how many 3 digit numbers are possible such that product of digits is a natural number less than 9
Let d1 , d2 and d3 be the digits. Notice that none of the digits is zero, otherwise we would not have a natural number for product.
Also notice that if (d1,d2,d3) is a solution, with distinct digits, there are at least 3! solutions, because switching these three digits will not affect the product.
We will assume d1≤d2≤d3 .
If d1=1 and:
d2=2 , then there are 3 possibilities for d3 ( 2 , 3 and 4 );
d2=3 , then there is only 1 possibility for d3 ( 3 );
d2>3 , there are no solutions.
If d1=2 and:
d2=2 , then there is only 1 possibility for d3 ( 2 );
d2>2 , there are no solutions.
If d1>2 , there are no solutions, since 33>9 .
Thus we obtained 14 solutions.
Now we consider the permutations (switching digits).
If d1=d2 or d1=d3 or d2=d3 (and the other digit different), we only have 3 permutations (counting with the original). So, for each solution in this case, we should add 2 more. We have 10 solutions in this case. We add 20 solutions to the total. Now we have a total of 34 solutions.
The final answer is 44.
Answers
Answer:296
Step-by-step explanation:Essentially, we just need to find values A, B, C such that 1<=ABC<=5
For ABC=1 , this is only possible with the values 1,1,1. This gives one option.
For ABC=2 , only if one of them is 2. This gives three options, as the 2 could be anywhere.
Likewise, =3 gives us 3 results, whereas =4 gives 6 results. =5 gives 3 as well.
So, in total, we have 1+3+3+6+3 options, or 16 total.
If, however, zero is considered a natural number (which, according to James Tiroli, is sometimes the case), then we have a new result.
Any three digits x, x, and 0 will satisfy your answer. And as with the solution of 2, there are three possible places for the zero to go: A, B, or C. In addition, one more of the x’s could be a zero. Sometimes. More on that in a moment.
If we assume 1<=x<=9 , then there are 3∗9∗9 options, or 243. But x can be zero, meaning 3∗10∗10 , or 300 options. Unless x is the first x, where it could only be 1 through 9.
That is the case in 2 of the 3 locations for 0. So then our answer becomes:
2∗9∗10+1∗10∗10 for the two different scenarios to be accounted for. Which means, if zero is a natural number in your mind, there are 280 ways to get 0. Add this to the previous 16 to get a grand total of 296 possibilities.
Answer:
222, 123, 321, 213, 231, 132
Step-by-step explanation: