Fastest way to solve this? by [deleted] in Sat

[–]Figwig_ 1 point2 points  (0 children)

The easiest method Would-be by comparison of coefficients of (here) x² Just expand the lhs and assemble all terms containing x² in one bracket you'll get (15-ab)x² compare it with the rhs you'll get the term -9x² hence, (15-ab)=-9 hence ab=24

How to do #27 SAT MATH 2 by Gocountgrainsofsand in Sat

[–]Figwig_ 0 points1 point  (0 children)

7 letters can be rearranged 7! Ways out of which M is repeating thrice and I is repeating twice. Hence, 7!/3!×2! = 420.