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Math Expert V
Joined: 02 Sep 2009
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How many positive odd factors does 768 have?  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 60% (01:16) correct 40% (01:19) wrong based on 139 sessions

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How many positive odd factors does 768 have?

A. zero
B. one
C. two
D. three
E. four

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How many positive odd factors does 768 have?  [#permalink]

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1
768 = $$2^8$$$$3^1$$

No of Odd-factors = $$2^0$$$$3^1$$ = (0+1)(1+1) = 1*2 = 2.

C is the answer.
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Senior SC Moderator V
Joined: 22 May 2016
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How many positive odd factors does 768 have?  [#permalink]

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Bunuel wrote:
How many positive odd factors does 768 have?

A. zero
B. one
C. two
D. three
E. four

To find the number of factors of an integer including itself and 1

1) prime factorize the integer:$$768=2^8*3^1$$

2) to find the number of odd factors only, use only the odd prime factor, 3. Take 3's exponent and ADD 1
$$(1+1)=2$$

3) for cases in which no other exponent is in play, step 2 is the answer.

768 contains two odd factors (1 and 3)

Different step #3 if there is more than one exponent. How many positive odd factors does 525 have? 1) $$525=3^15^27^1$$
2) Add 1 to each exponent: (2,3,2)
3) multiply those results: (2*3*2) = 12 factors of 525 including itself and 1

If $$n=a^p*b^q*c^r$$
Then the number of factors of $$n$$ including $$n$$ and 1 is expressed by the formula $$(p+1)(q+1)(r+1)$$
See Bunuel , HERE, "Finding the Number of Factors of an Integer" (scroll down)

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Re: How many positive odd factors does 768 have?  [#permalink]

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768 =2^8*3

So, there are only 2 odd factors i.e 1 and 3.
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Re: How many positive odd factors does 768 have?  [#permalink]

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2
Bunuel wrote:
How many positive odd factors does 768 have?

A. zero
B. one
C. two
D. three
E. four

First, let’s break 768 into its prime factors.

768 = 16 x 48 = 2^4 x 8 x 6 = 2^4 x 2^3 x 2^1 x 3^1 = 2^8 x 3^1

Thus, we see that 768 has only 2 odd factors, namely, 1 and 3.

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How many positive odd factors does 768 have?  [#permalink]

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1
generis

generis wrote:
Bunuel wrote:
How many positive odd factors does 768 have?

A. zero
B. one
C. two
D. three
E. four

To find the number of factors of an integer including itself and 1

1) prime factorize the integer:$$768=2^8*3^1$$

2) to find the number of odd factors only, use only the odd prime factor, 3. Take 3's exponent and ADD 1
$$(1+1)=2$$

3) for cases in which no other exponent is in play, step 2 is the answer.

768 contains two odd factors (1 and 3)

Different step #3 if there is more than one exponent. How many positive odd factors does 525 have? 1) $$525=3^15^27^1$$
2) Add 1 to each exponent: (2,3,2)
3) multiply those results: (2*3*2) = 12 factors of 525 including 525 and itself - I believe you mean to say 1 and itself.

If $$n=a^p*b^q*c^r$$
Then the number of factors of $$n$$ including $$n$$ and itself is expressed by the formula $$(p+1)(q+1)(r+1)$$
See Bunuel , HERE, "Finding the Number of Factors of an Integer" (scroll down)

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Joined: 22 May 2016
Posts: 3565
Re: How many positive odd factors does 768 have?  [#permalink]

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Harshgmat wrote:
generis

generis wrote:

Different step #3 if there is more than one exponent. How many positive odd factors does 525 have? 1) $$525=3^15^27^1$$
2) Add 1 to each exponent: (2,3,2)
3) multiply those results: (2*3*2) = 12 factors of 525 including 525 and itself - I believe you mean to say 1 and itself.

If $$n=a^p*b^q*c^r$$
Then the number of factors of $$n$$ including $$n$$ and itself is expressed by the formula $$(p+1)(q+1)(r+1)$$
See Bunuel , HERE, "Finding the Number of Factors of an Integer" (scroll down)

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Instructions for living a life. Pay attention. Be astonished. Tell about it. -- Mary Oliver Re: How many positive odd factors does 768 have?   [#permalink] 14 Oct 2018, 06:11
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