That can be solved by treating the digits of the number as two separate variables, and then expressing the question as an equation:
10A + B = 3AB
3AB - 10A - B = 0
A(3B - 10) - B = 0
A = B/(3B - 10)
Now, we know that A and B must both be positive single digit numbers (otherwise, our first equation wouldn't fit the problem). That means "3B - 10" must give us a positive (in order to get that positive result). Because of that, we can see that B must be greater than or equal to 4. So let's try plugging in some values:
if B = 4:
A = 4/(12 - 10) = 2
if B = 5:
A = 5/(15 - 10) = 4/5
As B goes up, A will get smaller and smaller. Since it is already a fraction, we know that the first answer is the only one. The answer then is 24.
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.
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:22111212112112121112122111212112112112211121211122Thus there are 10 numbers which meet the criteria.
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 .
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:22111212112112121112122111212112112112211121211122Thus there are 10 numbers which meet the criteria.
I'm sure there are more than 2 prime numbers that are 400 digits long.