Question

find the smallest number which when divided by 42 or 457 leaves a remainder of 5 in each case​

Answer 1

Answer:

First factor 42 into prime numbers;

42 = 2 x 3 x 7

Now factor 147 into prime numbers;

147 = 3 x 7 x 7

Now combine the primes as many times as they appear in both sets of factors;

2 x 3 x 7 x 7 = 294

Now add 5;

294 = 5 = 299

So 299 is the smallest number which leaves a remainder of 5 when divided by either 42 or 147