Question

Find three consecutive integers such that the sum of the first, twice the second, and three times the third is ?76.

Answer 1

Answer:

$\large\boxed{\text{These numbers do not exist}}$

Step-by-step explanation:

$n,\ n+1,\ n+2-\text{three consecutive integers such}\\\\\text{The equation:}\\\\n+2(n+1)+3(n+2)=76\qquad\text{use distributive property}\\\\n+2n+2+3n+6=76\qquad\text{combine like terms}\\\\(n+2n+3n)+(2+6)=76\\\\6n+8=76\qquad\text{subtract 8 from both sides}\\\\6n=68\qquad\text{divide both sides by 6}\\\\n=\dfrac{68}{6}\\\\n=\dfrac{34}{3}\\\\n=11\dfrac{1}{3}\notin\mathbb{Z}$