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 1 5 . 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 1 5 total solutions minus the solutions where x = 0 , so 1 5 - 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, 4 0 0 .
We can arrange ( 2 , 2 , 0 ) in two ways to get 3 digit numbers i.e, 2 2 0 , 2 0 2 .
We can arrange ( 2 , 1 , 1 ) in 3 ways i.e, 2 1 1 , 1 2 1 , 1 1 2 .
We cam arrange ( 1 , 3 , 0 ) in 4 ways i.e, 1 3 0 , 3 0 1 , 3 1 0 , 1 0 3 .
Hence, total number of 3 digit number having digit sum equal to 4 are 1 0 .
I think the pattern is that the total number is equal to the sum of all natural numbers upto 4 = 2 4 ( 4 + 1 ) = 1 0
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 1 0 + 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 x y z
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 1 0 of x y z