The "Digit Sum" of 456 would be 15, since 4+5+6=15. How many 3-digit numbers have a digit sum of 4? (The hundreds place cannot be 0, so an answer like 022 wouldn't count.)
Hint: There's a pattern in this problem!
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Use stars and bars/sticks and stones, where $x$ is the first digit (hundreds digit), $y$ is the second digit (tens), and $z$ is the third digit (ones).
$x$ + $y$ + $z$ = $4$
Since these are all nonnegative integers, none of them will be above $4$ , so definitely none of them will be above $9$ . Stars and bars states that the number of nonnegative integer solutions to this is $4$ plus the $2$ plus signs choose the $2$ plus signs. This is $6$ choose $2$ , which is $15$ . All of the solutions with $x$ = $0$ are not valid. If $x$ = $0$ , then $y$ + $z$ has to equal $4$ . Using stars and bars again, $y$ + $z$ = $4$ , and the number of solutions to this will come out to be $5$ choose $1$ , which is $5$ . The $15$ total solutions minus the solutions where $x$ = $0$ , so $15$ - $5$ = 10 .
If you allow negative numbers like -444 or -563, you can get 46 solutions, or more maybe so I don't think we should stay restricted at positive numbers only.
x+y+z=4,x>=1,y>=0,z>=0 or x+y+z=6,x>=1,y>=1,z>=1, C(5,2)=5*4/2=10
The way to approach this problem is simple. find all number combinations that sum up to 4 (dont forget that you can use 0 and you can use these numbers in different orders). these sums are 1+1+2, 1+3+0, 2+2+0 and 4+0+0. These are all the possible sums that you can use. out of 1+1+2 you can make 112, 211 or 121. out of 3+1+0 you can make 310, 103, 130 and 301. out of 2+2+0 you can make 220 and 202. and out of 4+0+0 you can make 400. That makes 3+4+2+1=10 three digit numbers that have a digit sum of 4.
First take all possible sets of 3 single digit numbers that add up 4
They are $(4,0,0),(2,2,0) , (2,1,1) , (1,3,0)$
We can arrange $(4,0,0)$ in only one way to get 3 digit number i.e, $400.$
We can arrange $(2,2,0)$ in two ways to get 3 digit numbers i.e, $220 , 202.$
We can arrange $(2,1,1)$ in 3 ways i.e, $211 , 121 , 112.$
We cam arrange $(1,3,0)$ in 4 ways i.e, $130 , 301 , 310 , 103.$
Hence, total number of 3 digit number having digit sum equal to 4 are $\boxed{10}.$
I think the pattern is that the total number is equal to the sum of all natural numbers upto 4 = $\dfrac {4(4+1)}{2} = 10$
That pattern looks good for small digit sums: 1 3-digit number has digit sum 1; 3 have digit sum 2; 6 have digit sum 3; 10 have digit sum 4; 15 have digit sum 5...but what about larger digit sums? As an extreme case, only 1 3-digit number has digit sum 27. Where does the pattern break down?
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I actually didn't clearly understand what you have stated, sorry for that. I didn't solve this based on pattern, so I don't know much about it. Sorry:(
This pattern works for all digit sums less than 10 because all ways to split them into 3 digits only involve single digits.
For a digit sum of 10, the sum $10+0+0$ doesn't correspond to a 3-digit number, so there are only 54 such numbers.
The average digit sum (see this problem ) of 3-digit numbers in base 10 is 14, so I think that the number of 3-digit numbers with fixed digit sum $n$ (let's call this $a_n$ ) is maximised for a digit sum of 14 (which is about half of the maximum digit sum)
After that, I don't know how it continues, but maybe the numbers decrease in the same way as they increased.
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Let the number be $xyz$
We have $x+y+z = 4$
Now we know that $4 = 1+3 = 2+2 = 4+0 = 2+1+1$
With 1 and 3 we can make 103, 130, 301, 310
With 2 and 2 : 202, 220
With 4 and 0 : 400
With 2,1,1 we have : 211, 121, 112
So total we have $\boxed{10}$ of $xyz$