Question

find the least number which when divisible by10,20 and 30 gives remainder 3,4 and 5​

Answer 1

Answer:

Answer:Here's how you approach such questions. First find the lcm of the numbers that the required number is divisible with, i.e., 3,5,6,8,10,12 which is 120

Answer:Here's how you approach such questions. First find the lcm of the numbers that the required number is divisible with, i.e., 3,5,6,8,10,12 which is 120Therefore the required number is of the form 120x+2 (since it leaves a reminder of 2)

Answer:Here's how you approach such questions. First find the lcm of the numbers that the required number is divisible with, i.e., 3,5,6,8,10,12 which is 120Therefore the required number is of the form 120x+2 (since it leaves a reminder of 2)Now the number is a multiple of 13.

Answer:Here's how you approach such questions. First find the lcm of the numbers that the required number is divisible with, i.e., 3,5,6,8,10,12 which is 120Therefore the required number is of the form 120x+2 (since it leaves a reminder of 2)Now the number is a multiple of 13.so 120x+2 is a multiple of 13

Answer:Here's how you approach such questions. First find the lcm of the numbers that the required number is divisible with, i.e., 3,5,6,8,10,12 which is 120Therefore the required number is of the form 120x+2 (since it leaves a reminder of 2)Now the number is a multiple of 13.so 120x+2 is a multiple of 13=> 117x + 3x+2 is a multiple of 13

Answer:Here's how you approach such questions. First find the lcm of the numbers that the required number is divisible with, i.e., 3,5,6,8,10,12 which is 120Therefore the required number is of the form 120x+2 (since it leaves a reminder of 2)Now the number is a multiple of 13.so 120x+2 is a multiple of 13=> 117x + 3x+2 is a multiple of 13The least value of x for which this holds true is 8 (3*8+2 = 26)

Answer:Here's how you approach such questions. First find the lcm of the numbers that the required number is divisible with, i.e., 3,5,6,8,10,12 which is 120Therefore the required number is of the form 120x+2 (since it leaves a reminder of 2)Now the number is a multiple of 13.so 120x+2 is a multiple of 13=> 117x + 3x+2 is a multiple of 13The least value of x for which this holds true is 8 (3*8+2 = 26)Therefore the least such number is 120*8+2 = 962