What is the sum of all two-digit numbers which when divided by 3 leave 2 as the remainder?
1635
Finding the Sum of Two-Digit Numbers with Remainder 2 When Divided by 3
The question asks for the sum of all two-digit numbers that leave a remainder of 2 when divided by 3. Let's break this down to find these specific two-digit numbers and then calculate their sum.
A two-digit number \(N\) leaves a remainder of 2 when divided by 3 if it can be expressed in the form \(N = 3k + 2\), where \(k\) is an integer. We are looking for two-digit numbers, which range from 10 to 99.
Identifying the Two-Digit Numbers
We need to find the values of \(k\) such that \(10 \le 3k + 2 \le 99\).
First, let's find the smallest two-digit number:
- \(3k + 2 \ge 10\)
- \(3k \ge 10 - 2\)
- \(3k \ge 8\)
- \(k \ge \frac{8}{3}\)
- Since \(k\) must be an integer, the smallest possible value for \(k\) is 3.
- The smallest two-digit number is \(3(3) + 2 = 9 + 2 = 11\).
Next, let's find the largest two-digit number:
- \(3k + 2 \le 99\)
- \(3k \le 99 - 2\)
- \(3k \le 97\)
- \(k \le \frac{97}{3}\)
- Since \(k\) must be an integer, the largest possible value for \(k\) is 32 (since \(97 \div 3 = 32.33\ldots\)).
- The largest two-digit number is \(3(32) + 2 = 96 + 2 = 98\).
So, the two-digit numbers that leave a remainder of 2 when divided by 3 start from 11 and end at 98. The sequence of these numbers is 11, 14, 17, 20, ..., 98.
Recognizing the Pattern: Arithmetic Progression
The sequence 11, 14, 17, ..., 98 is an arithmetic progression (AP) because the difference between consecutive terms is constant (14 - 11 = 3, 17 - 14 = 3, and so on). This constant difference is the common difference, which is 3.
In this AP:
- The first term (\(a_1\)) is 11.
- The last term (\(a_n\)) is 98.
- The common difference (\(d\)) is 3.
Finding the Number of Terms
To find the sum of an AP, we first need to know the number of terms (\(n\)). We can use the formula for the n-th term of an AP: \(a_n = a_1 + (n-1)d\).
- Substitute the known values: \(98 = 11 + (n-1)3\).
- Subtract 11 from both sides: \(98 - 11 = (n-1)3\).
- \(87 = (n-1)3\).
- Divide both sides by 3: \(\frac{87}{3} = n-1\).
- \(29 = n-1\).
- Add 1 to both sides: \(n = 29 + 1\).
- \(n = 30\).
There are 30 two-digit numbers that leave a remainder of 2 when divided by 3.
Calculating the Sum of the Arithmetic Progression
Now we can calculate the sum (\(S_n\)) of these 30 numbers using the formula for the sum of an AP:
\(S_n = \frac{n}{2} (a_1 + a_n)\)
- Substitute the values: \(S_{30} = \frac{30}{2} (11 + 98)\).
- \(S_{30} = 15 (109)\).
- \(S_{30} = 15 \times 109\).
Let's perform the multiplication:
\(15 \times 109 = 15 \times (100 + 9) = 15 \times 100 + 15 \times 9 = 1500 + 135 = 1635\).
Alternatively, using long multiplication:
| 1 | 0 | 9 | ||
|---|---|---|---|---|
| \(\times\) | 1 | 5 | ||
| 4 | 5 | |||
| 5 | 0 | 0 | ||
| 1 | 6 | 3 | 5 | |
The sum of all two-digit numbers which when divided by 3 leave 2 as the remainder is 1635.
Summary of Steps
- Identify the range of two-digit numbers (10 to 99).
- Determine the form of numbers leaving remainder 2 when divided by 3 (\(3k+2\)).
- Find the smallest and largest two-digit numbers of this form (11 and 98).
- Recognize the sequence as an arithmetic progression.
- Calculate the number of terms in the sequence (30).
- Use the sum formula for an arithmetic progression to find the total sum (1635).
Conclusion
The sum of all two-digit numbers which when divided by 3 leave 2 as the remainder is 1635.
Revision Table: Arithmetic Progression Concepts
| Concept | Description | Formula |
|---|---|---|
| Arithmetic Progression (AP) | A sequence where the difference between consecutive terms is constant. | \(a_1, a_1+d, a_1+2d, \ldots\) |
| Common Difference | The constant difference between consecutive terms. | \(d = a_{k+1} - a_k\) |
| n-th term of AP | The value of the term at position \(n\). | \(a_n = a_1 + (n-1)d\) |
| Sum of n terms of AP | The sum of the first \(n\) terms. | \(S_n = \frac{n}{2}(a_1 + a_n)\) or \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\) |
Additional Information: Numbers and Remainders
When an integer \(N\) is divided by a positive integer \(m\), the result can be expressed in the form \(N = qm + r\), where \(q\) is the quotient and \(r\) is the remainder. The remainder \(r\) must satisfy \(0 \le r < m\).
In this problem, we were concerned with numbers divided by 3, leaving a remainder of 2. This means the numbers are of the form \(3k + 2\), where \(k\) is an integer. Examples of such numbers include:
- If \(k=0\), \(3(0) + 2 = 2\) (not a two-digit number)
- If \(k=1\), \(3(1) + 2 = 5\) (not a two-digit number)
- If \(k=2\), \(3(2) + 2 = 8\) (not a two-digit number)
- If \(k=3\), \(3(3) + 2 = 11\) (the first two-digit number)
- ...
- If \(k=32\), \(3(32) + 2 = 98\) (the last two-digit number)
- If \(k=33\), \(3(33) + 2 = 99 + 2 = 101\) (not a two-digit number)
Understanding the relationship between numbers, divisors, quotients, and remainders is fundamental in number theory problems like this one.
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