Lovely Erasers

If you add $1$ eraser to the erasers that I had by far, the new number of erasers can be grouped into groups of $2,3,4,5$ .

The lowest common multiple(lcm) of $2,3,4,5=3\cdot4\cdot5=60$

So the minimum number of erasers that I can have $=60-1=59$

The next number that meet the requirements $=60\cdot2-1=119$

But $119>100$

So the answer we are looking for is indeed $\boxed{59}$

$\textbf{Generalisation:}$ The number of erasers that meet the requirements can be written as $60k-1$ for positive integers $k$ .

Proof：

See Modular Arithmetic

Let the number of erasers be $n$

$n-1\equiv0(mod2)$

$\Rightarrow n\equiv1(mod2)$

$\Rightarrow n\equiv2-1(mod2)$

$\Rightarrow n\equiv-1(mod2)$

$\Rightarrow n+1\equiv0(mod2)$

---

$n-2\equiv0(mod3)$

$\Rightarrow n\equiv2(mod3)$

$\Rightarrow n\equiv3-1(mod3)$

$\Rightarrow n\equiv-1(mod3)$

$\Rightarrow n+1\equiv0(mod3)$

---

$n-3\equiv0(mod4)$

$\Rightarrow n\equiv3(mod4)$

$\Rightarrow n\equiv4-1(mod4)$

$\Rightarrow n\equiv-1(mod4)$

$\Rightarrow n+1\equiv0(mod4)$

---

$n-4\equiv0(mod5)$

$\Rightarrow n\equiv4(mod5)$

$\Rightarrow n\equiv5-1(mod5)$

$\Rightarrow n\equiv-1(mod5)$

$\Rightarrow n+1\equiv0(mod5)$

---

$\therefore n+1\equiv0(mod60)$

$n+1=60k$ such that $k$ is positive integer.

Hence, we have $n=60k-1$

Relevant wiki: Chinese Remainder Theorem

Let the number of erasers be $n$ . We can solve the problem using the Chinese remainder theorem (CRT) as follows.

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