainders from the divisors, we get 16-7=9, 24-15=9, 32-23=9 and 40-31=9. This means that the number we are looking for is 9 less than a multiple of 960.

Step 6: So, the least number which satisfies these conditions is 960 - 9 = 951.