Kenny1000 wrote:

Can anyone help with this problem?

Its a fill in the box question, there are no options.

If 78 and 66 are both factors of x, what is the smallest number of factors x could have in total?

-----ASIDE---------------------

A lot of integer property questions can be solved using prime factorization.

For questions involving divisibility, divisors, factors and multiples, we can say:

If k is a factor of N, then k is "hiding" within the prime factorization of NConsider these examples:

3 is a factor of 24, because 24 = (2)(2)(2)

(3), and we can clearly see the

3 hiding in the prime factorization.

Likewise,

5 is a factor of 70 because 70 = (2)

(5)(7)

And

8 is a factor of 112 because 112 = (2)

(2)(2)(2)(7)

And

15 is a factor of 630 because 630 = (2)(3)

(3)(5)(7)

-----BACK TO THE QUESTION!---------------------

GIVEN: 78 is a factor of x

78 = (2)(3)(13)

This means 2, 3 and 13 must be in the prime factorization of x

In other words: x = (2)(3)(13)(?)(?)(?)

Please note that the (?)'s represent additional prime numbers that could also be in the prime factorization of x. However, at this point, all we know for certain is that 2, 3 and 13 must be in the prime factorization of x

GIVEN: 66 is a factor of x

66 = (2)(3)(11)

This means 2, 3 and 11 must be in the prime factorization of x

From the earlier information, we already know that x = (2)(3)(13)(?)(?)(?)

Since we already have a 2 and a 3 in the prime factorization of x, we don't need to add more 2's or 3's

But we do need to add 11 to the prime factorization

In other words: x = (2)(3)(13)(11)(?)(?)

So, the smallest possible value of x that meets both conditions is x = (2)(3)(13)(11)

As we can see, 78 is a factor of x because x =

(2)(3)(13)(11)

We can also see that 66 is a factor of x because x =

(2)(3)(13)

(11)Now that we know the smallest possible value of x, we can apply a nice rule for finding the total number of factors have a positive number.

-----ASIDE-----

If the

prime factorization of N = (p^

a)(q^

b)(r^

c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (

a+1)(

b+1)(

c+1)(etc) positive divisors.

Example: 14000 = (2^

4)(5^

3)(7^

1)

So, the number of positive divisors of 14000 = (

4+1)(

3+1)(

1+1) =(5)(4)(2) = 40

---------------------

In our case, x = (2^

1)(3^

1)(13^

1)(11^

1)

So, the number of positive divisors of x = (

1+1)(

1+1)(

1+1)(

1+1) =(2)(2)(2)(2) = 16

Answer: 16 (I have edited the official answer to reflect my solution above)

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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