Four friends A, B, C and D play a game with coins over four rounds. In each round exactly one player is the 'giver': the giver hands each of the other three players as many coins as that player holds at the start of the round (thereby doubling each of their holdings). Each of the four players is the giver in exactly one round. No coins enter or leave the game, and no player ever runs out of coins completely.
At the end of the fourth round, A, B, C and D held 32, 48, 80 and 32 coins respectively.
It is also known that at the start of the game A held 98 coins and D held 14 coins, and that B started with more coins than C.
At the end of the fourth round, A, B, C and D held 32, 48, 80 and 32 coins respectively.
It is also known that at the start of the game A held 98 coins and D held 14 coins, and that B started with more coins than C.
Question 1
Single correct option
Who was the giver in Round 2?
- A. B
- B. C
- C. A
- D. D
Show answer and solution
Answer: (A) B
Deduction: The total is always 32 + 48 + 80 + 32 = 192. In a round, every non-giver's holding doubles, and the giver pays out the sum of the other three.
A starts with 98. If A did not give in Round 1, A would hold 196 > 192 ✗. So A gives first and is left with 98 − 94 = 4; B, C and D double (D: 28).
D cannot give in Round 2 (D has 28 but would owe 4 + 2B + 2C = 164) nor in Round 3 (D would hold 56 and owe 136). So D gives in Round 4. Working back from the end: before Round 4, A, B, C hold 16, 24, 40 and D holds 32 + 80 = 112; before Round 3, A holds 8.
- If C gives in Round 3 and B in Round 2: going back gives start B = 51, C = 29 ✓ (B > C).
- If B gives in Round 3 and C in Round 2: start B = 27, C = 53 ✗ (B < C).
So the order is A, B, C, D.
| Stage | A | B | C | D |
|---|---|---|---|---|
| Start | 98 | 51 | 29 | 14 |
| After Round 1 (A gives) | 4 | 102 | 58 | 28 |
| After Round 2 (B gives) | 8 | 12 | 116 | 56 |
| After Round 3 (C gives) | 16 | 24 | 40 | 112 |
| After Round 4 (D gives) | 32 | 48 | 80 | 32 |
A starts with 98. If A did not give in Round 1, A would hold 196 > 192 ✗. So A gives first and is left with 98 − 94 = 4; B, C and D double (D: 28).
D cannot give in Round 2 (D has 28 but would owe 4 + 2B + 2C = 164) nor in Round 3 (D would hold 56 and owe 136). So D gives in Round 4. Working back from the end: before Round 4, A, B, C hold 16, 24, 40 and D holds 32 + 80 = 112; before Round 3, A holds 8.
- If C gives in Round 3 and B in Round 2: going back gives start B = 51, C = 29 ✓ (B > C).
- If B gives in Round 3 and C in Round 2: start B = 27, C = 53 ✗ (B < C).
So the order is A, B, C, D.
| Stage | A | B | C | D |
|---|---|---|---|---|
| Start | 98 | 51 | 29 | 14 |
| After Round 1 (A gives) | 4 | 102 | 58 | 28 |
| After Round 2 (B gives) | 8 | 12 | 116 | 56 |
| After Round 3 (C gives) | 16 | 24 | 40 | 112 |
| After Round 4 (D gives) | 32 | 48 | 80 | 32 |
Question 2
Numerical answer
How many coins did B hold at the start of the game?
Show answer and solution
Answer: 51
Deduction: The total is always 32 + 48 + 80 + 32 = 192. In a round, every non-giver's holding doubles, and the giver pays out the sum of the other three.
A starts with 98. If A did not give in Round 1, A would hold 196 > 192 ✗. So A gives first and is left with 98 − 94 = 4; B, C and D double (D: 28).
D cannot give in Round 2 (D has 28 but would owe 4 + 2B + 2C = 164) nor in Round 3 (D would hold 56 and owe 136). So D gives in Round 4. Working back from the end: before Round 4, A, B, C hold 16, 24, 40 and D holds 32 + 80 = 112; before Round 3, A holds 8.
