Question

Determine the latest number which when divided by 3,4 and 5 leaves a reminder 2 in each case

Answer 1

Let x be the number,

x = 4 * a + 2

x = 3 * b + 2

x = 5 * c + 2

Where a, b and c are integers

For x to satisfy all the three equations x has to the LCM of 3, 4 and 5.

LCM ( 3, 5, 7 ) = 3 * 4 * 5 (as 3, 4 and 5 are co prime numbers and hence their LCM is a product of the 3 numbers)

So x =( 3 * 4 * 5) * a + 2

x = 60 * a + 2

Hence plug in the non negative integral values of a and get the series of number.

The smallest number is obtained when a = 1

Hence the first number is

x = 60 + 2

x = 62

Answer 2

Answer:

62

Step-by-step explanation: