If you are not allowed to repeat the digits then the answer is clearly 0.
If you are allowed to repeat digits then the only way you can possibly reach a product of 4 is by using a combination of 3 "1's" and 2 "2's". The possible combinations are therefore:
22111
21211
21121
21112
12211
12121
12112
11221
11212
11122
Thus there are 10 numbers which meet the criteria.
There are 15 of them.
121
1348.
There are no 3-digit whole numbers whose digits sum to 3. The smallest 3-digit number is 100, and the largest is 999, but in neither case is the sum of the digits equal to 3.
I'm sure there are more than 2 prime numbers that are 400 digits long.
There are 15 of them.
21.
Two (or four) digits added together cannot equal 42. Two-digit numbers multiplied together cannot equal 82.
121
21
1348.
There are no 3-digit whole numbers whose digits sum to 3. The smallest 3-digit number is 100, and the largest is 999, but in neither case is the sum of the digits equal to 3.
This is only possible if one of the digits is equal to zero. There are 90 3-digit numbers with a zero in the 10's place, and 90 3-digit numbers with a zero in the 1's place - and 9 numbers that have both a zero in the 10's place and a zero in the 1's place; these would be counted double if you just add the first two. So, you get: 90 + 90 - 9 such numbers.
002+002+001=005
There are no three-digit numbers that equal 17. In fact, there are no numbers with more or less than two digits that equal 17. In fact, in the whole infinite supply of numbers, there is only one single number that equals 17. That number is . . . . . . . 17 .
I'm sure there are more than 2 prime numbers that are 400 digits long.
N squared would be used to find the square root of a number or numbers. In order to find the number of three digit numbers such that the sum of the square results of any two digits are equal to the third digit the use of the formula (HOE)squared=Hsquared*10000+2HE*100+Esquared is needed.