Problem 27 · 2014 Math Kangaroo
Stretch
Number Theory
factorizationcasework
The chain of equations \(k=(2014+m)^{ rac{1}{n}}=1024^{ rac{1}{n}}+1\) should hold for the positive whole numbers \(k\), \(m\), \(n\). How many different values can \(m\) take?
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Answer: C — 2
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Hint 1 of 2
From the right-hand side, 1024^{1/n} must be a whole number, so 1024 is a perfect n-th power.
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Hint 2 of 2
1024 = 2^{10}; which n make (k−1)^n = 2^{10} work, and then is m positive?
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Approach: force 1024 to be a perfect power, then check m > 0
- The equation needs (k−1)^n = 1024 = 2^{10}, so n must divide 10: n ∈ {1,2,5,10}.
- Then k = 2^{10/n} + 1 and m = k^n − 2014.
- n=1 and n=2 give negative m; n=5 gives m = 1111 and n=10 gives m = 57035 (both positive).
- So m can take 2 different values.
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