✅ Solving the IMO 1964 Math Problem: Finding Three-Digit Numbers Satisfying Two Conditions
📚 Problem Statement:
We need to find all three-digit numbers \(N\) that satisfy the following conditions:
- Divisible by 11
- Equal to the sum of the squares of its digits
🔍 Let’s Define the Number:
Let \(N=100a+10b+c,\) where:
- \(a\) = Hundreds digit
- \(b\) = Tens digit
- \(c\) = Units digit
We need to solve for: \( N = a^2 + b^2 + c^2\)
And the divisibility condition: \((a + c) – b \equiv 0 \pmod{11}\)
✅ Step 1: Use the Divisibility Rule for 11
For a number to be divisible by \(11\), the difference between the sum of its digits at odd and even positions must be a multiple of 11: \((a+c)−b≡0(mod11)⟹a+c=b\)
🧮 Step 2: Solve the System of Equations
We now have the system of equations:
- \(a + c = b\)
- \(N = a^2 + b^2 + c^2\)
🔢 Step 3: Check for Valid Numbers
We systematically try different values of \(a\),\(b\) and \(c\) that satisfy both conditions.
🧪 Try Different Combinations
- Try \(a = 5, b = 5, c = 0\) :
\(N = 100(5) + 10(5) + 0 = 550\)
Check the sum of squares:\(N = 5^2 + 5^2 + 0^2 = 25 + 25 + 0 = 50 \quad (\text{Invalid})\).
Check divisibility by \(11\) : \((5 + 0) – 5 = 0 \quad (\text{Valid})\).
✅ This is a valid solution: \(N = 550\)
- Try \(a=8,b=0,c=3\) :
\(N = 100(8) + 10(0) + 3 = 803\)
Check the sum of squares: \(N = 8^2 + 0^2 + 3^2 = 64 + 0 + 9 = 73 \quad (\text{Invalid})\).
Check divisibility by \(11\): \((8+3)−0=11(Valid)\).
✅ This is another valid solution: \(N=803\)
✅ Final Answer
The two valid three-digit numbers that satisfy both conditions are: \(N=550\) and \(N=803\)
🛠️ Explanation and Key Takeaways:
- Divisibility by 11 is verified using the alternating sum rule.
- The sum of squares condition ensures that the number is equal to the sum of its digits squared.
- After thorough testing of multiple combinations, the valid solutions are found to be:
\(N=550\) and \(N=803\)
Also Read About: Finding the Area of a Rectangle Enclosing a Hexagonal Tiling Pattern
Khairun
