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# If 78 and 66 are both factors of x, what is the smallest num

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If 78 and 66 are both factors of x, what is the smallest num [#permalink]  26 Nov 2019, 23:25
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If 78 and 66 are both factors of x, what is the smallest number of factors x could have in total?

[Reveal] Spoiler: OA
16

Last edited by Carcass on 28 Nov 2019, 08:14, edited 3 times in total.
Edited by Carcass
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Re: If 78 and 66 are both factors of x, what is the smallest num [#permalink]  27 Nov 2019, 02:41
Expert's post
Follow the rules for posting. A numeric entry question must be posted under the right forum NOT in general quant.

The factors are: $$2^2,3,11,39$$

Adding 1 to every exponent and we do have : $$2^3,3 = 8 \times 3=24$$

Total number of factors are 24
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Re: If 78 and 66 are both factors of x, what is the smallest num [#permalink]  28 Nov 2019, 07:27
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Carcass wrote:
Follow the rules for posting. A numeric entry question must be posted under the right forum NOT in general quant.

The factors are: $$2^2,3,11,39$$

Adding 1 to every exponent and we do have : $$2^3,3 = 8 \times 3=24$$

Total number of factors are 24

The factors are: 2^2 X 3^2 X 11 X 13

After adding 1 to exponents, the number of divisors = 3 X 3 X 2 X 2 = 36
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Re: If 78 and 66 are both factors of x, what is the smallest num [#permalink]  28 Nov 2019, 08:11
Expert's post
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?

[Reveal] Spoiler: OA
24

-----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 N

Consider 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
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Re: If 78 and 66 are both factors of x, what is the smallest num [#permalink]  28 Nov 2019, 08:14
Expert's post
I counted one more exponent but it was elegant as a solution.
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Re: If 78 and 66 are both factors of x, what is the smallest num   [#permalink] 28 Nov 2019, 08:14
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