- If C gives in Round 3 and B in Round 2: going back gives start B = 51, C = 29 ✓ (B > C).
- If B gives in Round 3 and C in Round 2: start B = 27, C = 53 ✗ (B < C).
So the order is A, B, C, D.
| Stage | A | B | C | D |
|---|---|---|---|---|
| Start | 98 | 51 | 29 | 14 |
| After Round 1 (A gives) | 4 | 102 | 58 | 28 |
| After Round 2 (B gives) | 8 | 12 | 116 | 56 |
| After Round 3 (C gives) | 16 | 24 | 40 | 112 |
| After Round 4 (D gives) | 32 | 48 | 80 | 32 |
B started with 51 coins (51 + 29 = 192 − 98 − 14 = 80 ✓).
A starts with 98. If A did not give in Round 1, A would hold 196 > 192 ✗. So A gives first and is left with 98 − 94 = 4; B, C and D double (D: 28).
D cannot give in Round 2 (D has 28 but would owe 4 + 2B + 2C = 164) nor in Round 3 (D would hold 56 and owe 136). So D gives in Round 4. Working back from the end: before Round 4, A, B, C hold 16, 24, 40 and D holds 32 + 80 = 112; before Round 3, A holds 8.
- If C gives in Round 3 and B in Round 2: going back gives start B = 51, C = 29 ✓ (B > C).
- If B gives in Round 3 and C in Round 2: start B = 27, C = 53 ✗ (B < C).
So the order is A, B, C, D.
| Stage | A | B | C | D |
|---|---|---|---|---|
| Start | 98 | 51 | 29 | 14 |
| After Round 1 (A gives) | 4 | 102 | 58 | 28 |
| After Round 2 (B gives) | 8 | 12 | 116 | 56 |
| After Round 3 (C gives) | 16 | 24 | 40 | 112 |
| After Round 4 (D gives) | 32 | 48 | 80 | 32 |
B started with 51 coins (51 + 29 = 192 − 98 − 14 = 80 ✓).
Question 3
Numerical answer
How many coins in all did C hand over to the other players in the round in which C was the giver?
Show answer and solution
Answer: 76
Deduction: The total is always 32 + 48 + 80 + 32 = 192. In a round, every non-giver's holding doubles, and the giver pays out the sum of the other three.
A starts with 98. If A did not give in Round 1, A would hold 196 > 192 ✗. So A gives first and is left with 98 − 94 = 4; B, C and D double (D: 28).
D cannot give in Round 2 (D has 28 but would owe 4 + 2B + 2C = 164) nor in Round 3 (D would hold 56 and owe 136). So D gives in Round 4. Working back from the end: before Round 4, A, B, C hold 16, 24, 40 and D holds 32 + 80 = 112; before Round 3, A holds 8.
- If C gives in Round 3 and B in Round 2: going back gives start B = 51, C = 29 ✓ (B > C).
- If B gives in Round 3 and C in Round 2: start B = 27, C = 53 ✗ (B < C).
So the order is A, B, C, D.
| Stage | A | B | C | D |
|---|---|---|---|---|
| Start | 98 | 51 | 29 | 14 |
| After Round 1 (A gives) | 4 | 102 | 58 | 28 |
| After Round 2 (B gives) | 8 | 12 | 116 | 56 |
| After Round 3 (C gives) | 16 | 24 | 40 | 112 |
| After Round 4 (D gives) | 32 | 48 | 80 | 32 |
C gave in Round 3. At the start of Round 3, A, B and D held 8, 12 and 56 coins, so C handed over 8 + 12 + 56 = 76 coins (116 → 40).
A starts with 98. If A did not give in Round 1, A would hold 196 > 192 ✗. So A gives first and is left with 98 − 94 = 4; B, C and D double (D: 28).
D cannot give in Round 2 (D has 28 but would owe 4 + 2B + 2C = 164) nor in Round 3 (D would hold 56 and owe 136). So D gives in Round 4. Working back from the end: before Round 4, A, B, C hold 16, 24, 40 and D holds 32 + 80 = 112; before Round 3, A holds 8.
