The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1

Q: Wait, why do we just ignore the range [100, 199]? Isn't that part of the 3-digit numbers?

Great observation! While [100, 199] are indeed 3-digit numbers, the problem explicitly states that the digit '1' is prohibited. Since every single number from 100 to 199 starts with the digit '1', they are all automatically disqualified. We save time by focusing only on the valid territory: 200 to 500.

Q: How do I know if I missed any numbers containing the digit '1'?

The best way to be sure is to be systematic. We first found all multiples of 11 in the range [200, 500]. Then, we checked each one for the digit '1'. Since we are only dealing with a small set of 27 numbers, checking them individually is the most robust way to ensure no 'imposter' slips through.

Q: Is there a faster way to find the multiples of 11 without listing them?

You can use modular arithmetic! A number $n$ is a multiple of 11 if $n \equiv 0 \pmod{11}$. By finding the smallest multiple of 11 greater than or equal to 200 (which is $11 \times 19 = 209$) and the largest less than or equal to 500 (which is $11 \times 45 = 495$), you define your boundaries perfectly.

Q: Why did we subtract the sum of the excluded numbers instead of just adding the valid ones?

It is a matter of efficiency. Calculating the sum of an Arithmetic Progression is very fast using the formula $S_n = \frac{n}{2}(a + l)$. It is much easier to calculate the total sum of all multiples and then subtract the few that violate the 'no-1' rule than it is to manually identify and sum every single valid number one by one.

Q: What if I accidentally included a number like 121? Would that be counted?

Actually, 121 is in the range [100, 199], which we already excluded in our first step! By defining our search range as [200, 500] from the start, we effectively filtered out all numbers starting with '1'. We only had to worry about '1' appearing in the tens or units place thereafter.

Question

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 .

Sigma Percentile · Accepted Answer

### Analyzing the Battlefield

Our range is $100 \le n \le 500$. However, the constraint is that the digit '1' cannot appear in the number.

Consider the range $[100, 199]$. Every single number in this interval starts with the digit '1'. Consequently, all these numbers are disqualified immediately.

This narrows our search space significantly to the interval $[200, 500]$. By observing this structure, we eliminate a large portion of the search space before performing any complex calculations.

### The Arithmetic Progression

We now identify multiples of $11$ within the valid range $[200, 500]$. The first multiple of $11$ after $200$ is $209$ (since $11 	imes 19 = 209$), and the last multiple before $500$ is $495$ (since $11 	imes 45 = 495$).

This forms an Arithmetic Progression where the first term $a = 209$, the last term $l = 495$, and the common difference $d = 11$.

To find the number of terms $n$, we use the formula:
$$l = a + (n-1)d$$

Substituting our values:
$$495 = 209 + (n-1)11$$
$$286 = (n-1)11$$
$$n-1 = 26 \Rightarrow n = 27$$

There are exactly $27$ multiples of $11$ in our range.

### The Total Sum

With $n=27$, $a=209$, and $l=495$, we calculate the total sum $S_{27}$ using the formula:
$$S_n = \frac{n}{2}(a + l)$$

Substituting the values:
$$S_{27} = \frac{27}{2}(209 + 495) = \frac{27}{2}(704)$$
$$S_{27} = 27 	imes 352 = 9504$$

This value represents the sum of all multiples of $11$ between $200$ and $500$, regardless of whether they contain the digit '1'.

### Identifying the Imposters

We must now remove the 'imposters'—the numbers that contain the digit '1'. Since we already excluded numbers starting with '1', we only need to check for the digit '1' in the tens or units place.

Checking our multiples, we find:
- Multiples ending in '1': $11 	imes 21 = 231$, $11 	imes 31 = 341$, and $11 	imes 41 = 451$.
- Multiples with '1' in the tens place: $11 	imes 29 = 319$ and $11 	imes 38 = 418$.

These five numbers ($231, 341, 451, 319, 418$) are the only ones that violate our rule. Their sum is:
$$S_{ex} = 231 + 341 + 451 + 319 + 418 = 1760$$

### The Final Victory

The final step is to subtract the sum of the forbidden numbers from the total sum:
$$9504 - 1760 = 7744$$

The sum of all 3-digit multiples of $11$ that do not contain the digit '1' is **7744**.

Animated Solution for Mathematics - Sequence and Series: 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 .

Enter Numerical Value:

Visualized Solution

Defining the Search Range

Multiples of $11$ in $[200, 500]$

Number of Terms ( $n$ )

Sum of the AP Series

Exclusions: '1' at Units Place

Exclusions: '1' at Tens Place

Total Excluded Sum

Final Required Sum

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Battlefield

The Arithmetic Progression

The Total Sum

Identifying the Imposters

The Final Victory

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