Maths olympiad question · P5–P6★★★★ Expert
Sum and multiple: olympiad question 8
An expert question on sum and multiple for P5–P6, with a step-by-step solution, diagrams and the answer. Sum-and-multiple and difference-and-multiple questions appear in the Hua Xia Cup's P3 to P5 first rounds, qualifying rounds and finals.
The question
Three whole numbers on the board add up to 2113.
Rub out the last two digits of the largest, and what is left is exactly the middle one; rub out the last digit of the middle one, and what is left is exactly the smallest.
(Digits rubbed out may be 0.)
What is the largest number?
Show the hints (3)
- Hint 1: Write the other two in terms of the smallest: for example 5307, 53, 5.
- Hint 2: Sum = smallest × 1011 + the middle one's last digit × 101 + the largest's last two digits. How big can the last two parts be?
- Hint 3: The last two parts are at most 101 × 9 + 99 = 1008, less than 1011; the largest's tail is at most 99, less than 101. Work out 2113 ÷ 1011, then divide the remainder by 101.
Worked solution
Have a go first, then read the solution step by step.
- 1.Let the middle one's last digit be b and the largest's last two digits t. Middle = smallest × 10 + b; largest = middle × 100 + t = smallest × 1000 + b × 100 + t.
- 2.Sum = smallest × (1000 + 10 + 1) + b × (100 + 1) + t = smallest × 1011 + b × 101 + t = 2113.
- 3.b is at most 9 and t at most 99, so b × 101 + t is at most 1008, less than 1011: it is the remainder of 2113 ÷ 1011 = 2 remainder 91. The smallest number is 2.
- 4.t is at most 99, less than 101, so 91 ÷ 101 = 0 remainder 91: b = 0, t = 91.
- 5.Middle = 2 × 10 + 0 = 20; largest = 20 × 100 + 91 = 2091.
- 6.Check: 2091 + 20 + 2 = 2113 ✓
- ⚠️Common mistake: ignoring the rubbed-out digits and taking the largest as exactly 1000 × the smallest; or, with the quotient of 2113 ÷ 1011 found, forgetting the remainder 91 still splits into b × 101 + t.
Answer: 2091
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