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How do you solve digit word problems?

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How do you solve digit word problems?

by Osirus@VeritasPrep » Mon Jun 01, 2009 6:04 pm
Here is a sample problem from Barrons (data sufficiency)

If x is a two-digit (so x=ba with b and a digits), what is the last digit a of x?

1) The number 3x is a three-digit number whose last digit is a.

2) The digit a is less than 7.

The Original answer is E

In addition to explaining the answer if anyone could direct me to any websites that explain how to solve these type of problems that would be greatly appreciated. Thanks in advance.
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by dmateer25 » Mon Jun 01, 2009 6:28 pm
Here is how I went about this problem.

You know that we are working with the units digit so that is all we need to worry about essentially.

List out all the digits and multiply each by 3 and see which one has the same units digit.

0 x 3 = 0 same as 0
1 x 3 = 3 different than 1
2 x 3 = 6 different than 2
3 x 3 = 9 different than 3
4 x 3 = 12 (units digit would be 2, which is different than 4)
5 x 3 = 15 (units digit would be 5, which is the same as 5)
6
7
8
9
10

For statement 1 we can stop after 5 because we can see that the answer can be 0 or 5.

statement 2 alone tells us nothing valuable.

Together we know that a is less than 2. But as proven above a can be either 0 or 5, therefore the answer is E.
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by Osirus@VeritasPrep » Tue Jun 02, 2009 5:21 am
dmateer25 wrote:Here is how I went about this problem.

You know that we are working with the units digit so that is all we need to worry about essentially.

List out all the digits and multiply each by 3 and see which one has the same units digit.

0 x 3 = 0 same as 0
1 x 3 = 3 different than 1
2 x 3 = 6 different than 2
3 x 3 = 9 different than 3
4 x 3 = 12 (units digit would be 2, which is different than 4)
5 x 3 = 15 (units digit would be 5, which is the same as 5)
6
7
8
9
10

For statement 1 we can stop after 5 because we can see that the answer can be 0 or 5.

statement 2 alone tells us nothing valuable.

Together we know that a is less than 2. But as proven above a can be either 0 or 5, therefore the answer is E.
Thanks for the reply, I appreciate it. That is how I approached the problem as well. What I was unclear on was the way the book explained it. Here is the book's explanation:

If Statement(1) is true, then since x = ba, 3x = 3(10b + a) = 30b + 3a. Now, because b is multiplied by 10 in the expression for 3x, the final digit of 3x must be the final digit of 3a. Since a is a digit, 0 <= a <= 9, which implies 0 <= 3a <= 27. So for the last digit of 3a to be equal to a, 3a must equal a or 10 + a or 20 + a. If a = 3a, then a = 0. If 10 + a = 3a, then 10 = 2a or a = 5. If 20 + a = 3a then 20 = 2a or a = 10, but since 10 is not a digit this is not possible. So if (1) is true, then a is 0 or 5, and (1) alone is not sufficient. Thus the answer to question I is NO, and the only possible choices are B, C, or E.

Now since 26 and 25 are both two-digit numbers whose last digits are less than 7, Statement(2) alone is not sufficient. So the answer to question II is NO, and the only possible choices are C or E. Also, since (2) does not allow us to choose between 0 and 5, Statements(1) and (2) together are not sufficient, so the correct choice is E.
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