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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 $1000=24K-1$
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 $41$.
