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How many positive even integers are factors of 72ˆ3?

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cinthiaph
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How many positive even integers are factors of 72ˆ3? [#permalink] New post   13 Oct 2019, 13:38
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How many positive even integers are factors of 72ˆ3?

a. 9
b. 10
c. 31
d. 63
e. 67
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chetan2u
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Re: How many positive even integers are factors of 72ˆ3? [#permalink] New post   01 Nov 2019, 20:07
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cinthiaph wrote:
How many positive even integers are factors of 72ˆ3?

a. 9
b. 10
c. 31
d. 63
e. 67



Remember, it is always better to find Odd factors and then subtract from total factors....
\(72^3=(2^33^2)^3=2^93^6\)

(I) Total factors ----- \((9+1)(6+1)=10*7=70\)

(II) Odd factors -- take just the odd prime numbers =\(3^6.....(6+1)=7\)

(III) Even factors -- Total - odd = \(70-7=63\)

D
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nick1816
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Re: How many positive even integers are factors of 72ˆ3? [#permalink] New post   13 Oct 2019, 14:08
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\(72^3= 2^9*3^6\)
Factors of 72*3 can be written as \(2^a*3^b\)

If factors are even, a can take 9 values (1 to 9) and b can take 7 values (0 to 6)
Total positive even factors= 9*7=63

cinthiaph wrote:
How many positive even integers are factors of 72ˆ3?

a. 9
b. 10
c. 31
d. 63
e. 67
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Mo2men
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How many positive even integers are factors of 72ˆ3? [#permalink] New post   13 Oct 2019, 14:28
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cinthiaph wrote:
How many positive even integers are factors of 72ˆ3?

a. 9
b. 10
c. 31
d. 63
e. 67


\(72^3= 2^9*3^6\)
Then total factors = (9+1) * (6+1) = 70

Odd factors = (6+1) =7

even factors = 70 - 7 = 63

Answer: D
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How many positive even integers are factors of 72ˆ3? [#permalink] New post   14 Oct 2019, 05:09
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\(72^3=2^9∗3^6\)
Thus, total number of factors= (9+1)*(6+1)=70
So, number of odd factors=(6+1)=7
Therefore, number of even factors=70-7=63
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Re: How many positive even integers are factors of 72ˆ3? [#permalink] New post   03 Nov 2019, 11:28
(9+1) (6+1) i dont get it

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Re: How many positive even integers are factors of 72ˆ3? [#permalink] New post   07 Jun 2020, 00:07
cinthiaph wrote:
How many positive even integers are factors of 72ˆ3?

a. 9
b. 10
c. 31
d. 63
e. 67


Asked: How many positive even integers are factors of 72ˆ3?

72 = 2^3 * 3^2
72^3 = 2^9*3^6

Even integer factors = {2 + 2^2 + ... + 2^9)(1 + 3 + 3^2 + .... + 3^6)

Number of even integer factors = 9*7 = 63

IMO D
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Re: How many positive even integers are factors of 72ˆ3? [#permalink] New post   19 Jul 2021, 09:53
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\(72^3 = (2^3 * 3^2)^3 = (2^9) * (3^6)\)

Total factors : \((9 + 1) (6 + 1) = 10 * 7 = 70\)

Odd factors : \(3^6 = 6 + 1 = 7\)

Even factors : \(70 - 7 = 63\)

Answer D
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AbhaGanu
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Re: How many positive even integers are factors of 72ˆ3? [#permalink] New post   21 Jul 2021, 06:18
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Why is this done : (9+1)(6+1) ?
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Re: How many positive even integers are factors of 72ˆ3? [#permalink] New post   11 Aug 2021, 10:54
cinthiaph wrote:
How many positive even integers are factors of 72ˆ3?

a. 9
b. 10
c. 31
d. 63
e. 67

=>72^3=2^9*3^6
=>there are in total (9+1)*(6+1)= 70
=>the total no of odd integers = 7
=>therefore the total no of even integers = 70-7 = 63
Therefore IMO D
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Re: How many positive even integers are factors of 723? [#permalink] New post   09 May 2023, 19:20
chetan2u wrote:
cinthiaph wrote:
How many positive even integers are factors of 72ˆ3?

a. 9
b. 10
c. 31
d. 63
e. 67



Remember, it is always better to find Odd factors and then subtract from total factors....
\(72^3=(2^33^2)^3=2^93^6\)

(I) Total factors ----- \((9+1)(6+1)=10*7=70\)

(II) Odd factors -- take just the odd prime numbers =\(3^6.....(6+1)=7\)

(III) Even factors -- Total - odd = \(70-7=63\)

D


Hi. Can you please explain the logic behind (9+1)(6+1)?
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Re: How many positive even integers are factors of 723? [#permalink]
09 May 2023, 19:20
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