Find the least number that leaves remainder 17 when divided by 2 4, 5

Given:

The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3

The least number when divided by 9 remainder = 0

Concept Used:

For the least number take LCM and also check that all conditions are satisfying or not

For same remainder in each case add the remainder in the LCM 

Calculation:

First we have to find the LCM of 5, 6, 7, 8

5 = 51

6 = 21 × 31 

7 = 71 

8 = 23 

LCM of 5, 6, 7, 8 is 23 × 31 × 51 × 71 = 840

For same remainder we have to add 3 to the LCM

⇒ 840 + 3 

Now check weather it is divisible by 9 or not

843/9 = Not divisible by 9

Now we have to take next multiple of LCM and then add 3

840 × 2 + 3

⇒ 1683

Divide it by 9

1683/9 = 187

⇒ 1683 is divisible by 9 

⇒ 1683 is the least number when divided by 5, 6, 7, 8 gives remainder 3

The least number must be divisible by 9 

For divisibility of 9 sum of digits of the number must be divisible by 9 

Checking it by options:

Option 1) 1677

1 + 6 + 7 + 7 = 21

⇒ 1677 is not divisible by 9

Option 2) 1683

1 + 6 + 8 + 3 = 18

⇒ 1683 is divisible by 9

Option 3) 2523

2 + 5 + 2 + 3 = 12

⇒ 2523 is not divisible by 9

Option 4) 3363

3 + 3 + 6 + 3 = 15

⇒ 3363 is not divisible by 9

∴ The least number is 1683 which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.

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.