How many five-digit numbers of the form XXYXX is/are divisible by 33?
3
The question asks us to find the number of five-digit numbers that have the specific form XXYXX and are divisible by 33.
A five-digit number of the form XXYXX can be written mathematically as:
\(\text{XXYXX} = X \times 10000 + X \times 1000 + Y \times 100 + X \times 10 + X \times 1\)
Simplifying this expression, we get:
\(\text{XXYXX} = 10000X + 1000X + 100Y + 10X + X = (10000+1000+10+1)X + 100Y = 11011X + 100Y\)
Here, X is the first digit, so X must be a digit from 1 to 9 (to make it a five-digit number). Y can be any digit from 0 to 9.
Understanding Divisibility by 33
For a number to be divisible by 33, it must be divisible by its prime factors, which are 3 and 11. So, we need to check the divisibility rules for both 3 and 11 for the number XXYXX.
Applying Divisibility Rule for 11 to XXYXX
The rule for divisibility by 11 states that the alternating sum of the digits of the number, starting from the rightmost digit and moving left, must be divisible by 11.
For the number XXYXX, the digits are X, X, Y, X, X (from left to right).
The alternating sum is:
\(+X - X + Y - X + X = Y\)
For the number to be divisible by 11, this alternating sum, Y, must be divisible by 11.
Since Y is a single digit from 0 to 9, the only value for Y that is divisible by 11 is 0.
Therefore, Y must be 0.
Applying Divisibility Rule for 3 to XXYXX
The rule for divisibility by 3 states that the sum of the digits of the number must be divisible by 3.
For the number XXYXX, the digits are X, X, Y, X, X.
The sum of the digits is:
\(X + X + Y + X + X = 4X + Y\)
For the number to be divisible by 3, this sum, \(4X + Y\), must be divisible by 3.
We already found that Y must be 0. Substituting Y = 0 into the sum of digits:
\(4X + 0 = 4X\)
So, \(4X\) must be divisible by 3.
Since 4 and 3 are relatively prime (they have no common factors other than 1), for \(4X\) to be divisible by 3, X must be divisible by 3.
Finding Possible Values for X
We know that X must be a digit from 1 to 9 (because XXYXX is a five-digit number, and the first digit X cannot be 0). We also know that X must be divisible by 3.
The digits from 1 to 9 that are divisible by 3 are 3, 6, and 9.
So, the possible values for X are 3, 6, and 9.
Determining the Numbers
We found that Y must be 0 and X can be 3, 6, or 9.
Let's list the possible five-digit numbers of the form XXYXX:
- If X = 3 and Y = 0, the number is 33033.
- If X = 6 and Y = 0, the number is 66066.
- If X = 9 and Y = 0, the number is 99099.
Let's quickly verify these numbers are indeed divisible by 33. We can check if they are divisible by 3 and 11.
For 33033:
- Sum of digits: \(3+3+0+3+3 = 12\) (Divisible by 3)
- Alternating sum of digits: \(3-3+0-3+3 = 0\) (Divisible by 11)
For 66066:
- Sum of digits: \(6+6+0+6+6 = 24\) (Divisible by 3)
- Alternating sum of digits: \(6-6+0-6+6 = 0\) (Divisible by 11)
For 99099:
- Sum of digits: \(9+9+0+9+9 = 36\) (Divisible by 3)
- Alternating sum of digits: \(9-9+0-9+9 = 0\) (Divisible by 11)
All three numbers satisfy the conditions.
Therefore, there are exactly 3 five-digit numbers of the form XXYXX that are divisible by 33.
Revision Table: Analyzing XXYXX Divisibility by 33
| Condition | Requirement | Applicability to XXYXX | Result for Digits |
|---|---|---|---|
| Five-digit number | First digit ≠ 0 | X ≠ 0 | X ∈ \(\{1, 2, ..., 9\}\) |
| Divisible by 33 | Divisible by 3 AND Divisible by 11 | Check both rules | Both conditions must be met |
| Divisibility by 11 | Alternating sum of digits is divisible by 11 | \(X - X + Y - X + X = Y\) | Y must be divisible by 11, so Y = 0 |
| Divisibility by 3 | Sum of digits is divisible by 3 | \(X+X+Y+X+X = 4X+Y\) | \(4X+Y\) must be divisible by 3 |
| Combining Y=0 with Divisibility by 3 | \(4X+0\) must be divisible by 3 | \(4X\) must be divisible by 3 | X must be divisible by 3 |
| Possible X values (X ≠ 0) | X ∈ \(\{1, ..., 9\}\) and X is divisible by 3 | X can be 3, 6, or 9 | Possible numbers are 33033, 66066, 99099 |
Additional Information on Number Divisibility Rules
Understanding divisibility rules can greatly simplify problems involving factors of numbers. Here's a quick recap of the rules used:
- Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. For example, 123 is divisible by 3 because \(1+2+3=6\), and 6 is divisible by 3.
- Divisibility by 11: A number is divisible by 11 if the alternating sum of its digits (starting from the rightmost digit with a positive sign) is divisible by 11. For example, for 132, the alternating sum is \(+2 - 3 + 1 = 0\), which is divisible by 11, so 132 is divisible by 11. For 121, the alternating sum is \(+1 - 2 + 1 = 0\), which is divisible by 11, so 121 is divisible by 11.
When a number needs to be divisible by a composite number like 33, which is the product of coprime numbers (3 and 11), the number must be divisible by each of its coprime factors individually.
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