Problem 29 · 2015 Math Kangaroo
Stretch
Number Theory
divisibilitycasework
Maria divides 2015 by 1. Then she divides 2015 by 2, and then in order by 3, 4 and so on up to and including 1000. For each division she writes down the remainder. What is the biggest remainder she has noted down?
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Answer: C — 671
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Hint 1 of 2
The remainder is always less than the divisor, so big remainders need a divisor a bit above a half (or third) of 2015.
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Hint 2 of 2
With the divisor capped at 1000, try n just above 2015/3 so the quotient is 2 and the remainder 2015 - 2n is large.
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Approach: maximise the remainder under the divisor cap of 1000
- The remainder of 2015 divided by n is at most n - 1, and n only runs up to 1000.
- For n between 672 and 1007 the quotient is 2, so the remainder is 2015 - 2n, which is largest when n is smallest in that range.
- At n = 672 the remainder is 2015 - 1344 = 671, and nothing under the cap beats it.
- So the biggest remainder is 671 (C).
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