How many numbers are there that appear in at least one of the arithmetic sequences $10,16,22,28,\ldots,1000$ and $10,21,32,43,\ldots,1000?$

Question

How many numbers are there that appear in at least one of the arithmetic sequence $10, 16, 22, 28, \ldots, 1000$ and the arithmetic sequence $10, 21, 32, 43, \ldots, 1000?$

How can I count all the numbers in both sequences and keep count of the over counting? How do I use casework here?

Answer 1

There are $\dfrac{1000-10}6+1=166$ numbers in the first sequence

and $\dfrac{1000-10}{11}+1=91$ numbers in the second sequence.

There are $\dfrac{1000-10}{66}+1=16$ numbers in both sequences.

Can you take it from here?

Answer 2

The numbers in the first sequence are $6k+10$ for $0\leq k \leq \frac{1000-10}{6}=165$. There are $166$ numbers in this sequence.

The numbers in the second sequence are $11k+10$ for $0\leq k \leq \frac{1000-10}{11}=90$. There are $91$ numbers in this sequence.

The numbers in both sequences are $nk+10$ for $0\leq k \leq \frac{1000-10}{n}$ where $n=\text{lcm}(6,11)=66$. There are $16$ numbers in both sequences.

Using inclusion/exclusion we have

$$166+91-16=241$$

numbers that appear in at least one sequence.