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11 Oct 2018, 02:43
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
59% (01:19) correct
41%
(01:23)
wrong
based on 679
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive odd factors does 768 have?
A. zero
B. one
C. two
D. three
E. four
A. zero
B. one
C. two
D. three
E. four
11 Oct 2018, 02:50
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768 = \(2^8\)\(3^1\)
No of Odd-factors = \(2^0\)\(3^1\) = (0+1)(1+1) = 1*2 = 2.
C is the answer.
No of Odd-factors = \(2^0\)\(3^1\) = (0+1)(1+1) = 1*2 = 2.
C is the answer.
11 Oct 2018, 10:18
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Bunuel
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)
Answer C
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)
