find three consecutive positive integers such that 4 times the first 13 more than the third.
The two end digits cannot be even because subtracting one from the other would be an even difference. Explore the odd choices.
123, 345, 567, 789
n-1 , n , n+1
4(n-1)-13 = n+1
4 n - 4 - 13 = n+1
3 n = 18
n = 6
5, 6, 7
To find three consecutive positive integers such that 4 times the first is 13 more than the third, we can follow these steps:
Let's assume the first positive integer is x.
The second consecutive positive integer would be x + 1.
The third consecutive positive integer would be x + 2.
According to the problem, we have the following equation:
4 times the first (4x) = the third (x + 2) + 13.
Now we can set up the equation and solve for x:
4x = x + 2 + 13
Simplify the right side: 4x = x + 15
Subtract x from both sides: 4x - x = 15
Combine like terms: 3x = 15
Divide both sides by 3: x = 5
So the first positive integer (x) is 5.
The second positive integer (x + 1) is 5 + 1 = 6.
The third positive integer (x + 2) is 5 + 2 = 7.
Therefore, the three consecutive positive integers are 5, 6, and 7.