Three-digit numbers P, Q, and R, QQP and PQQ are formed using the single digits P, Q and R such that:
PQR + 2 = PQQ
and
PQR * 2 = QQP
Determine all possible combinations of values of P, Q and R that satisfy the two equations.
Here is my work so far:
What we know:
R+2=Q, PQ remains the same meaning there were no carrying needed. R+2<=9
The second equation, R*2=P. R cannot be 0 as anything multiplied by 0 is 0.
What we know about R:
Ranges from 1 – 7
What we know about P:
Cannot be 0 as it is in the hundred digits.
P*2 = Q
What we know about Q:
Q*2 = the first digit of QR.
Hint: The first equation gives us $ R + 2 = Q$.
Hint: The second equation then gives us $199 P - 4 = 88Q$
Hence conclude that $P = 4, Q = 9, R = 7$.