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26 Nov 2019, 00:50
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D
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Difficulty:
55%
(hard)
Question Stats:
58% (01:32) correct
42%
(01:29)
wrong
based on 371
sessions
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If a positive integer is equal to the following product: \(2^5b^3c^4\), where b and c are distinct prime numbers greater than 2, how many distinct even factors does the integer have?
A. 1
B. 5
C. 50
D. 100
E. 120
Are You Up For the Challenge: 700 Level Questions
A. 1
B. 5
C. 50
D. 100
E. 120
Are You Up For the Challenge: 700 Level Questions
20 Feb 2020, 02:25
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Bunuel
Solution:
- • \(b\) and \(c\) are distinct prime numbers greater than \(2\).
- o We know that any prime number greater \(2\) is an odd number.
- \(b\) and \(c\) are odd prime numbers.
Method 1: \(2^5*b^3*c^4\)
- • Total factors = \((power\ of\ 2 +1)*(power\ of\ b+1)*(power\ of\ c+1)=(5+1)*(3+1)*(4+1)=6*4*5=120\)
• Odd factors = \((Power\ of\ b+1)*(power\ of\ c+1)=(3+1)*(4+1)=4*5=20\)
- o Even factors = \(120-20 = 100\).
Note:- while finding odd factors we completely ignore the power of even prime, but we can’t ignore the power of odd prime while calculating the even factors because \(even*odd = even\)
- • Even factors = \((power\ of\ 2)*(power\ of\ b +1)*(power\ of\ c + 1)=(5)*(3+1)*(4+1)=5*4*5=100\).
General Discussion
26 Nov 2019, 01:03
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Bunuel
Let \(x\) = \(2^5*b^3*c^4\) = \(2*2^4*b^3*c^4\) = \(2y\)
--> \(y = 2^4*b^3*c^4\)
Note that, even factors of \(x\) is same as all positive factors of \(y\), Since \(x = 2y\)
--> Positive factors of y = (4 + 1)*(3 + 1)*(4 + 1) = 100
--> Even factors of x = All factors of y = 100
IMO Option D
