Question

Find the smallest number which
when divided by 42 at and 147
leaves a remainder of 5 in each
each case.

Please answere it fast it's very urgent. And on more thing that please don't give error answer only if you know then only answer and I want correct answer ​

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

Answer 2

Answer:

please refer to attachment

Step-by-step explanation: