GMAT Strategy

AB
x CA
OEBC
In the multiplication above, each letter stands for a different non-zero digit,
with A x B < 10. What is the two-digit number AB?
(A) 23 (B) 24 (C) 25 (D) 32 (E) 42

hI GUYS NEED SOME EXPLANATION ON THESE TYPE OF QUESTIONS

Wow - what a terrific question! On any of these questions in which they're testing you on a calculation with variables at the digit level, the key invariably becomes this:

1) Start with what you know, and
2) Use that to limit the number of possibilities

Now, that's pretty vague but follow me here. Here's what we know:

- B*A is less than 10, and B*A = C
- A is the tens digit of the top number, and multiplying that by another 2-digit number gives you a 4-digit product.

Already we know that:

- B*A has to be either 2*3 or 2*4. Why? Because it has to be less than 10 (and anything 4*3 or higher is bigger than 10), and we can't repeat digits. 2*2 and 3*3 are out, as is 1*anything because we need a new third unique digit for C.

Here, I guess we at least knew that answer choice C is out, since we can't have B = 5 or we'll violate those rules.

- So C is either 6 or 8. And we know that, in doing the multiplication, we'll need to end up with BC as the tens and units digits. So if we try the different combinations of:

A = 2, B = 3, C = 6
A = 3, B = 2, C = 6
A = 2, B = 4, C = 8
A = 4, B = 2, C = 8

We need to make sure that B ends up in the tens place. If we don't end up repeating B then we can stop the math. Now, the point is kind of moot if you do the others but my initial thought was that A should be bigger, since you need to get over 1,000 in the product. (Now, that's moot since C has to be so big, but at first I didn't want to start with A = 2 since so many 20-something * up-to-40-something calculations end up less than 1,000 and would give us only 3 digits) So I may have gotten lucky but I started with A = 4, B = 2 to make sure that I got a 4-digit product right away:

42
*84

first line: 168 (doing multiplication by hand)
second line: 3360
total 3528

And so here we ended up with B in the tens place and C in the units place, and O and E are unique digits, so the parameters are all taken care of. I ended up a little lucky trying 42 first, since my logic wasn't entirely foolproof, but that worked.

If you had started with the others, though, you'd find pretty quickly that you don't end up with B in the tens digit of the others, so you don't have do do all 4 multiplication problems to completion, at least...


And, more importantly, in terms of takeaway, if you look at the limitations you can cut down the number of trial-and-error steps you need to take. Limiting B and A to 2/3/4 helped quite a bit, and knowing that we needed the two 2-digit numbers to multiply to something over 1,000 didn't quite come into play here but could have in another circumstance. The key is leaning on the handful of constraints that they give you so that you can start throwing out the possibilities that will never work.

A rectangular box has the dimensions 12 inches x 10 inches x 8 inches. What is the
largest possible volume of a right cylinder that is placed inside the box?

Guys need help plzz

Brian@VeritasPrep wrote:Now, that's pretty vague but follow me here. Here's what we know:

- B*A is less than 10, and B*A = C

Hi Brian... Without options given, we cant make this guess right?.. Lets say a=4 and b =3, then a*b is 12 and c is 2 in this case.. It applies for any other combinations too.. It would av been challenging if the choices were not given or if it is a DS type.. Is it really falling under gmat category quest?..

Hi Brian,

There is the first step I cant follow. why is A times B equals C? could you please explain

Thanks

Brian@VeritasPrep wrote:Wow - what a terrific question! On any of these questions in which they're testing you on a calculation with variables at the digit level, the key invariably becomes this:

1) Start with what you know, and
2) Use that to limit the number of possibilities

Now, that's pretty vague but follow me here. Here's what we know:

- B*A is less than 10, and B*A = C
- A is the tens digit of the top number, and multiplying that by another 2-digit number gives you a 4-digit product.

Already we know that:

- B*A has to be either 2*3 or 2*4. Why? Because it has to be less than 10 (and anything 4*3 or higher is bigger than 10), and we can't repeat digits. 2*2 and 3*3 are out, as is 1*anything because we need a new third unique digit for C.

Here, I guess we at least knew that answer choice C is out, since we can't have B = 5 or we'll violate those rules.

- So C is either 6 or 8. And we know that, in doing the multiplication, we'll need to end up with BC as the tens and units digits. So if we try the different combinations of:

A = 2, B = 3, C = 6
A = 3, B = 2, C = 6
A = 2, B = 4, C = 8
A = 4, B = 2, C = 8

We need to make sure that B ends up in the tens place. If we don't end up repeating B then we can stop the math. Now, the point is kind of moot if you do the others but my initial thought was that A should be bigger, since you need to get over 1,000 in the product. (Now, that's moot since C has to be so big, but at first I didn't want to start with A = 2 since so many 20-something * up-to-40-something calculations end up less than 1,000 and would give us only 3 digits) So I may have gotten lucky but I started with A = 4, B = 2 to make sure that I got a 4-digit product right away:

42
*84

first line: 168 (doing multiplication by hand)
second line: 3360
total 3528

And so here we ended up with B in the tens place and C in the units place, and O and E are unique digits, so the parameters are all taken care of. I ended up a little lucky trying 42 first, since my logic wasn't entirely foolproof, but that worked.

If you had started with the others, though, you'd find pretty quickly that you don't end up with B in the tens digit of the others, so you don't have do do all 4 multiplication problems to completion, at least...


And, more importantly, in terms of takeaway, if you look at the limitations you can cut down the number of trial-and-error steps you need to take. Limiting B and A to 2/3/4 helped quite a bit, and knowing that we needed the two 2-digit numbers to multiply to something over 1,000 didn't quite come into play here but could have in another circumstance. The key is leaning on the handful of constraints that they give you so that you can start throwing out the possibilities that will never work.

Good question "Why does A times B equal C?"

Remember, the initial question is:

AB
x CA
OEBC

Let's say we were doing this with actual numbers. If we're multiplying, say, 21 * 32, we'd start by"

21
x 32

First, you'd multiply the units digits together, 1 * 2 to get 2 in the units place. Then when you multiply the bottom units digit (2) by the top tens digit (2), you'll put that result (4) in the tens place. Next, you'll put a placeholder 0 in the units place while you multiply the tens digit (3) by the top number:

21
x 32
42
_0 (and work leftward from here)

So what you're doing is multiplying the units digits togetehr to form the units digit of the product, then moving to the left. You know that the product of the units digits will become the new units digit. In the abstract calculation, the units digits are B * A, and the product is C. *AND* we know that A * B is less than 10 - otherwise (say it was 4 * 7...the units digit would be 8 but we'd have to carry the 2) this is a lot more complicated.


I hope that helps...

Does GMAT ask these kinda questions ?