Problem 8 · 2019 Math Kangaroo
Medium
Number Theory
divisibilityfactorization
How many of the numbers from \(2^{10}\) to \(2^{13}\) (including these two numbers) are divisible by \(2^{10}\)?
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Answer: D — 8
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Hint 1 of 2
A number in this range divisible by \(2^{10}\) must equal \(2^{10}\) times a whole number.
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Hint 2 of 2
Count the whole numbers k with \(2^{10} \le k\cdot 2^{10} \le 2^{13}\), i.e. \(1 \le k \le 8\).
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Approach: count the multiples of \(2^{10}\) in the range
- A multiple of \(2^{10}\) in the range is \(k\cdot 2^{10}\) with \(2^{10} \le k\cdot 2^{10} \le 2^{13}\).
- Dividing through by \(2^{10}\) gives \(1 \le k \le 2^{3} = 8\).
- So k = 1, 2, …, 8, giving 8 such numbers.
- Answer (D) 8.
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