Quick LCM Problem

Notice that the remainder is basically negative one for all the numbers.

Therefore, the smallest number that works is just the $lcm(5,4,3,2)-1$ , or just $60-1=59$ . Thus, any positive numbers that are equivalent to $59 \pmod {60}$ work. However, we want a number less than 100, so our answer is $\boxed{59.}$

$\text{Let } n \text{ be the number of oranges. Then we have , } \\$ $n \equiv 1 \text{ (mod 2) } \implies n - 1 \equiv 0 \text{ (mod 2) } \implies n + 1 \equiv 0 \text{ (mod 2) } \\ n \equiv 2 \text{ (mod 3) } \implies n - 2 \equiv 0 \text{ (mod 3) } \implies n + 1 \equiv 0 \text{ (mod 3) } \\ n \equiv 3 \text{ (mod 4) } \implies n - 3 \equiv 0 \text{ (mod 4) } \implies n + 1 \equiv 0 \text{ (mod 4) } \\ n \equiv 4 \text{ (mod 5) } \implies n - 4 \equiv 0 \text{ (mod 5) } \implies n + 1 \equiv 0 \text{ (mod 5) } \\$

$\text{ So we have to find the number } n+1 \text{ which is multiple of } 2 , 3 , 4 \text{ and } 5 \text{ which are } 60 , 120 , 180 , \cdots \text{ and so on } \\ \text{ But } n+1 \text{ should be } \leq 100 \\ \implies n+1 = 60 \\ \implies n=\boxed{59}$

Relevant wiki: Chinese Remainder Theorem

We can solve the problem using Chinese remainder theorem. Let the number be $n$ .

$\begin{aligned} n & \equiv 1 \text{ (mod 2)} & \small \color{#3D99F6} \implies n \equiv 2a+1 \text{, where }a \text{ is an integer.} \\ \implies 2a+1 & \equiv 2 \text{ (mod 3)} \\ 2a & \equiv 1 \text{ (mod 3)} & \small \color{#3D99F6} \implies a=2 \text{ and } n \equiv 2\times 3b + 2a+1 = 6b+5 \\ \implies 6b+5 & \equiv 3 \text{ (mod 4)} \\ 2b & \equiv 2 \text{ (mod 4)} & \small \color{#3D99F6} \implies b=1 \text{ and } n \equiv 24c + 6b + 5 = 24c+11 \\ \implies 24c+11 & \equiv 4 \text{ (mod 5)} \\ 4c & \equiv 3 \text{ (mod 5)} & \small \color{#3D99F6} \implies c=2 \\ \implies n & \equiv 24c + 11 \\ & \equiv \boxed{59} \end{aligned}$

The first statement precise that we have an odd number of oranges.

With the last one we know that this number ends by 9 (indeed, any multiple of 5 ends by 5 or 0 but we know that the number is odd. Thus 5 + 4 = 9).

Then, the seconde statement tells us that a number < 100 (which ends by 9) - 2 is divible by 3. The only three possibles numbers are 29, 59 and 89.

Finally the last statement allows us to find the answer : 59.