Find three positive consecutive integers such that the product of the first and the second is two more than three times the third

3 consecutive integers: a+2=b+1=c

a*b+2>3*c

substitute a and b definitions with c:
(c-2)*(c-1)+2>3c
c^2-c-2c+2+2>3c
c^2-3c+4>3c
c^2-6c+4>0
factoring
c^2-6c+4=(c-3)^2-5

(c-3)^2-5>0
(c-3)^2>5
c-3>sqrt(5)
c>sqrt(5)+3

c is an integer so we round up and don't have to calculate it exactly:
c=6, because they are consecutive a=4, b=5
4*5+2>3*6
20+2>18

Wednesday1.) < 2.) 1961015.)6.486.)68.761 can be rounded up to 69i can't truly see the rest and my next comment will be thursday

72 is the mode because it shows up the most number of times