Question 1210210: How many positive integers are there whose digits strictly decrease from left to right, and the sum of the digits is 6?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let the positive integer have digits $d_1 d_2 \dots d_k$ such that $9 \ge d_1 > d_2 > \dots > d_k \ge 0$ and $d_1 + d_2 + \dots + d_k = 6$.
Since the digits are strictly decreasing and non-negative, all the digits must be distinct.
Also, since the sum of the digits is 6, the number of digits cannot be too large. The maximum possible number of digits occurs when the digits are $5, 1$, or $4, 2$, or $3, 2, 1$, or $6$.
We consider the possible sets of distinct digits whose sum is 6:
Case 1: One digit
The only possibility is the digit 6. The number is 6. (1 integer)
Case 2: Two digits
We need two distinct digits $d_1 > d_2$ such that $d_1 + d_2 = 6$. The possible pairs are (5, 1) and (4, 2). The corresponding integers are 51 and 42. (2 integers)
Case 3: Three digits
We need three distinct digits $d_1 > d_2 > d_3$ such that $d_1 + d_2 + d_3 = 6$. The only possible set of digits is {3, 2, 1}. The corresponding integer is 321. (1 integer)
Case 4: Four digits
We need four distinct digits $d_1 > d_2 > d_3 > d_4$ such that $d_1 + d_2 + d_3 + d_4 = 6$. The only possible set of digits is {3, 2, 1, 0}. The corresponding integer is 3210. (1 integer)
Case 5: More than four digits
If there are 5 distinct non-negative digits, the minimum sum would be $4 + 3 + 2 + 1 + 0 = 10$, which is greater than 6. So, we cannot have more than 4 digits.
The positive integers satisfying the conditions are 6, 51, 42, 321, and 3210.
The number of such integers is $1 + 2 + 1 + 1 = 5$.
Final Answer: The final answer is $\boxed{5}$
Answer by ikleyn(52777)  (Show Source):
You can put this solution on YOUR website! .
How many positive integers are there whose digits strictly decrease from left to right,
and the sum of the digits is 6?
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The answer in the post by @CPhill is incorrect.
The numbers satisfying the imposed conditions are
6, 60, 51, 510, 42, 420, 321, 3210.
The number of such numbers is 8.
Solved.
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