Vtyjkyuk

A number when divided by 2 leaves a remainder of 1. When it is divided by 3 leaves a remainder 2. When it is divided by 4 it leaves a remainder of 3. And when it is divided by 5 it leaves remainder of 4. What should be the number ?

Note : I already formulated some formula but I think it might not work:

x = 2a + 1

x = 3b + 2

x = 4c + 3

x = 5d + 4

This is a classic problem in number theory solved by the Chinese remainder theorem. You can look up this algorithm many places, but indeed the (smallest) solution (as posted by @metamorphy) is $59$.

Note that if $x equiv 3 pmod4$ you get 'for free' that $x equiv 1 pmod2$. So these two conditions toghether are equivalent to the first one. Thus, you're looking at

$$
cases
x equiv 3 pmod4 \
x equiv 2 pmod3 \
x equiv 4 pmod5
$$

By the chinese remainder theorem, this system of equations has a unique solution modulo $4cdot3cdot5 = 60$. The only numbers in $1,dots,60$ that are congruent to $4$ modulo $5$ are

$$
0, 4,9,14,19,24,29,34,39,44,49,54,59.
$$

Now, our number is odd by the first observation, so we can reduce the former to check only these

$$
9,19,29,39,49,59
$$

and finally, only $59$ here is $2$ modulo $3$. Thus, the solutions have the form

$$
60k + 59
$$

or since $59 equiv -1 pmod60$,

$$
60k -1 quad (k in mathbbZ)
$$

Continuing your thought:
$$begincasesx=2a+1\x=3b+2\x=4c+3\x=5d+4endcases Rightarrow begincasesx=2a+1\2a+1=3b+2\2a+1=4c+3\2a+1=5d+4 endcasesRightarrow
begincasesx=2a+1\a=frac12(3b+1)\3b+2=4c+3\3b+2=5d+4 endcasesRightarrow \
begincasesx=2a+1\a=frac12(3b+1)\b=frac13(4c+1)\
4c+3=5d+4 endcasesRightarrow
begincasesx=2a+1\a=frac12(3b+1)\b=frac13(4c+1)\
c=frac14(5d+1) endcases$$
Now you must solve the last Diophantine equation (or by trial and error) and find backwards:
$$d=11,c=14,b=19,d=29,x=59.$$
Alternatively, using modular arithmetic:
$$begincases
xequiv 1pmod2\
xequiv 2pmod3\
xequiv 3pmod4\
xequiv 4pmod5endcases Rightarrow
begincases
x=2a+1\
2a+1=2pmod3\
2a+1=3pmod4\
2a+1=4pmod5 endcasesRightarrow
begincases
x=2a+1\
aequiv 2pmod3\
aequiv 1pmod2\
aequiv 4pmod5 endcasesRightarrow \
begincases
x=2a+1\
a=3b+2\
3b+2equiv1pmod2\
3b+2equiv 4pmod5 endcasesRightarrow
begincases
x=2a+1\
a=3b+2\
bequiv-1pmod2\
bequiv 4pmod5 endcasesRightarrow
begincases
x=2a+1\
a=3b+2\
b=2c-1\
2c-1equiv 4pmod5 endcasesRightarrow \
begincases
x=2a+1\
a=3b+2\
b=2c-1\
cequiv 0pmod5 endcases Rightarrow
begincases
colorredx=2(30n-1)+1=colorred60n-1\
a=3(10n-1)+2=30n-1\
b=2(5n)-1=10n-1\
c=5n, nin mathbb Z endcases$$

If we increase $a,b,c,d$ by one from your system of equations, observe that these equations now tell us that $x=2a-1=3b-1=4c-1=5d-1$ with the new values. Thus $x+1=3b=4c=5d$, so $60|(x+1)$; thus $x=60n-1$ for any integer $n$.