The least number which when divided by 2, 3, 4, 5 and 6

The least number which when divided by 4, 5 and 6 leaves remainder 1, 2 and 3 respectively, is : [A]57 [B]59 [C]61 [D]63

57 Here 4 – 1 = 3, 5 – 2 = 3, 6 – 3 = 3 ∴ The required Number = LCM of (4, 5, 6) – 3 = 60 – 3 = 57.

Hence option [A] is correct answer.

Below are a few different approaches (besides the standard Extended Euclidean Algorithm).

${\rm mod}\ 60\!:\ x\equiv 1\equiv 7n\!\iff\! n\equiv\dfrac{1}7\equiv \dfrac{-59}7\equiv \dfrac{-119}7\equiv -17\,$

therefore $\smash[t]{\,x = 7(\overbrace{-17\!+\!60k}^{\large n}) = -119+420k}$

Alternatively $\:\! $ mod $\,60\!:\ \color{#c00}{7^4\equiv 1}\,$ (by true mod $3,4,5)\,$ so $\smash[b]{\,\color{#c00}{7^{-1}\equiv 7^3}\equiv 7(\underbrace{-11}_{\Large 7^2})\equiv -17}$

Alternatively $ $ we can employ $ $ Inverse Reciprocity

$\qquad\!\! \dfrac{1}7\ {\rm mod}\ 60\ \equiv\ \dfrac{1-60\overbrace{\left(\color{#c00}{\dfrac{1}{60}}\ {\rm mod}\ 7\right)}^{\Large \color{#0a0}{\equiv\, 2}}}7\,\equiv\, \dfrac{-119}7 \,\equiv\, -17\ $

where we've used: $\ \ {\rm mod}\ 7\!:\,\ \color{#c00}{\dfrac{1}{60}}\equiv \dfrac{8}4\color{#0a0}{\equiv 2}\,\ $ (or recurse on $\,\dfrac{1}{60}\bmod 7 \,\equiv\, \dfrac{1}4 \bmod 7\,)$

See here and its links for further methods and elaboration.

Remark $ $ In the first method we found a numerator $\,-119\equiv 1\pmod{60}$ that's also divisible by $7$ by brute force, i.e. we tested $\,1-60k\,$ for $\,k=1,2\ldots$ But now we see that the solution $\,\color{#0a0}{ k\equiv 2}\,$ is simply $\, k = \color{#c00}{\dfrac{1}{60}}\,\bmod\, 7,\,$ a "reciprocal" of our sought $\,\dfrac{1}7\bmod 60\,$ (i.e. swap $\,7, 60)$.

Beware $\ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.