What is the largest 5-digit number, which leaves the remainder 7, when divided by 18 as well as by 11 ?
99997
Finding the Largest 5-Digit Number with Specific Remainders
The question asks for the largest 5-digit number that gives a remainder of 7 when divided by 18 and also when divided by 11.
Let the required number be \( N \). According to the problem, when \( N \) is divided by 18, the remainder is 7. This can be written as:
\( N \equiv 7 \pmod{18} \)
Also, when \( N \) is divided by 11, the remainder is 7. This can be written as:
\( N \equiv 7 \pmod{11} \)
A number that leaves the same remainder \( r \) when divided by two different numbers, say \( a \) and \( b \), can be expressed in the form \( N = k \times \text{LCM}(a, b) + r \), where \( k \) is an integer and \( \text{LCM}(a, b) \) is the Least Common Multiple of \( a \) and \( b \).
In this problem, the divisors are 18 and 11, and the remainder is 7. So, the required number \( N \) must be of the form:
\( N = k \times \text{LCM}(18, 11) + 7 \)
Calculating the Least Common Multiple (LCM) of 18 and 11
First, we need to find the LCM of 18 and 11.
- Factorize 18 into its prime factors: \( 18 = 2 \times 3 \times 3 = 2^{1} \times 3^{2} \)
- Factorize 11 into its prime factors: 11 is a prime number. \( 11 = 11^{1} \)
To find the LCM, we take the highest power of all prime factors involved in the factorization of the numbers.
- Prime factor 2: highest power is \( 2^{1} \)
- Prime factor 3: highest power is \( 3^{2} \)
- Prime factor 11: highest power is \( 11^{1} \)
\( \text{LCM}(18, 11) = 2^{1} \times 3^{2} \times 11^{1} = 2 \times 9 \times 11 = 18 \times 11 = 198 \)
So, the numbers that leave a remainder of 7 when divided by both 18 and 11 are of the form:
\( N = 198k + 7 \)
where \( k \) is a non-negative integer.
Finding the Largest 5-Digit Number of This Form
We are looking for the largest such number that is a 5-digit number. The largest 5-digit number is 99999.
We need to find the largest value of \( k \) such that \( 198k + 7 \le 99999 \).
\( 198k \le 99999 - 7 \)
\( 198k \le 99992 \)
To find the largest integer value of \( k \), we divide 99992 by 198:
\( k \le \frac{99992}{198} \)
\( k \le 505.0101... \)
Since \( k \) must be an integer, the largest possible integer value for \( k \) is 505.
Calculating the Required Number
Now we substitute \( k = 505 \) back into the formula \( N = 198k + 7 \):
\( N = 198 \times 505 + 7 \)
\( N = 99990 + 7 \)
\( N = 99997 \)
This number, 99997, is a 5-digit number. Since we used the largest possible integer value for \( k \) such that \( 198k + 7 \le 99999 \), 99997 is the largest 5-digit number that satisfies the condition.
Verification
Let's quickly verify if 99997 leaves a remainder of 7 when divided by 18 and 11.
Dividing by 18: \( 99997 \div 18 \)
\( 99997 = 18 \times 5555 + 7 \). The remainder is 7.
Dividing by 11: \( 99997 \div 11 \)
\( 99997 = 11 \times 9090 + 7 \). The remainder is 7.
The number 99997 satisfies both conditions.
Comparing with Options
Let's look at the given options:
- 99981
- 99988
- 99997
- 99999
Our calculated number, 99997, matches option 3.
| Number | Remainder when divided by 18 | Remainder when divided by 11 | Satisfies Conditions? |
|---|---|---|---|
| 99981 | \( 99981 = 18 \times 5554 + 9 \) (Remainder is 9) | \( 99981 = 11 \times 9089 + 2 \) (Remainder is 2) | No |
| 99988 | \( 99988 = 18 \times 5554 + 16 \) (Remainder is 16) | \( 99988 = 11 \times 9089 + 9 \) (Remainder is 9) | No |
| 99997 | \( 99997 = 18 \times 5555 + 7 \) (Remainder is 7) | \( 99997 = 11 \times 9090 + 7 \) (Remainder is 7) | Yes |
| 99999 | \( 99999 = 18 \times 5555 + 9 \) (Remainder is 9) | \( 99999 = 11 \times 9090 + 9 \) (Remainder is 9) | No |
The table confirms that 99997 is the correct number.
