The positive integer 200 has how many factors?

Question

A. 2
B. 10
C. 12
D. 15
E. 24

Bunuel · Accepted Answer

Finding the Number of Factors of an Integer First make prime factorization of an integer n=a^p*b^q*c^r , where a , b , and c are prime factors of n and p , q , and r are their powers. The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1) . NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2 Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors. BACK OT THE ORIGINAL QUESTION: 200=2^3*5^2 , so it has (3+1)(2+1)=12 different factors. Answer: C. For more check Number Theory chapter of Math Book: math-number-theory-88376.html Hope it helps.

Bunuel · Answer

No. A factor is a positive divisor.

ENGRTOMBA2018 · Answer

The OA can not be E . It should be C

Direct formula for number of factors of a number N = a^p*b^q... = (p+1)(q+1)... And these include 1 and N as well.

Thus for 200= 2^3*5^2 -----> 4*3=12

kamathimanshu · Answer

Can negative numbers also be factors.
If so the answer turns out to be 12*2 = 24.

Please help?

ENGRTOMBA2018 · Answer

As far as I have seen , "factors" in GMAT refers to positive factors only.

karanchamp1 · Answer

Hi Bunuel As of 2016 do we consider negative factors also in the GMAT , as in factors of 200=(2)^3*(5)^2-->(3+1)(2+1)positive factors-->2*12(both positive and negative factors)=24? I appreciate your input :-D

Thanks in advance :-D

karanchamp1 · Answer

Many many thanks for the lightning fast reply :-D

You're the best :-D

BrentGMATPrepNow · Answer

APPROACH #1 - FORMULA
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

Now onto the question...
 Example : 200 = (2^ 3 )(5^ 2 )
So, the number of positive divisors of 200 = ( 3 +1)( 2 +1)
= (4)(3) 
= 12
= C

APPROACH #2 - LIST
We can quickly list all of the factors of 200
I suggest we do so in PAIRS of values whose product is 200
We get:
1 and 200
2 and 100
4 and 50
5 and 40
8 and 25
10 and 20
DONE!

Total = 12
= C

RELATED VIDEO FROM OUR COURSE
 https://www.youtube.com/watch?v=lVZ9GXd6ZBE

ScottTargetTestPrep · Answer

We can break 200 into primes, then add 1 to each exponent and find the product of all the sums. That product will give us the number of total factors.

200 = 20 x 10 = 2^2 x 5^1 x 2^1 x 5^1 = 2^3 x 5^2

Thus, 200 has (3 + 1)(2 + 1) = 4 x 3 = 12 factors.

Answer: C

Kushchokhani · Answer

Dear Bunuel, -2*-1 = 2. So -2 is a negative divisor. We have learnt in school that product of two negatives makes a positive. So is it that for the limited purpose of GMAT, negative factors are ignored? What might be the ideology behind the same

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The positive integer 200 has how many factors?

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The positive integer 200 has how many factors?

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