How many positive even integers are factors of 72ˆ3?

Question

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

chetan2u · Accepted Answer

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

nick1816 · Answer

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

Mo2men · Answer

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

siddharthkapoor · Answer

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

Rhododendron · Answer

(9+1) (6+1) i dont get it

Posted from my mobile device

Kinshook · Answer

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

MathRevolution · Answer

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

AbhaGanu · Answer

Why is this done : (9+1)(6+1) ?

Crytiocanalyst · Answer

=>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

RenB · Answer

Relevant reference post- https://blog.gmatwhiz.com/2022/09/28/nu ... n-factors/

jonfuckingsnow · Answer

Bunuel  pls attach more questions like these!

bumpbot · Answer

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How many positive even integers are factors of 72ˆ3?

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How many positive even integers are factors of 72ˆ3?

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