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How many factors of 1500 are not divisible by 15? [#permalink]
1500 can be expressed as 2^2*3^1*5^3
#factors = 3*2*4 = 24
15 can be expressed as 3^1*5^1
Divide 1500/15, we can prime factorize as 2^2*5*2 with #factors = 9. All these factors are divisible by 15.

#Factors not divisible by 15 = 24 - 9 = 15

Originally posted by sandman13 on 16 Jun 2018, 03:02.
Last edited by sandman13 on 16 Jun 2018, 05:07, edited 1 time in total.
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Re: How many factors of 1500 are not divisible by 15? [#permalink]
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CAMANISHPARMAR wrote:
How many factors of 1500 are not divisible by 15?

A) 24
B) 12
C) 11
D) 9
E) 15


The above approaches are great. However, if we didn't think of those approaches, we can likely solve it by brute force in well under 2 minutes.
When we scan the answer choices (ALWAYS scan the answer choices before solving the problem!), we see that the answer choices are relatively small, which suggests we might just LIST the factors of 1500 and then cross out the factors that are divisible by 15.

We'll find the factors IN PAIRS of values that have a product of 1500.
We get: 1 & 1500, 2 & 750, 3 & 500, 4 & 375, 5 & 300, 6 & 250, 10 & 150, 12 & 125, 15 & 100, 20 & 75, 25 & 60, 30 & 50

Now, cross out the factors that ARE divisible by 15:
1 & 1500, 2 & 750, 3 & 500, 4 & 375, 5 & 300, 6 & 250, 10 & 150, 12 & 125, 15& 100, 20 & 75, 25 & 60, 30 & 50

There are 15 factors remaining.

Answer: E

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Re: How many factors of 1500 are not divisible by 15? [#permalink]
sandman13 wrote:
1500 can be expressed as 2^2*3^1*5^3
#factors = 3*2*4 = 24
15 can be expressed as 3^1*5^1
Divide 1500/15, we can prime factorize as 2^2*5*2 with #factors = 9. All these factors are divisible by 15.

#Factors not divisible by 15 = 24 - 9 = 15


For the factors, could you explain why you did 3*2*4 instead of 3*2*5

2,3 & 5 are the prime numbers
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Re: How many factors of 1500 are not divisible by 15? [#permalink]
CAMANISHPARMAR wrote:
Let's write 1500 in prime factorization form, \(1500=2^2*3^1*5^3\)

To find the number of factors of 1500 is a straightforward application of number of factors formula:-
(p+1)(q+1)(r+1)... [where p,q,r are exponents of each prime factor]

Hence the no of factors of 1500 without any restriction = (2+1)(1+1)(3+1) = 3*2*4 = 24

The no factors of 1500 not divisible by 15 = Total of no factors of 1500 without any restriction - Total no of factors of 1500 which are divisible by 15

Therefore first we must find Total no of factors of 1500 which are divisible by 15:-

Now we can write, 1500 = 15 * 100 = 15 * \(5^2*2^2\)

Since 15 * \(5^2*2^2\) is multiple of 15, and every factor of \(5^2*2^2\) is also a factor of 1500 and is divisible by 15 if we are multiplying each factor of \(5^2*2^2\) by 15.

Hence the no of factors which are divisible by 15 are all the factors of \(5^2*2^2\); that is (2+1)(2+1) = 3*3 = 9

Therefore, the no factors of 1500 not divisible by 15 = 24 - 9 = 15

Correct Answer is option (E)


For the factors, could you explain why you did 3*2*4 instead of 3*2*5

2,3 & 5 are the prime numbers
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Re: How many factors of 1500 are not divisible by 15? [#permalink]
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CAMANISHPARMAR wrote:
How many factors of 1500 are not divisible by 15?

A) 24
B) 12
C) 11
D) 9
E) 15


1500 = 15 x 100 = 15 x 5 x 5 x 2 x 2 = 15 x 5^2 x 2^2

Thus, the factors that are divisible by 15 are:

15, 15 x 5, 15 x 2, 15 x 5^2, 15 x 2^2, 15 x 5 x 2, 15 x 5^2 x 2, 15 x 5 x 2^2, 15 x 5^2 x 2^2

Thus, there are 9 factors that are divisible by 15.

Now let’s determine the total number of factors of 1500:

1500 = 15 x 100 = 2^2 x 5^3 x 3^1

So 1500 has (2 + 1)(3 + 1)(1 + 1) = 3 x 4 x 2 = 24 total factors.

Thus, 24 - 9 = 15 factors are not divisible by 15.

Answer: E
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Re: How many factors of 1500 are not divisible by 15? [#permalink]
Total Count of Distinct Factors - No. of Factors Divisible by 15 = No. of Factors NOT Divisible by 15


Prime Factorization of 1, 500 = 2'2nd * 3'1st * 5'3rd

Total Count of Distinct Factors = (2 + 1) * (1 + 1) * (3 + 1) = 3 * 2 * 4 = 24 Total Distinct Factors


Any Factor of 1, 500 that is Divisible by 15 will take the FORM = 15k ---- where k = +Pos. Integer

From the Prime Factorization of 1, 500, we can Take as Common - (3 * 5) = 15


1, 500 = 2'2nd * 3'1st * 5'3rd

(3'1 * 5'1) * [ 2'2 * 5'2]

The Different Combinations of Factors inside the Bracket will be the K that fulfills the Requirement of being a Divisible by 15.

Total Combinations of Prime Bases in the Bracket = (2 + 1) * (2 + 1) = 3 * 3 = 9

9 Factors that are Evenly Divisible by 15


24 - 9 = 15 Factors that are NOT Divisible by 15

-E-
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Re: How many factors of 1500 are not divisible by 15? [#permalink]
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