Find the number of distinct integers in the collection of first $500$ terms of each of the arithemetic progressions $\{1,4,...\}$ and $\{9,14,...\}$

Question

Let $1,4,...$ and $9,14,...$ be two arithmetic progressions. Then the number of distinct integers in the collection of first $500$ terms of each of progressions is $$A. 833,\space B. 835,\space C.837 ,\space D. 901$$

My solution goes like this:

The two A.P.'s general term might be represented as $$1+(n_1-1)3=9+(n_2-1)5\implies 3n_1-5n_2=6\implies 3(n_1-2)=5n_2.$$ Hence, $3|n_2\implies n_2=3k$ and so, $3(n_1-2)=5n_2=15k\implies n_1-2=5k\implies n_1=5k+2.$ Now, $5k+2\leq 500\implies 5k\leq 498\implies k\leq 99.$ Due to this, we may conclude, $99$ terms of both the A.P.'s are similar. So, total number of common terms in both the A.P's is : $99+99=198$ and the number of distinct terms in both the A.P's in total is: $1000-198=802.$

However, I feel something's wrong with my solution, as my solution, is not matching with any of the options. But I don’t understand where is the solution going wrong, if that's the case. I am not quite getting it...

Answer 1

You've done a great job, and almost have it. However, note that determining the number of distinct terms involves removing just the additional repeated terms, not all of the terms that have been repeated (i.e., the "total number of common terms" as you call it). As JonathanZ supports MonicaC's comment states, you have

answered the question "how many numbers are there that show up in only one of the series". And I don't think anyone could argue that that's what "distinct" means.

In particular, you've instead determined the number of unique terms, i.e., how many terms that appear just once.

You've found there were $99$ terms that were repeated so, since each was repeated only once, the total needs to be reduced by just $99$. This then leaves one copy of each of those terms, which leads to the result being

$$1000 - 99 = 901$$

i.e., answer $D$.