In the multiplication problem below, each capital letter represents a

hi..
ans D 8...
initially it becomes a bit confusing as we have almost 16 letters for 10 digits...
but then since the question stems asks us to concentrate only at B, the method is as follows...
HB9
x 9
S0W
it means w=1 as 9*9=81 and 9*B+8=a0..
B=8 as only this value gives us a 0 as units digit for eq

hi desigmat u have wrongly written 9*9 as 63 it is 81

My solution revolves around the trick of testing the anwser choices, however I would love to see an algebraic approach for this problem.

The information provided is that 9*9 = 81, the result of this is that the 8 is carried one place to the left. The problem asks you to seek for a solution that adds 2 to the tens digit place.

By testing the answer choices, what in my opinion is a tedious process, we can figure out the answer.

A) 29*9 = (9*9) + (9*20) = 81 + 180 = 261 - incorrect
B) 49*9 = (9*9) + (9*40) = 81 + 360 = 441 - incorrect
C) 79*9 = (9*9) + (9*70) = 81 + 630 = 711 - incorrect
D) 89*9 = (9*9) + (9*80) = 81 + 720 = 801 - correct
E) 99*9 = (9*9) + (9*90) = 81 + 810 = 891 - incorrect (for completeness)

Answer: D

Again, I would love to see an algebraic approach less tedious than this one.

hi i dont think you have to get into such calculations..
you have rightly said 9*9=81..
and next step= 9*b+8=a0... this means a0 -8 =9*B...only b as 8 gives 2 inunits digit

Answer = D = 8

9*9 = 81, which gives w = 1 & creates a carry of 8

Setting up the equation (For tens digit)

9*B + 8 = 0

To give a "0" at the end, we require 9*B to give "2" in the units place

Only 9*8 = 72 satisfies the condition

This question works toward helping you develop the "Math instinct" we talk about.
You get B*9 + 8 = ..0
So you know that B*9 should end with 2. Immediately 8*9 = 72 will come to your mind. Along with it, what should also come to your mind is that in the multiplication table of 9, every product ends with a different digit.

9*1 = 9
9*2 = 18
9*3 = 27
9*4 = 36
9*5 = 45
9*6 = 54
9*7 = 63
9*8 = 72
9*9 = 81
9*10 = 90

This happens for tables of 1, 3, 7 and 9 - i.e. all odd numbers except 5. So if this were a DS question, you would know that this is sufficient information without running through the table of 9.

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