Two finite arithmetic sequences are given: 2, 5, 8, 11, ... up to the 1000th term, and -1, 6, 13, 20, ... up to the 1000th term. How many numbers appear in both?

Step-by-step solution

1. 1

   Describe each sequence by a congruence. $(a_n)$ has $d=3$ and starts at $2$, so its terms are exactly the integers $x\equiv 2\pmod 3$ in its range. $(b_n)$ has $d=7$ and starts at $-1$, so its terms are the integers $x\equiv -1\equiv 6\pmod 7$.

2. 2

   Find the last term of each. $a_{1000}=2+999\cdot 3=2999,\qquad b_{1000}=-1+999\cdot 7=6992$ The overlap can therefore only live in $[2,\;2999]$, the shorter range.

3. 3

   Solve the pair of congruences. We need $x\equiv 2\pmod 3$ and $x\equiv 6\pmod 7$. Testing $x=6,13,20$ against the first congruence, $20=3\cdot 6+2$ works, so by the Chinese Remainder Theorem $x\equiv 20\pmod{21}$ (the modulus is $\operatorname{lcm}(3,7)=21$).

4. 4

   Count the solutions inside the range. The shared terms are $20,41,62,\dots\le 2999$, an AP with difference $21$: $\left\lfloor\frac{2999-20}{21}\right\rfloor+1=\left\lfloor\frac{2979}{21}\right\rfloor+1=141+1=142$

5. 5

   Confirm and answer. Building both 1000-term sets explicitly and intersecting them also returns $142$ elements, so the answer is C.