- If C gives in Round 3 and B in Round 2: going back gives start B = 51, C = 29 ✓ (B > C).
- If B gives in Round 3 and C in Round 2: start B = 27, C = 53 ✗ (B < C).
So the order is A, B, C, D.
| Stage | A | B | C | D |
|---|---|---|---|---|
| Start | 98 | 51 | 29 | 14 |
| After Round 1 (A gives) | 4 | 102 | 58 | 28 |
| After Round 2 (B gives) | 8 | 12 | 116 | 56 |
| After Round 3 (C gives) | 16 | 24 | 40 | 112 |
| After Round 4 (D gives) | 32 | 48 | 80 | 32 |
C gave in Round 3. At the start of Round 3, A, B and D held 8, 12 and 56 coins, so C handed over 8 + 12 + 56 = 76 coins (116 → 40).
Question 4
Single correct option
At the end of which round did A hold the fewest coins?
- A. Round 2
- B. Round 1
- C. Round 4
- D. Round 3
Show answer and solution
Answer: (B) Round 1
Deduction: The total is always 32 + 48 + 80 + 32 = 192. In a round, every non-giver's holding doubles, and the giver pays out the sum of the other three.
A starts with 98. If A did not give in Round 1, A would hold 196 > 192 ✗. So A gives first and is left with 98 − 94 = 4; B, C and D double (D: 28).
D cannot give in Round 2 (D has 28 but would owe 4 + 2B + 2C = 164) nor in Round 3 (D would hold 56 and owe 136). So D gives in Round 4. Working back from the end: before Round 4, A, B, C hold 16, 24, 40 and D holds 32 + 80 = 112; before Round 3, A holds 8.
- If C gives in Round 3 and B in Round 2: going back gives start B = 51, C = 29 ✓ (B > C).
- If B gives in Round 3 and C in Round 2: start B = 27, C = 53 ✗ (B < C).
So the order is A, B, C, D.
| Stage | A | B | C | D |
|---|---|---|---|---|
| Start | 98 | 51 | 29 | 14 |
| After Round 1 (A gives) | 4 | 102 | 58 | 28 |
| After Round 2 (B gives) | 8 | 12 | 116 | 56 |
| After Round 3 (C gives) | 16 | 24 | 40 | 112 |
| After Round 4 (D gives) | 32 | 48 | 80 | 32 |
A's holdings: 98 → 4 → 8 → 16 → 32. The minimum (4) occurs at the end of Round 1, when A was the giver; afterwards A's holding doubles every round.
A starts with 98. If A did not give in Round 1, A would hold 196 > 192 ✗. So A gives first and is left with 98 − 94 = 4; B, C and D double (D: 28).
D cannot give in Round 2 (D has 28 but would owe 4 + 2B + 2C = 164) nor in Round 3 (D would hold 56 and owe 136). So D gives in Round 4. Working back from the end: before Round 4, A, B, C hold 16, 24, 40 and D holds 32 + 80 = 112; before Round 3, A holds 8.
- If C gives in Round 3 and B in Round 2: going back gives start B = 51, C = 29 ✓ (B > C).
- If B gives in Round 3 and C in Round 2: start B = 27, C = 53 ✗ (B < C).
So the order is A, B, C, D.
| Stage | A | B | C | D |
|---|---|---|---|---|
| Start | 98 | 51 | 29 | 14 |
| After Round 1 (A gives) | 4 | 102 | 58 | 28 |
| After Round 2 (B gives) | 8 | 12 | 116 | 56 |
| After Round 3 (C gives) | 16 | 24 | 40 | 112 |
| After Round 4 (D gives) | 32 | 48 | 80 | 32 |
A's holdings: 98 → 4 → 8 → 16 → 32. The minimum (4) occurs at the end of Round 1, when A was the giver; afterwards A's holding doubles every round.
Saved on this device. Sign in or create a free account to keep it on every device.