AkshdeepS wrote:

Which of the following integers has the greatest number of factors?

1. 8

2. 51

3. 75

4. 118

5. 121

To find the number of factors of an integer I will use answer (D) 118 as an example

Find the prime factorization of the integer:

\(118 = 2^159^1\)

(1) List the

exponents -- in this case {1,1}

(2) add \(1\) to each exponent in the list -- here, then, those exponents become

(1 + 1) = 2

(1 + 1) = 2

Now we have these results: {2,2}

(3) MULTIPLY the results: \((2*2)=4\)

The integer \(118\) has \(4\) factors

including 118 itself and 1

A) \(8=2^3\)

List exponents. In this case: {3}

Add 1 to the exponent of 3. \((3+1)=4\)

-- In the case of a single exponent, Step 2 is the answer

-- # of factors in 8, including 1 and 8 =

4 (See small type for those 4 factors)

B) \(51\) = \(3^117^1\)

Total # of factors including 1 and 51 itself is

\((1+1)*(1+1)=(2*2)\) =

4 C). \(75=3^15^2\)

Total number of factors is

\((1+1)*(2+1)=(2*3)\) =

6D)\(118=2^159^1\)

Total # of factors is

\((1+1)*(1+1)=(2*2)\) =

4E) \(121=11^2\)

--Add 1 to each exponent: \((2+1)=3\). Stop. (Only one exponent, like A)

-- Total # of factors is (2+1) =

3Answer C

PKN , not sure how you got that \(75=2^3*5^1\)

Theory from

Bunuel , link below

"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."

HERE, scroll down to Finding the Number of Factors of an IntegerExample: \(360=2^33^25^1\)

Total number of factors including 360 itself and 1 is

\((3+1)*(2+1)*(1+1)=(4*3*2)=24\)

**You can also just "muscle" each answer's factors. List them. Listing for large numbers is difficult.

A. 8: 1, 2, 4, 8 = 4 factors

B. 51: 1, 3, 17, 51 = 4 factors

C. 75: 1, 3, 5, 15, 25, 75 = 6 factors

D. 118: 1, 2, 59, 118 = 4 factors

E. 121: 1, 11, 121 = 3 factors