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

To find the least number that leaves a remainder of 3 when divided by 5, 6, 7, and 8, but is exactly divisible by 9, we follow these steps:

Step-by-Step Calculation
1. Find the L.C.M. of 5, 6, 7, and 8:

Prime factors:

5 = 5

6 = 2 * 3

7 = 7

8 = 2 * 2 * 2 = 2^3

L.C.M. = 2^3 * 3 * 5 * 7 = 8 * 3 * 5 * 7 = 840.

2. Formulate the expression for the number:
Since the number leaves a remainder of 3 when divided by 840, we can express it as:
Number = 840k + 3, where k is a positive integer (1, 2, 3...).

3. Apply the divisibility by 9 condition:
We need the number (840k + 3) to be exactly divisible by 9.

Divide 840 by 9: 840 = 9 * 93 + 3.

So, (840k + 3) = (9 * 93k + 3k + 3).

For this to be divisible by 9, (3k + 3) must be divisible by 9.

Test values for k:

If k = 1: 3(1) + 3 = 6 (not divisible by 9)

If k = 2: 3(2) + 3 = 9 (divisible by 9)

4. Calculate the final number:
Using k = 2:
Number = 840 * 2 + 3 = 1680 + 3 = 1683.