SAT Math : Least Common Multiple

Study concepts, example questions & explanations for SAT Math

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Example Questions

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Example Question #11 : Least Common Multiple

What is the least common multiple of ?

Possible Answers:

Correct answer:

Explanation:

We need to ensure that all the numbers share a common factor of  are divisible by . We get  leftover along with the  that doesn't divide evenly with . Now that all these numbers share a common factor of , we multiply them all out including the  we divided out. We get  or 

Example Question #11 : Least Common Multiple

What is the least common multiple of ?

Possible Answers:

Correct answer:

Explanation:

We need to ensure that all the numbers share a common factor of  are divisible by . We get  leftover along with the  that doesn't divide evenly with . Next,  are divisible by . We also get  leftover. Then, we can divide the s out to get . Now that all these numbers share a common factor of , we multiply them all out including the  we divided out. We get  or 

Example Question #11 : Least Common Multiple

What is the least common multiple of the first six positive integers?

Possible Answers:

Correct answer:

Explanation:

Let's divide the even numbers first. We will divide them by .

 

Next, we have two s, so let's divide them by  to get . So far we have factors of  remaining from the original six integers with factors of  been used. Now that they have a common factor of , we multiply everything out. We get  or 

Example Question #14 : Least Common Multiple

Which can be a group of remainders when four consecutive integers are divided by ?

Possible Answers:

Correct answer:

Explanation:

If you divide a number by , you cannot have a remainder of   You can either have  in that order.

Example Question #711 : Arithmetic

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

Possible Answers:

67

11

33

121

132

Correct answer:

67

Explanation:

We are told that a, b, and c are integers, and that 4a = 6b = 11c. Because a, b, and c are positive integers, this means that 4a represents all of the multiples of 4, 6b represents the multiples of 6, and 11c represents the multiples of 11. Essentially, we will need to find the least common multiples (LCM) of 4, 6, and 11, so that 4a, 6b, and 11c 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 4a = 6b = 12c = 132.

4a = 132

Divide each side by 4.

a = 33

Next, let 6b = 132.

6b = 132

Divide both sides by 6.

b = 22

Finally, let 11c = 132.

11c = 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.

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