If a, b and c are positive integers for which (a, b,c) is the only ordered triple such that 6a +9b + 20c = 61, then what is
a+b+c?

If a, b, and c are all positive integers and 6a + 9b + 20c = 61, then right away you know c < 3, because if c = 3, then 20c = 60, and there are no solutions for a and b.

So consider two cases:

• If c = 2, then we're left with 6a + 9b = 21.

• If c = 1, then 6a + 9b = 41.

We have gcd(6, 9) = 3, but 41 is prime and 3 does not divide 41. So we throw out the second case.

This leaves us with 6a + 9b = 21. Notice that if a = b = 1, then 6a + 9b = 15. Add 6 to both sides to get 21. So a = 2, b = 1, and c = 2, and hence a + b + c = 5.

1. 22

2. 1

3. 7

4. 18

step-by-step explanation

e2020 got it right

12

step-by-step explanation:

12 + 5 = 17

12x2=24

24-7=17