Explanation:
So this below is a solution I devised myself when I was around 11 yrs old doing math competitions with my friend so… (I don’t have a name for this solution, and can’t give any reference material)
alright so notice that when you subtract the remainder from the divisor in this question, all of their the values equate to 4:
6–2 = 4
9–5 = 4
11–7 = 4
I am going to use this property to devise a method from the problem.
The property above means
(n+ 4) % 6 = 0
(n+ 4) % 9 = 0
(n+ 4) % 11 = 0
where n is the dividend
notice that for n to be in accordance with the restriction of the question, n+4 must be a multiple of 6, 9 and 11 simultaneously (common multiple)
Since the question asks for the number of “n”s that are three-digit positives…
100 <= n <= 999 (which means)
104<= n+4 <= 1003
ok, so now we have to find the common multiples of 6,9,11 within the range 104~ 1003
the least common multiple of 6,9,11 is 198.
the smallest multiple of 198 that is larger than or equal to 104 is 198, which is 1*198… the largest multiple of 198 smaller than 1003 is 990, which is 5*198
so that means we have all the way from the first multiple of 198 to the fifth multiple of 198, inclusive, which is (5–1) + 1 = 5
and there we have it!!!
the answer is 5
P.S. we could also just count the multiples, but the (5–1)+1 is there because it may not always be countable, and the +1 is there to account for the first number that was subtracted