### All PSAT Math Resources

## Example Questions

### Example Question #1 : Factors / Multiples

How many positive integers less than ten thousand are multiples of both eight and eighteen?

**Possible Answers:**

139

72

555

70

138

**Correct answer:**

138

In order to find all of the numbers that are multiples of both 8 and 18, we need to find the least common mutliple (LCM) of 8 and 18. The easiest way to do this would be to list out the multiples of 8 and 18 and determine the smallest one that is common to both.

First, let's list the first several multiples of eight:

8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88 . . .

Next, we list the first several multiples of eighteen:

18, 36, 54, 72, 90, 108, 126, 144 . . .

By comparing the multiples of eight and eighteen, we can see that the smallest one that they share is 72. Thus, the LCM of 8 and 18 is 72.

Because the LCM is 72, this means that every multiple of 72 is also a multiple of both 8 and 18. So, in order to find all of the multiples less than ten thousand that are both multiples of 8 and 18, we simply need to find how many multiples of 72 are less than 10000, and to do this, all we have to do is to divide 10000 by 72.

When we divide 10000 by 72, we get 138 with a remainder of 64; therefore, 72 will go into ten thousand 138 times before it exceeds ten thousand. In other words, there are 138 numbers less than 10000 that are multiples of 72 and, by extension, also multiples of both 8 and 18.

The answer is 138.

### Example Question #1 : Factors / Multiples

If is divisible by 2, 4 and 7, which of the following is a possible value of ?

**Possible Answers:**

114

168

172

136

154

**Correct answer:**

168

The simplest way to solve this question is to go through each answer choice and check to see if it is divisible by 2, 4 and 7. When we do this, we find that only 168 is evenly divisible by each of these numbers. 114 is not divisible by 4 or 7, 172 is not divisibly by 7, 136 is not divisble by 7, and 154 is not divisible by 4.

### Example Question #2 : Factors / Multiples

If *a*, *b*, and *c* are positive integers such that 4*a* = 6*b* = 11*c*, then what is the smallest possible value of *a *+ *b *+ *c*?

**Possible Answers:**

121

33

11

132

67

**Correct answer:**

67

We are told that *a*, *b*, and *c* are integers, and that 4*a* = 6*b* = 11*c*. Because *a*, *b*, and *c* are positive integers, this means that 4*a* represents all of the multiples of 4, 6*b* represents the multiples of 6, and 11*c* represents the multiples of 11. Essentially, we will need to find the least common multiples (LCM) of 4, 6, and 11, so that 4*a*, 6*b*, and 11*c* are all equal to one another.

First, let's find the LCM of 4 and 6. We can list the multiples of each, and determine the smallest multiple they have in common. The multiples of 4 and 6 are as follows:

4: 4, 8, 12, 16, 20, ...

6: 6, 12, 18, 24, 30, ...

The smallest multiple that 4 and 6 have in common is 12. Thus, the LCM of 4 and 6 is 12.

We must now find the LCM of 12 and 11, because we know that any multiple of 12 will also be a multiple of 4 and 6.

Let's list the first several multiples of 12 and 11:

12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, ...

11: 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, ...

The LCM of 12 and 11 is 132.

Thus, the LCM of 4, 6, and 12 is 132.

Now, we need to find the values of *a*, *b*, and *c*, such that 4*a* = 6*b* = 12*c* = 132.

4*a* = 132

Divide each side by 4.

*a* = 33

Next, let 6*b* = 132.

6*b* = 132

Divide both sides by 6.

*b* = 22

Finally, let 11*c* = 132.

11*c* = 132

Divide both sides by 11.

*c* = 12.

Thus, *a* = 33, *b* = 22, and *c* = 12.

We are asked to find the value of *a* + *b* + *c*.

33 + 22 + 12 = 67.

The answer is 67.

### All PSAT Math Resources

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