In the addition below, each of the letters A and $B$ stands for one digit, and the sum as printed is correct: $A9+4B=132$. What is the value of $A\times B$?

Begin at the units column: $9+B$ must end in 2, so $B=3$ and a ten is carried. Now the tens column reads $A+4+1$, and it must make 13 so that the 1 carries on to give the hundreds digit of 132. So $A+5=13$ and $A=8$. Checking: 89 + 43 = 132. The product is $8\times 3=24$. The traps: 27 comes from forgetting the carry into the tens and writing $A+4=13$, giving $A=9$ — but 99 + 43 is 142, not 132; 11 is $A+B$, the sum where the product was asked for; 32 multiplies A by the printed 4 instead of by $B$.