Nothing like a lil' SAT prep math

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Nothing like a lil' SAT prep math

Postby RubyJewelStone » Thu Feb 01, 2007 3:37 pm

I am thoroughly confused.

In the correctly worked multiplication problem below, each letter represents a distinct digit from 0 through 9. What digit must D represent?

Image

A 0
B 2
C 4
D 5
E 9


The answer they gave...
Looking at the rightmost side we see that A * A = A (plus perhaps something to carry over to the next column). There are only three possibilities: 5*5 = 5 (carry 2), 6*6 = 6 (carry 3), or 1 * 1 = 1. But if A = 1, then D = B, which isn't allowed. Therefore A = 5 or 6, and the only possible value for D is 0.


Helpsies? :sweat:
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Postby Aileen Kailum » Thu Feb 01, 2007 5:15 pm

Ugh. The SAT. There's something I don't want to relive...

The solution gave a pretty good example of how they narrowed down the choices for A to 5 or 6. I'll go from there.

Let's say A equals 5. Your choices for D are 0, 2, 4, 5, and 9. If you plug in 0 for D, you get:
(remember, we're saying A=5 and D=0)

505
x 5
----
B5B5

B's not important, so you don't have to worry about it. Well, 5x5 is 25, so you drop the 5 and carry the 2. Next, 5x0 is 0. 0+2 is 2, so drop the 2. After that, multiply 5x5 for 25 and finish the problem.

By working this out, you see that 0 is a possible answer for D.

Now, let's say that D equals 2:

525
x 5
----
B5B5

5x5 is 25. Drop the 5 and carry the 2. So far, so good. But a problem comes up because 5x2 is 10. 10+2 is 12. You would have to drop the 2 and carry the 1, but that would cause the next step--5x5=25, 25+2=27--to come out incorrect. So D cannot equal 2.

The next steps would be to work out the problem when D equals 4, 5, and 9. But since you already know that D equals 0, you don't have to.

Anyways, I hope that helped. I wasn't entirely sure what you were asking, so if I just gave you a hopelessly off-topic answer, sorry. :)
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Postby Icarus » Thu Feb 01, 2007 5:19 pm

Well, you have:

ADA
* A
-------
BABA

The first thing they mention is that looking at the ones column, we have that A*A=xA. The only digits where that is the case are 1, 5, 6, and 0. Since DA*A isn't just A, 0 is out. Similarly, as 1 times a number yields that number, 1 is out as well.

Thus, the only possibilities are 5 and 6. Now, if we look at the third column, we see once again that A squared plus what ever was carried from D*A is equal to A. Since A = 5 or 6, and the choices for D are 0,2,4,5, and 9; the only one that fits is 0, because for all numbers greater than or equal to 2, when multiplying by 5 or 6 you have to carry.
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Postby Mithrandir » Sat Feb 03, 2007 6:02 am

Did that help clarify, or do you need more explinations?
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Postby RubyJewelStone » Sun Feb 04, 2007 12:53 pm

I get it now. Thanksies. <3
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Postby Mithrandir » Sun Feb 04, 2007 7:19 pm

Good. Feel free to keep these kinda questions coming. We've got quite a few math geeks who love to help out on this kinda stuff. :thumb:
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