Question

1. a passcode is made up of five digits. work it out from the following information: the fourth digit is three times the second digit. the third digit is six times the fifth digit. all the digits are different. the first digit is one less than the fourth digit

1. five teams play in a hockey tournament. each team plays every other team once. team a has played four games team b has played three games team c has played four games team d has played two games team e has played three games

which teams still need to play each other?

Explanation:

Step1: Analyze passcode digits

Let digits be \(d_1,d_2,d_3,d_4,d_5\) (1-5th).

• \(d_4=3d_2\): \(d_2\) can be 0,1,2,3 (digits 0-9). \(d_4\) valid only if \(d_2=1\) (\(d_4=3\)) or \(d_2=2\) (\(d_4=6\)) or \(d_2=3\) (\(d_4=9\)).
• \(d_3=6d_5\): \(d_5\) can only be 1 (\(d_3=6\)) (since \(d_5=0\) makes \(d_3=0\), duplicate).
• \(d_1=d_4-1\): If \(d_4=3\), \(d_1=2\); if \(d_4=6\), \(d_1=5\); if \(d_4=9\), \(d_1=8\).
• All digits different: \(d_3=6\), so \(d_4≠6\). Thus \(d_4=3\) or 9. If \(d_4=3\), \(d_2=1\), \(d_1=2\) (digits 2,1,6,3,1 → duplicate 1). If \(d_4=9\), \(d_2=3\), \(d_1=8\) (digits 8,3,6,9,1 → all unique).

Step2: Analyze hockey tournament

Total possible games: \(\binom{5}{2}=10\).
Total played games: \(\frac{4+3+4+2+3}{2}=8\) (each game counted twice).
Remaining games: 2.
Team A played all (vs B,C,D,E). Team C played all (vs A,B,D,E). Team D played 2 games (only vs A and C). So unplayed games: B vs E, and check others: B played 3 (vs A,C,?), E played 3 (vs A,C,?). B and E haven’t played each other; D has no unplayed games. Remaining games: B vs E.

Answer:

1. 83691
2. Team B and Team E