1. ## Verbal Problem

The sum of 3 consecutive integers is 75 less than 5 times the smallest integer. Find the LARGEST of the integers.

thx

2. $n + (n + 1) + (n + 2) = 5n - 75$
Solve for n; the largest integer would be n + 2.

3. Let a, b, c be three consecutive integers with a < b < c. So we have:

The sum of 3 consecutive integers: a + b + c

(Is) 75 less than 5 times the smallest integer: 5a - 75

So we have: a + b + c = 5a - 75.

If a, b and c are consecutives, and as we choose "a" as the smallest integer, then b = a+1 and c = a+2.
Thus: a + b + c = 5a - 75 => a + (a+1) + (a+2) = 5a - 75

3a + 3 = 5a - 75
-2a = -78
a = 39

The largest integer is c, wich is a + 2, so c = 41.

4. Originally Posted by bondesan
(Is) 75 less than 5 times the smallest integer: 75 - 5a
When translating the phrase "less than" for subtraction, you have to change the order of numbers, so it's $5a - 75$ and not $75 - 5a$.

5. Originally Posted by eumyang
When translating the phrase "less than" for subtraction, you have to change the order of numbers, so it's $5a - 75$ and not $75 - 5a$.
Indeed, thanks for the advice. I'll correct it.