2020 AMC 10A Problem 22

Attempt Problem 22 of the 2020 AMC 10A below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10A solutions, or check the answer key.

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22.

For how many positive integers n1000n \le 1000 is998n+999n+1000n\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloornot divisible by 3?3? (Recall that x\lfloor x \rfloor is the greatest integer less than or equal to x.x.)

2222

2323

2424

2525

2626

Answer: A
Concepts:floor and ceiling functionsdivisibilityfactor counting
Difficulty rating: 2380
Video solution:
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Written solution:

Write 1000=qn+r1000=qn+r, where 0r<n0\le r<n. Then 1000n=q\left\lfloor\dfrac{1000}{n}\right\rfloor=q. The other two floors are usually also qq, except that subtracting 11 or 22 from 10001000 crosses a multiple of nn when rr is small.

For n>1n>1, the sum is not divisible by 33 exactly when r=0r=0 or r=1r=1. The case r=0r=0 gives divisors of 10001000, excluding 11, for 1515 values. The case r=1r=1 gives divisors of 999999, excluding 11, for (3+1)(1+1)1=7(3+1)(1+1)-1=7 values. The total is 2222. Thus, A is the correct answer.

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