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