Find the least number which when divided leaves following remainders

Question

The question is, find the least number when divided by $36, 48$ and $64$ leaves remainder $25, 37$ and $53$ respectively. The answer is said to be $($the $\operatorname{lcm}$ of the three numbers $-11),$ where $36- 25= 48- 37= 64- 53= 11.$

Can anyone please give me the logic for doing this ? I am not familiar with the modular operator method. So any answer not using that would be helpful. Thanks

Answer 1

So let us assume that the required number is $x$.
Now by the given conditions, \begin{align} x\equiv 25\quad (mod\quad 36) \tag{1}\\ x\equiv 37 \quad(mod \quad 48)\tag{2}\\ x\equiv 53\quad (mod \quad 64)\tag{3}\\ \end{align}
$(1)\implies$ $x=36u+25$ for some $u\in\Bbb{Z}$ $\tag{4}$
Now $x=36u+25$ satisfies $(2)$
So $(2)\implies$ \begin{align} &36u+25\equiv37\quad (mod \quad 48)\\ \text{i.e} \quad & 36u\equiv 12 \quad(mod \quad 48)\\ \text{i.e} \quad & 3u\equiv 1 \quad(mod \quad 4)\\ \end{align} Solution of the above congruence is $u\equiv -1 \quad(mod\quad 4)\implies u\equiv 3 \quad(mod\quad 4)$
So $u=4v+3$ for some $v\in\Bbb{Z}$.
Thus from $(4),\quad x=144v+133$, $\tag{5}$

Again $(5)$ satisfies $(3)$. So
\begin{align} & 144v+133\equiv 53\quad (mod \quad 64)\\ \text{i.e} \quad & 144v\equiv -80 \quad(mod \quad 64)\\ \text{i.e} \quad & 9v\equiv -5 \quad(mod \quad 4)\\ \end{align} Solution of the above congruence is $v\equiv -1 \quad(mod\quad 4)\implies v\equiv 3 \quad(mod\quad 4)$
So $v=4w+3$ for some $w\in\Bbb{Z}$.
Thus from $(5),$$ \quad x=576w+565 \tag{6}$

Hence the required solution is $x\equiv 565 \quad(mod \quad 576)$

Since the least integer (obviously positive) is required, so putting $w=-1$ in $(6)$ yields $x=11$

---

PS: You could also use Chinese Remainder Theorem to find this value.