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Bunuel
What is the range of the prime factors of m, if \(m = 2^5*3^{11} - 9^{6} - 3^{11}\) ?

A. 0
B. 1
C. 2
D. 4
E. 5


\(m = 2^5*3^{11} - 9^{6} - 3^{11}\)

3^11(2^5-3-1)
3^11(28)
3^11(2^2*7)

range would be 7-2 = 5 IMO E


Where does the 3-1 come from?
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Bunuel
What is the range of the prime factors of m, if \(m = 2^5*3^{11} - 9^{6} - 3^{11}\) ?

A. 0
B. 1
C. 2
D. 4
E. 5


\(m = 2^5*3^{11} - 9^{6} - 3^{11}\)

3^11(2^5-3-1)
3^11(28)
3^11(2^2*7)

range would be 7-2 = 5 IMO E


Where does the 3-1 come from?

Archit,

We can factor out 3^11 from 2^5*3^11 -(3^2)^6-3^11 ----> 2^5*3^11-3^12-3^11-----> 3^11(2^5-3-1)---->3^11(32-4)--->3^11(28) ---->3^11(2^2*7).
So the range will be 7 - 2 = 5
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What is the range of the set consisting of prime factors of m, if \(m = 2^5*3^{11} - 9^{6} - 3^{11}\) ?

A. 0
B. 1
C. 2
D. 4
E. 5

Official Solution:

What is the range of the set consisting of prime factors of \(m\), if \(m = 2^5*3^{11} - 9^{6} - 3^{11}\) ?

A. \(0\)
B. \(1\)
C. \(2\)
D. \(3\)
E. \(4\)


We can start by simplifying the expression for \(m\):

\(2^5* 3^{11} - 9^6 - 3^{11} = \)

\(=2^5*3^{11} - (3^2)^6 - 3^{11} = \)

\(=2^5 *3^{11} - 3^{12} - 3^{11}\)

Next, we can factor out \(3^{11}\) to obtain:

\(3^{11}(2^5 - 3 - 1) = \)

\(=3^{11} *28=\)

\(=3^{11} *2^2*7\)

Therefore, the set consisting of prime factors of \(m\) is {2, 3, 7}. The range of this set is (the largest term) - (the smallest term) = 7 - 2 = 5.


Answer: E
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