Divisible by 3 or 7

Find the number of four-digit positive integers divisible by 3 or 7.
Source: NCTM Mathematics Teacher, August 2006

Solution
Consider A=\{1002,1005,1008,\cdots,9996,9999\}, the set of four-digit positive integers divisible by 3 and B=\{1001,1008,1015,\cdots,9989,9996\}, the set of four-digit positive integers divisible by 7. The count of A=(9999-1002)/3+1=3000 and the count of B=(9996-1001)/7+1=1286.
Some integers like 1008 and 9996 appear in both sets so we  subtract them as duplicates from the total count. Since 1008+(7\times 3)=1008+(3\times 7)=1029, starting from 1008 the duplicates occur  every 7th integer in set A or every 3rd integer in set B. If we use set A, the count of duplicates = (9996-1008)/(7\times 3)+1=429. The number of four-digit positive integers divisible by 3 and/or 7 is 3000+1286-429=3857.

Answer: 3857

Alternative solution
If we divide 9999 by 3, we get 3333 which means that there are 3333 integers \leq 9999 that are divisible by 3. But, if we divide 9999 by 7, we get 1428.43 which is not an integer. So we say there are 1428 integers \leq 9999 that are divisible by 7. To  avoid this problem we use the floor function \lfloor x\rfloor = the largest integer less than or equal to x for some real number x. In our example, \lfloor 1428.43\rfloor=1428.
Number of four-digit integers divisible by 3
\lfloor 9999/3\rfloor-\lfloor 999/3\rfloor=3000
Number of four-digit integers divisible by 7
\lfloor 9999/7\rfloor-\lfloor 999/7\rfloor=1286
Number of four-digit integers divisible by 21
\lfloor 9999/21\rfloor-\lfloor 999/21\rfloor=429
There are 3000+1286-429=3857 four-digit integers divisible by 3 and/or 7.

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , . Bookmark the permalink.

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