# How many numbers are there between $0$ and $1000$ which on division by $2, 4, 6, 8$ leave remainders $1, 3, 5, 7$ resp?

How many numbers are there between $$0$$ and $$1000$$ which on division by $$2, 4, 6, 8$$ leave remainders $$1, 3, 5, 7$$ resp?

What I did:- Observe the difference between divisor and remainder. $$2-1=1$$ $$4-3=1$$ $$LCM(2,4,6,8)=24K-1$$ Between $$0$$ and $$1000$$ there are $$1001$$ numbers $$n=24K-1$$ $$1001=24K-1$$ $$1002=24K$$ But in the solution it has been given $$1000=24K-1$$

• Every such number $n$ corresponds to a number $n+1$ in $\{1,\dots,1001\}$ which is divisible by all of $2,4,6,8$. Mar 20, 2022 at 8:11

You want the number of $n$ between $0$ and $1000$ such that $n \equiv -1 \mod 24$. $1000 = 41 \times 24 + 16$, so there is one of these in each of the $41$ intervals $[0, 23],\; [1 \cdot 24, 1 \cdot 24 + 23],\; \ldots [40 \cdot 24, 40 \cdot 24 + 23]$ where $40 \cdot 24 + 23 = 983$. The next one would be $982+24 = 1006$, which is too big. So the answer is $41$.
Let $$x$$ be all of the numbers that could happen. We get that $$x \equiv 23 \pmod {24}$$.
As $$1000$$ is not congruent to $$23 \pmod {24}$$, in total, this gives us $$\lfloor\frac{1000}{24}\rfloor$$ possibilities, which gives us the answer of $$41$$.