2020 AMC 10A Problem 22

Question

For how many positive integers

n \le 1000

is

\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor

not divisible by

3?

(Recall that

\lfloor x \rfloor

is the greatest integer less than or equal to

x.

)

22

23

24

25

26

Answer: A

Concepts:floor and ceiling functions divisibility factor counting

Difficulty rating: 2380

Small hint:

Let

1000=qn+r

Big hint:

The sum of floors fails to be a multiple of

3

only when subtracting

1

or

2

crosses a multiple of

n

Video solution:

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Written solution:

Write

1000=qn+r

, where

0\le r<n

. Then

\left\lfloor\dfrac{1000}{n}\right\rfloor=q

. The other two floors are usually also

q

, except that subtracting

1

or

2

from

1000

crosses a multiple of

n

when

r

is small.

For

n>1

, the sum is not divisible by

3

exactly when

r=0

or

r=1

. The case

r=0

gives divisors of

1000

, excluding

1

, for

15

values. The case

r=1