A 3-digit number ranges from 100 to 999. We need to find how many of these numbers are divisible by 11. A number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is a multiple of 11. - IQnection
How Many 3-Digit Numbers Between 100 and 999 Are Divisible by 11?
How Many 3-Digit Numbers Between 100 and 999 Are Divisible by 11?
When exploring three-digit numbers—those ranging from 100 to 999—a common question arises: How many of these numbers are divisible by 11? Understanding this not only helps in number theory but also enhances practical skills in divisibility rules. A key insight lies in the divisibility rule of 11, which states that a number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is a multiple of 11.
In a 3-digit number, abc (where a, b, c are individual digits), position 1 (leftmost) is odd, position 2 is even, and position 3 is odd. Thus, the rule becomes:
Understanding the Context
(a + c) − b must be divisible by 11.
Since a ranges from 1 to 9, and b and c range from 0 to 9, the maximum value of (a + c) is 9 + 9 = 18, and the minimum is 0 + 0 = 0. So, (a + c − b) ranges from –9 (if a=1, c=0, b=9) to 18 (if a=9, c=9, b=0). The only multiples of 11 within this range are −11, 0, and 11. But since −11 < –9, we exclude it. Thus, valid differences are:
🔹 (a + c − b) = 0 or (a + c − b) = 11
We now count all 3-digit numbers (100 to 999) satisfying either condition.
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Key Insights
Case 1: (a + c − b) = 0
This means:
a + c = b
But since b is a single digit (0–9), a + c ≤ 9 (because b ≤ 9). Also, a ≥ 1, so a ranges from 1 to 9.
We iterate over all valid (a, c) pairs where a + c ≤ 9, and set b = a + c. Since b must be a digit (0–9), and a ≥ 1, a + c can range from 1 to 9.
For each valid a from 1 to 9:
- The smallest c is 0, largest is min(9 − a, 9)
But since a + c ≤ 9 → c ≤ 9 − a
And c must be ≥ 0
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So, number of valid c values = (9 − a + 1) = 10 − a
But only if 10 − a ≥ 1 → a ≤ 9 (always true).
Sum over a from 1 to 9:
Total = Σₐ₌₁⁹ (10 − a) = (9+8+...+1) = 45
Each such (a, c) gives exactly one b = a + c (since b must be digit, and a + c ≤ 9 → valid).
So, Case 1 yields 45 numbers.
Example: a=3, c=5 → b=8 → number 385 → (3+5)−8 = 8−8 = 0 → divisible by 11 ✅
Case 2: (a + c − b) = 11
This means:
a + c = b + 11 → b = a + c − 11
Here, b must be between 0 and 9 inclusive. So:
0 ≤ a + c − 11 ≤ 9 →
11 ≤ a + c ≤ 20
But max a + c = 9 + 9 = 18 → so a + c ranges from 11 to 18
For each sum s = a + c from 11 to 18, we compute number of (a, c) pairs where a ∈ [1,9], c ∈ [0,9], and a + c = s.