Revision Table: Key Concepts for Remainder Problems
| Concept | Explanation | Relevance to Problem |
|---|---|---|
| Division Algorithm | For integers \( a \) and \( b \) with \( b > 0 \), there exist unique integers \( q \) and \( r \) such that \( a = bq + r \) and \( 0 \le r < b \). Here, \( a \) is the dividend, \( b \) is the divisor, \( q \) is the quotient, and \( r \) is the remainder. | Used to express the problem condition \( N \div 18 \) gives remainder 7 and \( N \div 11 \) gives remainder 7. |
| Congruence Modulo n | \( a \equiv b \pmod{n} \) means that \( a - b \) is divisible by \( n \). Equivalently, \( a \) and \( b \) have the same remainder when divided by \( n \). | The condition "leaves remainder 7 when divided by 18" is \( N \equiv 7 \pmod{18} \). The condition "leaves remainder 7 when divided by 11" is \( N \equiv 7 \pmod{11} \). |
| Numbers with same remainder by multiple divisors | If a number \( N \) leaves the same remainder \( r \) when divided by \( a \) and \( b \), then \( N - r \) is divisible by both \( a \) and \( b \). Therefore, \( N - r \) must be a multiple of \( \text{LCM}(a, b) \). So, \( N - r = k \times \text{LCM}(a, b) \), which means \( N = k \times \text{LCM}(a, b) + r \). | This is the core principle used to formulate the expression \( N = 198k + 7 \). |
| Largest N-digit number | The largest N-digit number is \( 10^N - 1 \). For 5 digits, it is \( 10^5 - 1 = 100000 - 1 = 99999 \). | This sets the upper limit for finding the required number. |
Additional Information: Exploring Related Number Properties
This problem involves finding a number based on its remainders when divided by different numbers. This is a fundamental concept in number theory, related to the Chinese Remainder Theorem (CRT).
- Chinese Remainder Theorem (CRT): The CRT deals with finding a number that satisfies several congruence relations with different moduli (divisors). A simple case is when the moduli are pairwise coprime, like 18 and 11 (since their greatest common divisor is 1). In such cases, a unique solution exists within a certain range (modulo the product of the moduli). Our problem is a direct application of finding numbers congruent to the same remainder modulo coprime numbers.
- Divisibility Rules: Knowing divisibility rules can sometimes help verify results or quickly check options.
- Divisibility by 11: The alternating sum of the digits is divisible by 11. For 99997: \( 7 - 9 + 9 - 9 + 9 = 7 \). \( 7 \div 11 \) gives remainder 7. So \( 99997 \equiv 7 \pmod{11} \).
- Divisibility by 18: A number is divisible by 18 if it is divisible by both 2 and 9. To leave a remainder of 7 when divided by 18, a number must be odd and leave a remainder of 7 when divided by 9. For 99997: It's odd. Sum of digits: \( 9+9+9+9+7 = 43 \). \( 43 \div 9 \) gives remainder 7 (\( 43 = 4 \times 9 + 7 \)). So \( 99997 \equiv 7 \pmod{9} \). Since 99997 is odd, \( 99997 \equiv 1 \pmod{2} \). Numbers of the form \( 18k+7 \) will always be odd (\( 18k \) is even, \( 18k+7 \) is odd) and leave a remainder of 7 mod 9 (\( 18k+7 = 2 \times 9k + 7 \equiv 7 \pmod{9} \)). Both checks align for 99997.
- Finding Smallest/Largest Numbers: The method used here can be adapted to find the smallest positive number, the largest/smallest number within a specific range (like 5-digit numbers), or the \( n \)-th such number by varying the value of \( k \). For the smallest positive number in this case, \( k \) would be 0, giving \( N = 198 \times 0 + 7 = 7 \). For the largest 5-digit number, we found the maximum \( k \) value.
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