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Find the number of positive integers that satisfy both the following conditions:

Each digit is a 1 or a 3

The sum of the digits is 12

 Apr 20, 2024
 #1
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We can solve this problem by considering the number of 1s in the integer. Let x represent the number of 1s. Then the number of 3s is 4−x (since there are 4 digits total).

 

The sum of the digits is then x+3(4−x)=12, which combines like terms to 2x=12. Solving for x, we get x=6.

 

Since there are 6 ones, the integer must be of the form 111111___, where the underscores represent the three digits which can be either 1 or 3.

 

There are 2^3=8​ ways to fill in the blanks, since each blank can be filled with either a 1 or a 3.

 

For example, the valid integers include 111111, 111333, and 311131.

 

Therefore, the answer is 8.

 Apr 22, 2024

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