Question

Find the largest 4-digit number when divided by 40,48,60 leaves remainder 3 in each case.​

Answer 1

SOLUTION :

$\bigstar$ Firstly, we get L.C.M of 40,48 & 60.

$\begin{array}{r|l} 2 & 40,48,60 \\ \cline{2-2} 2 & 20,24,30 \\ \cline{2-2} 2 & 10,12,15 \\ \cline{2-2} 2 & 5 , 6 , 15 \\ \cline{2-2} 3& 5,3,15\\ \cline{2-2} 5 & 5,1,5 \\ \cline{2-2} & 1,1,1 \end{array}$

∴ L.C.M = 2 × 2 × 2 × 2 × 3 × 5 = 240

Now,

We know greatest four digit = 9999

240) 9999( 41

        960

______________

             399

             240

_______________

              159

_______________

∴ 9999 - 159 = 9840 which is divisible.

Thus;

The required number = 9840 - 3 = 9837

Answer 2

Answer:

YOUR ANSWER IS HERE :

A. 9996 is divisible by all 3 number.

Step-by-step explanation:

A. The greatest four digit number is 9999 .

Now find the HCF of 40 , 48 , 60 -

40 = 2 × 2 × 5

48 = 2 × 2 × 2 × 2 × 3

60 = 2 × 2 × 3 × 5

HCF = 4

Divide 9999 by 4

9999 ÷ 4 = 2499 and remainder is 3

2499 × 4 + 3

9999 - 3 = 9996

The answer is 9996 and the numbers remainder is 3 .

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