Question

find the smallest number which can be divided by 3, 4, 5 and 6 and least the remainder 1, 2, 3 and 4 in each case but exactly divided by 13​

Answer 1

Answer:

The required number is 301.

Step-by-step explanation: we have to find a number that is divisible by 2, 3, 4, 5, and 6 (meaning that it is a multiple of those numbers) and that number + 1 being divisible by

I believe you want the smallest or least number which leaves a remainder of 1 when divided by any of the numbers 2, 3, 4, 5, 6, or 7. We just need to find the Least Common Multiple (LCM) of the divisors and add 1. The LCM, by definition, is the smallest number which would be evenly divisible by all of those divisors; adding the 1 will provide the desired remainder. To find the LCM we simply express each divisor as the product of its prime factors, then multiply together as many of those prime factors as were needed to create the divisors. Thus 2, 3, 5, and 7, each being prime, is already properly expressed; in addition, 4 = 2 x 2, and 6 = 2 x 3. The LCM then is given by 2 x 2 x 3 x 5 x 7 = 420. Adding 1 (for the remainders, remember?), gives us 421 as the answer to the question.