Let q the second number.

Either you know the following theorem:

`LCM(x,y)*HCF(x,y)=x*y`

and apply it to q and 42:

`42q=840*14`

**Therefore** q=840*14/42=280

or we find a second method

Use the prime factorization:

840=5*7*2^3*3

42=2*3*7

q is a divisor of 840 therefore q is the product of some of the terms 5, 7, 2^3, 3

Let's find which ones.

3 is a divisor of 42, 3 is not a divisor of LCM(q,42) therefore 3 is not a divisor of q.

5 is a divisor of HCF(q,42) and 5 is not a divisor of 42 therefore 5 is a divisor of q.

7 is a divisor of LCM(q,42) therefore 7 is a divisor of q and 42

2^3 is a divisor of 42 or q and is not a divisor of 42 therefore 2^3 is a divisor of q.

**In conclusion,** q=7*2^3*5=280