Question

Find the smallest number which on being divided by 8, 12 and 15 leaves remainder 6, 10 and 13 respectively. But when divided by 23 no reminder is left

Answer 1

Answer:

598

Step-by-step explanation:

Since (8-6) =(12-10) =(15-13) =2,

There fore, the required number =(LCM of 8,12,15)-2

= 120-2  =118

But by the given condition the number must be exactly divided by 23

The number =(multiple of 120)-2 = (120*5) - 2 =598 (since 598/23 = 6)