Number N Conditions
The problem asks us to find the largest 5-digit number, let's call it N, that satisfies certain conditions related to division and remainders.
Specifically, when N is divided by 6, 12, 15, and 24, the remainders are 2, 8, 11, and 20, respectively. We need to find the greatest possible value for N.
Remainder Pattern Analysis
Let's write down the given conditions using modular arithmetic:
- \(N \equiv 2 \pmod{6}\)
- \(N \equiv 8 \pmod{12}\)
- \(N \equiv 11 \pmod{15}\)
- \(N \equiv 20 \pmod{24}\)
We can observe a pattern here. Let's look at the difference between the divisor and the remainder for each case:
- \(6 - 2 = 4\)
- \(12 - 8 = 4\)
- \(15 - 11 = 4\)
- \(24 - 20 = 4\)
Since the difference is the same (4) in all cases, it means that if we add 4 to the number N, the result (\(N+4\)) will be perfectly divisible by 6, 12, 15, and 24.
In other words:
- \(N + 4 \equiv 0 \pmod{6}\)
- \(N + 4 \equiv 0 \pmod{12}\)
- \(N + 4 \equiv 0 \pmod{15}\)
- \(N + 4 \equiv 0 \pmod{24}\)
This means \(N+4\) must be a common multiple of 6, 12, 15, and 24.
LCM Calculation
To find the number N, we first need to find the Least Common Multiple (LCM) of the divisors: 6, 12, 15, and 24.
Let's find the prime factorization of each number:
- \(6 = 2 \times 3\)
- \(12 = 2^2 \times 3\)
- \(15 = 3 \times 5\)
- \(24 = 2^3 \times 3\)
The LCM is found by taking the highest power of each prime factor present in the numbers:
LCM\((6, 12, 15, 24) = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120\).
So, \(N+4\) must be a multiple of 120.
Expressing N via LCM
We can express the relationship as:
\(N + 4 = k \times \text{LCM}(6, 12, 15, 24)\)
Where k is a positive integer.
\(N + 4 = 120k\)
Therefore, the number N can be represented as:
\(N = 120k - 4\)
Greatest 5-Digit N Determination
We are looking for the greatest 5-digit number N. The largest 5-digit number is 99,999.
We need to find the largest integer k such that N is less than or equal to 99,999.
\(N \le 99999\)
\(120k - 4 \le 99999\)
\(120k \le 99999 + 4\)
\(120k \le 100003\)
Now, we solve for k:
\(k \le \frac{100003}{120}\)
\(k \le 833.358...\)
Since k must be an integer, the largest possible integer value for k is 833.
Final N Calculation
Now, we substitute the value \(k=833\) back into the equation for N:
\(N = 120k - 4\)
\(N = 120 \times 833 - 4\)
\(N = 99960 - 4\)
\(N = 99956\)
This value, 99956, is the greatest 5-digit number that satisfies all the given remainder conditions.
Similar Questions
- If HCF of 768 and \(x^6y^2\) is \(32xy\) for natural numbers \(x \ge 2, y \ge 2\), then what is the value of \((x + y)\)?
- What is the HCF of \(2^{36}-1\) and \(2^{45}-1\)?
The HCF of \(x\) and \(y\) is \(H\). Consider the following statements in respect of the HCF of \(p=\frac{x^3 + y^3}{x^2-xy+y^2}\) and \(q=\frac{x^3-y^3}{x^2+xy+y^2}\) :
I. The HCF of \(p\) and \(q\) can be \(H\).
II. The HCF of \(p\) and \(q\) can be \(2H\).
Which of the statements given above is/are correct?
- What is the HCF of \(x^3 + y^3 + 3xy-1\) and \((x + y)^2 - 1\)?
- A number N is such that when divided by 4, 6, 7 or 9, it leaves 3 as remainder. What is the smallest 4-digit number that satisfies this property?
- A plank of wood 4·25 m long and 3·4 m wide is to be cut into square pieces of equal size. How many square pieces of largest size can be cut from the plank, if no wastage is allowed ?
- What is the HCF of \(x^4 - 13x^2y^2 - 300y^4\), \(x^3 - 4x^2y - 4xy^2 - 5y^3\) and \(x^3 - 125y^3\)?
Let \(N\) be the least positive multiple of 11 that leaves a remainder of 5 when divided by 6, 12, 15, 18. Which one of he following is correct?
Let \(n\) be a natural number. The HCF of \(n, n + 10\) is 10. If the LCM is \(x\) (a 2-digit number), then how many values of \(x\) are possible?
Consider the following statements in respect of prime numbers p and q:
- I. Their LCM is always an odd number.
- II. Sum of their LCM and HCF is always an even number.
Which of the statements given above is/are correct?
Important Questions from LCM and HCF
The HCF and LCM of two numbers are 12 and 72, respectively. If the ratio of the two numbers is 2 ∶ 3, then the larger of the two numbers is:
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Joseph visits the club on every 5 th day, Harsh visits on every 24 th day, while Sumit visits on every 9 th day. If all three of them met at the club on a Sunday, then on which day will all three of them meet again?
What is the least number which when divided by 12,20 and 24 leaves in each case a remainder of 8?
Which of the following is a pair of co-primes?
