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

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Which of the following integers has the greatest number of factors?  [#permalink]

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15 Aug 2018, 10:07
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Which of the following integers has the greatest number of factors?

1. 8

2. 51

3. 75

4. 118

5. 121

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Which of the following integers has the greatest number of factors?  [#permalink]

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Updated on: 17 Aug 2018, 09:26
1
AkshdeepS wrote:
Which of the following integers has the greatest number of factors?

1. 8

2. 51

3. 75

4. 118

5. 121

The no of factors of a number depends upon the power of the prime factors of that number on prime factorization.
$$75=3^1*5^2$$

No of factors=(1+1)*(2+1)=6, so no of factors is the highest.

Ans. (C)

P.S:-
$$8=2^3$$; Sum of powers=3
$$51=17^1*3^1$$; Sum of powers=2
$$118=2^1*59^1$$; Sum of powers=2
$$121=11^2$$; Sum of powers=2
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Originally posted by PKN on 15 Aug 2018, 10:14.
Last edited by PKN on 17 Aug 2018, 09:26, edited 1 time in total.
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Which of the following integers has the greatest number of factors?  [#permalink]

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15 Aug 2018, 10:41
Prime factorizing all the options.
1) 8 = $$2^3$$. Number of factors = 4.
2) 51 = $$3^1$$ x $$17^1$$. Factors = 2*2 = 4.
3) 75 = $$3^1$$ x $$5^2$$. Factors = 2*3 = 6.
4) 118 = $$2^1$$ x $$59^1$$. Factors = 2*2 = 4.
5) 121 = $$11^2$$. Factors = 3.

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Which of the following integers has the greatest number of factors?  [#permalink]

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17 Aug 2018, 09:07
2
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
$$(1+1)*(2+1)=(2*3)$$ = 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

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

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Re: Which of the following integers has the greatest number of factors?  [#permalink]

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17 Aug 2018, 09:23
generis wrote:
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

-- 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

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
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Re: Which of the following integers has the greatest number of factors?  [#permalink]

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17 Aug 2018, 18:19

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

75=5^2*3 so, 3*2=6
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Re: Which of the following integers has the greatest number of factors?  [#permalink]

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20 Aug 2018, 11:02
AkshdeepS wrote:
Which of the following integers has the greatest number of factors?

1. 8

2. 51

3. 75

4. 118

5. 121

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.

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Re: Which of the following integers has the greatest number of factors?   [#permalink] 20 Aug 2018, 11:02
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