Question

Find the least number which when divided by 20 25 35 and 40 leaves remainders 14 19 29 and 34 respectively​

Answer 1

(20×35×25×40)÷4

30×4=120+14+19+29+34

Answer 2

       

Simple method!!!

   

In such questions, when we see the divisors and the remainders deeper, we can get an interesting fact that the divisors and the corresponding remainders differ each other by a same integer.

20 - 14 = 6

25 - 19 = 6

35 - 29 = 6

40 - 34 = 6

So we can find the answer by subtracting this 6 from the LCM of the given divisors.

The LCM of the divisors 20, 25, 35 and 40 is 1400.

So the answer is 1400 - 6 = 1394.

Complicated method!!!

     

Let the least number be N.

If N leaves remainder 14 on division by 20, let N = 20a + 14.

If N leaves remainder 19 on division by 25, let N = 25b + 19.

If N leaves remainder 29 on division by 35, let N = 35c + 29.

If N leaves remainder 34 on division by 40, let N = 40d + 34.

So that,

N  =  20a + 14  =  25b + 19  =  35c + 29  =  40d + 34

Add all sides with 6, so we get,

N + 6  =  20a + 14 + 6  =  25b + 19 + 6  =  35c + 29 + 6  =  40d + 34 + 6

⇒ N + 6  =  20a + 20  =  25b + 25  =  35c + 35  =  40d + 40

⇒ N + 6  =  20(a + 1)  =  25(b + 1)  =  35(c + 1)  =  40(d + 1)

Here it seems that  N + 6  is a multiple of the divisors 20, 25, 35 and 40.

As the question is to find the least number,  N + 6 has to be the least multiple (LCM) of the divisors.

∴ N + 6 = LCM(20, 25, 35, 40)

⇒ N + 6 = 1400

⇒ N = 1400 - 6

⇒ N = 1394

So the answer is 1394.

Plz ask me if you have any doubt on my answer.

Thank you. :-))

$\mathfrak{\#adithyasajeevan}$