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Find the number of common terms in the two series S1 and S2

$S1= 1,3,6,10.......200$ terms $S2= 3,6,9,12,15......200$ terms

Is there any easy approach for finding the common terms in these two series?

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  • $\begingroup$ any thoughts? As a suggestion, write out the first dozen common terms. Try to spot a pattern. $\endgroup$ – lulu Aug 17 '17 at 20:53
  • $\begingroup$ okay @lulu...got it $\endgroup$ – Sakuzi Markel Aug 17 '17 at 20:55
  • $\begingroup$ 6,15,21....@lulu $\endgroup$ – Sakuzi Markel Aug 17 '17 at 20:56
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    $\begingroup$ left off $3$... Hint: look at what positions in the first series these are in. $\endgroup$ – lulu Aug 17 '17 at 21:02
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HINT

After the first 5 terms, $S_1 < S_2$ and $S_2$ is really a set of all multiples of 3. Now $S_1(n) = 1 +2 +\ldots + n = n(n+1)/2$ and you seek the number of these up to $600$ which is divisible by 3.

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  • $\begingroup$ How to fnd the number of terms in S1 that are divisible by 3? @gt6989b $\endgroup$ – Sakuzi Markel Aug 18 '17 at 18:35
  • $\begingroup$ @SakuziMarkel Remember $S_1(n) = n(n+1)/2$. When is a product of integers divisible by 3? $\endgroup$ – gt6989b Aug 18 '17 at 22:14
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S2 is a pretty simple pattern, so the easiest way to me seems to find all elements of $S1$ that are divisible by 3 (and less than the largest element of $S2$, 600). A good way to do that is to only look at the remainders after division by 3. Notice now that S1 is periodic, because you are adding the terms 2,0,1, repeatedly to get the pattern 1,0,0,1,0,0,1,0,0,1. Each 0 is going to be divisible by 3, so we can generate these by plugging in the index of the 0s in the list into the equation n(n+1)/2. (for example the first 0 is the second in the list and corresponds to 2*3/3=3).

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By brute force:

"Steal" the triangular numbers in range $[3,600]$ from Wikipedia and select the multiples of $3$:

$$\color{green}{3}, \color{green}{6}, 10, \color{green}{15}, \color{green}{21}, 28, \color{green}{36}, \color{green}{45}, 55, \color{green}{66}, \color{green}{78}, 91, \color{green}{105}, \color{green}{120}, 136, \color{green}{153}, \color{green}{171}, 190, \color{green}{210}, \color{green}{231}, 253, \color{green}{276}, \color{green}{300}, 325, \color{green}{351}, \color{green}{378}, 406, \color{green}{435}, \color{green}{465}, 496, \color{green}{528}, \color{green}{561}, 595.$$

For the same "price", you get the pattern.

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