All about divisibility and counting
How many positive integers from 1 to 8644 inclusive are divisible by EXACTLY TWO of 2,3 and 5?
The answer is 2016.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Haha Yes, that is much simpler than my "high tech" approach ....
Let N ( k ) be the number of integers between 1 and 8 6 4 4 inclusive divisible by k .
Then the desired value is N ( 2 ∗ 3 ) + N ( 2 ∗ 5 ) + N ( 3 ∗ 5 ) − 3 ∗ N ( 2 ∗ 3 ∗ 5 ) =
⌊ 6 8 6 4 4 ⌋ + ⌊ 1 0 8 6 4 4 ⌋ + ⌊ 1 5 8 6 4 4 ⌋ − 3 ∗ ⌊ 3 0 8 6 4 4 ⌋ = 1 4 4 0 + 8 6 4 + 5 7 6 − 3 ∗ 2 8 8 = 2 0 1 6 .
The reason we subtract 3 ∗ N ( 3 0 ) is that N ( 3 0 ) is "nested" in each of N ( 6 ) , N ( 1 0 ) and N ( 1 5 ) , so since we wish to exclude integers divisible by 2 ∗ 3 ∗ 5 = 3 0 we need to subtract this value three times from the sum of the other N values.
Log in to reply
In German we call this "mit Kanonen auf Spatzen schiessen" ;) Merry Christmas, Brian! More fun in 2016!
Log in to reply
Hahaha Yes, it is indeed a "sledgehammer" approach, but it did crack the nut, (I relied on my niece for the translation, (she's the linguist in the family), so I hope that was the correct interpretation). :) Merry Christmas to you too, Dr. Bretscher, and may 2016 be full of good will and good math!
I used a "low tech" approach: We count seven such numbers <30, and then the pattern is repeated 288 times up to 3 0 × 2 8 8 = 8 6 4 0 , for a grand total of 7 × 2 8 8 = 2 0 1 6 . Happy 2016!