Question

Find the smallest number which increased by 7 is exactly divisible by both 247 and 130? With procedure

Answer 1

Take LCM of 247 and 130 is 13×19×10=2470

this is the smallest number divisible by 247 and 130
but the required number is 2470-7 = 2463
Here is your answer... if you like it mark as brainly

Answer 2

Answer:

Required number is 2470-7 = 2463

Step-by-step explanation:

Given

To find out the smallest number, determine LCM

LCM of 247 and 130= 13×19×10=2470

The smallest number which is exactly divisible by 247 and 130  when increase by 7, so the required number is obtained by subtracting 7 from the LCM

2470  - 7=  2463

The required number is 2470-7 = 2463