Oh, dude, that's an easy one. So, you can write any number less than 1000 as the product of 3 consecutive numbers, right? Well, except for the numbers at the beginning and end because they don't have 3 numbers before or after them to multiply with. So, that leaves us with 997 numbers. Easy peasy lemon squeezy.
The numbers are 30, 31 and 32.
1 and 2, whose product is 2.
All the odd numbers.
They are: 32*33 = 1056
There are two consecutive odd, whole numbers. The numbers are 39 and 41.
The numbers are 30, 31 and 32.
There is no such number.
1 and 2, whose product is 2.
All the odd numbers.
For the product to be zero, one of the numbers must be 0. So the question is to find the maximum sum for fifteen consecutive whole numbers, INCLUDING 0. This is clearly achived by the numbers 0 to 14 (inclusive), whose sum is 105.
They are: 32*33 = 1056
No whole number would fit (their product would be even for starters).
10 and 12
Consecutive whole numbers have no other whole numbers between them.
There are two consecutive odd, whole numbers. The numbers are 39 and 41.
The average of 33 consecutive whole numbers is 58, what is the smallest of these whole numbers? The answer is 42
21/4 is not a whole number and it cannot be represented as consecutive whole numbers.