A number is divisible by 9 if the sum of its digits is divisible by 9. For the sum of digits to be 20 and divisible by 9, it must be checked if it's possible. There are exactly 5 such combinations that satisfy the condition.
The sides are 9, 12, and 15 (since 3x + 4x + 5x = 36). This is a right triangle with legs 9 and 12 1/2 × 9 × 12=54
The area is 24 because the original problem likely had a different calculation method; otherwise, it’s good to check for any errors.
The solutions are (x,y)=(8,1),(6,2),(4,3),(2,4),(0,5). Counting the positive solutions, we get 5 solutions.
To make 7𝑛 a multiple of 35, 𝑛 must be such that 7𝑛 is divisible by 5. Thus, 𝑛 needs to be a multiple of 5. The smallest such 𝑛 is 5.
The integers are 7, 14, 21, 28, 35, and 42. Their sum is 7+14+21+28+35+42= 147.
