PSAT Math - Factors / Multiples

$\textup{If }a\textup{ is the greatest prime factor of 34 and }b\textup{ is the greatest}$

$\textup{prime factor of 20, what is the value of }a-b\textup{ ?}$

Explanation

Factor each number into prime roots:

$34=17\times2$

$20=5\times2\times2$

$\textup{If }a\textup{ is the greatest prime factor of 34 and }b\textup{ is the greatest}$

$\textup{prime factor of 20, what is the value of }a-b\textup{ ?}$

Explanation

Factor each number into prime roots:

$34=17\times2$

$20=5\times2\times2$

The sum of the first seven prime numbers divided by two is

Explanation

The first seven primes are 2, 3, 5, 7, 11, 13, and 17. Don't forget about 2, the smallest prime number, and also the only even prime! Adding these seven numbers gives a sum of 58, and 58/2 = 29.

The sum of the first seven prime numbers divided by two is

Explanation

The first seven primes are 2, 3, 5, 7, 11, 13, and 17. Don't forget about 2, the smallest prime number, and also the only even prime! Adding these seven numbers gives a sum of 58, and 58/2 = 29.

What is the product of the distinct prime factors of 24?

$\dpi{100} \small 6$

$\dpi{100} \small 8$

$\dpi{100} \small 9$

$\dpi{100} \small 24$

$\dpi{100} \small 5$

Explanation

The prime factorization of 24 is (2)(2)(2)(3). The distinct primes are 2 and 3, the product of which is 6.

What is the product of the distinct prime factors of 24?

$\dpi{100} \small 6$

$\dpi{100} \small 8$

$\dpi{100} \small 9$

$\dpi{100} \small 24$

$\dpi{100} \small 5$

Explanation

The prime factorization of 24 is (2)(2)(2)(3). The distinct primes are 2 and 3, the product of which is 6.

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?

121

132

Explanation

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.

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?

121

132

Explanation

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.

If the variable x is an integer divisible by the numbers 2 and 3, which of the following is necessarily divisible by 2, 3 and 5?

5x + 2

x + 30

2x + 30

6x + 30

5x + 30

Explanation

For this question, use the fact that a sum of two multiples is a multiple. In other words:

if x is a multiple of 3 and y is a multiple of 3: (x + y) is a multiple of 3.

Thus in this question, x is a multiple of 2 and 3. We need to find a number that is a multiple of 2, 3, and 5.

Take 5x + 30:

\[x is divisible by 2. 5 times x is still divisble by 3. 30 is divisible by 2.\] -> divisible by 2.

\[x is divisible by 3. 5 times x is still divisble by 3. 30 is divisible by 3.\] -> divisible by 3.

\[5x is divisible by 5. 30 is divisible by 5.\] -> divisible by 5.

Thus 5x + 30 is divisible by 2, 3 and 5.

If is divisible by 2, 3 and 15, which of the following is also divisible by these numbers?

Explanation

Since v is divisible by 2, 3 and 15, v must be a multiple of 30. Any number that is divisible by both 2 and 15 must be divisible by their product, 30, since this is the least common multiple.

Out of all the answer choices, v + 30 is the only one that equals a multiple of 30.

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