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

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
\((1+1)*(2+1)=(2*3)\) = 6

D)\(118=2^159^1\)
Total # of factors is
\((1+1)*(1+1)=(2*2)\) = 4

E) \(121=11^2\)
--Add 1 to each exponent: \((2+1)=3\). Stop. (Only one exponent, like A)
-- Total # of factors is (2+1) = 3

Answer 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 Integer

Example: \(360=2^33^25^1\)

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

To find the number of factors of an integer

-- using 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 number in the list -- here {2,2}

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

The integer \(118\) has \(4\) factors including 118 itself and 1

A) \(8=2^3\)
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 is 4

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
\((1+1)*(2+1)*=(2*3)\) = 6

D)\(118=2^159^1\)
Total # of factors is
\((1+1)*(1+1)*=(2*2)\) = 4

E) \(121=11^2\)
--Add 1 to each exponent: \((2+1)=3\). Stop. (Only one exponent, like A)
-- Total # of factors in 75 is (2+1) = 3

Answer C

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

This was a typo error..did correctly in worksheet. Thank you.

Posted from my mobile device

answer - 75.

number of factors = PFs raised to power. Add 1 and multiply

75=5^2*3 so, 3*2=6

We can determine the total number of factors of a number by breaking a number into primes, adding 1 to the exponent of each unique prime and then multiplying. Let’s follow the same process for each answer choice:

A) 8 = 2^3

So 8 has 3 + 1 = 4 factors.

B) 51 = 17^1 x 3^1

So 51 has (1 + 1)(1 + 1) = 4 factors.

C) 75 = 5^2 x 3^1

75 has (2 + 1)(1 + 1) = 6 factors.

D) 118 = 2^1 x 59^1

118 has (1 + 1)(1 + 1) = 4 factors.

E) 121 = 11^2

121 has 2 + 1 = 3 factors.

Answer: C

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