Problem:
For how many positive integers n≤1000 is
⌊n998​⌋+⌊n999​⌋+⌊n1000​⌋
not divisible by 3? (Recall that ⌊x⌋ is the greatest integer less than or equal to x.)
Answer Choices:
A. 22
B. 23
C. 24
D. 25
E. 26
Solution:
Clearly, n=1 fails. Except for the special case of n=1,
⌊n1000​⌋−⌊n998​⌋
equals either 0or1. If it equals 0, this implies that ⌊n998​⌋=⌊n999​⌋=⌊n1000​⌋, so their sum is clearly a multiple of 3, so this will always fail. If it equals 1, the sum of the three-floor terms is 3[n999​]±1, so it is never a multiple of 3. Thus, we are looking for all nî€ =1 such that
⌊n1000​⌋−⌊n998​⌋=1
This implies that either
⌊n998​⌋+1=⌊n999​⌋
or
⌊n999​⌋+1=⌊n1000​⌋
Let's analyze the first equation of these two. This equation is equivalent to the statement that there is a positive integer a such that
n998​<a≤n999​⟹998<an≤999⟹an=999⟹a=n999​⟹n∣∣∣∣∣​999.
Analogously, the second equation implies that
n∣1000
So our only n that satisfy this condition are nî€ =1 that divide 999 or 1000. Using the method to find the number of divisors of a number, we see that 999 has 8 divisors and 1000 has 16 divisors. Their only common factor is 1, so there are 8+16−1=23 positive integers that divide either 999 or 1000. Since the integer 1 is a special case and does not count, we must subtract this from our 23, so our final answer is 23−1=(A) 22​.
OR
Counting down n from 1000,999,998… we notice that ⌊n998​⌋+⌊n999​⌋+⌊n1000​⌋ is only not divisible by 3 when n is a divisor of only 1000 or 999(1000,999,500,333…). Notice how the factors of 998:1,2,499, and 998 , do not work.
The prime factorization of 999 is 33â‹…37, so 4â‹…2=8 factors in total. The prime factorization of 1000 is 23â‹…53, so 4â‹…4=16 factors in total. However, 1 obviously does not work, so we have to subtract 2 (1 is counted twice) from the total.
8+16−2=(A) 22​
The problems on this page are the property of the MAA's American Mathematics Competitions