The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1" and they all are multiple of 11, is ___________.
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For "sum of numbers" problems under multiple constraints, a systematic listing of all possible numbers is the most reliable method. Use the strongest constraint first (e.g., divisibility rule) to generate candidates, then filter them using the other constraints (allowed digits, range). To sum them, adding the place values (sum of all units digits, sum of all tens digits, etc.) is often less prone to calculation errors than adding the full numbers directly.
Step 1: Understanding the Conditions
We are looking for 3-digit numbers \(N = 100a + 10b + c\) that satisfy the following conditions:
\(100 \le N \le 500\).
The digits \(a, b, c\) must be from the set D = \{0, 2, 3, 4, 5, 6, 7, 8, 9\}. The digit '1' is not allowed.
\(N\) is a multiple of 11.
From conditions 1 and 2, the first digit \(a\) can be 2, 3, or 4. The number 500 itself is not a multiple of 11, so we do not need to consider \(a=5\). Step 2: Applying the Divisibility Rule of 11
For a 3-digit number \(abc\), the divisibility rule for 11 states that the alternating sum of its digits, \(a - b + c\), must be a multiple of 11 (i.e., 0, \(\pm\)11, \(\pm\)22, ...).
Let's find the possible range for \(a-b+c\):
Maximum value: With \(a=4\), \(c=9\), \(b=0\), we get \(4 - 0 + 9 = 13\).
Minimum value: With \(a=2\), \(c=0\), \(b=9\), we get \(2 - 9 + 0 = -7\).
Thus, the only possible values for \(a-b+c\) are 0 and 11. Step 3: Finding all Possible Numbers
We enumerate the numbers based on the two cases for the divisibility rule. Case 1: \(a - b + c = 0 \implies b = a + c\)
For \(a=2\): \(b=2+c\). The pairs (c,b) must not contain '1'. Possible pairs are (0,2), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9). This gives the numbers: 220, 242, 253, 264, 275, 286, 297.
For \(a=3\): \(b=3+c\). Possible pairs (c,b) are (0,3), (2,5), (3,6), (4,7), (5,8), (6,9). This gives the numbers: 330, 352, 363, 374, 385, 396.
For \(a=4\): \(b=4+c\). Possible pairs (c,b) are (0,4), (2,6), (3,7), (4,8), (5,9). This gives the numbers: 440, 462, 473, 484, 495. Case 2: \(a - b + c = 11 \implies b = a + c - 11\)
For \(a=2\): \(b=c-9\). Since \(b \ge 0\), \(c\) must be 9. This gives \(b=0\). Number: 209.
For \(a=3\): \(b=c-8\). \(c\) can be 8 or 9. If \(c=8\), \(b=0\). Number: 308. If \(c=9\), \(b=1\), which is not allowed.
For \(a=4\): \(b=c-7\). \(c\) can be 7, 8, or 9. If \(c=7\), \(b=0\). Number: 407. If \(c=8\), \(b=1\), not allowed. If \(c=9\), \(b=2\). Number: 429. Step 4: Calculating the Sum
We have found all the required numbers. We will sum them directly.
Numbers starting with 2: \(220+242+253+264+275+286+297+209 = 2046\).
Numbers starting with 3: \(330+352+363+374+385+396+308 = 2508\).
Numbers starting with 4: \(440+462+473+484+495+407+429 = 3190\).
Total Sum = \(2046 + 2508 + 3190 = 7744\).
Alternatively, using the place value method for a more structured calculation:
List of numbers: 209, 220, 242, 253, 264, 275, 286, 297, 308, 330, 352, 363, 374, 385, 396, 407, 429, 440, 462, 473, 484, 495.
Sum of units digits (c): (9+0+2+3+4+5+6+7) + (8+0+2+3+4+5+6) + (7+9+0+2+3+4+5) = 36 + 28 + 30 = 94. Sum of tens digits (b): (0+2+4+5+6+7+8+9) + (0+3+5+6+7+8+9) + (0+2+4+6+7+8+9) = 41 + 38 + 36 = 115. Sum of hundreds digits (a): There are 8 numbers starting with 2, 7 numbers starting with 3, and 7 numbers starting with 4. Sum = \(8 \times 2 + 7 \times 3 + 7 \times 4 = 16 + 21 + 28 = 65\).
Total Sum = (Sum of hundreds digits) \(\times\) 100 + (Sum of tens digits) \(\times\) 10 + (Sum of units digits)
Total Sum = \(65 \times 100 + 115 \times 10 + 94\)
Total Sum = \(6500 + 1150 + 94 = 7744\). Final Answer: The sum is 7744.
