Unraveling the Mystery: Finding a Special Number
Have you ever come across a math puzzle that makes you scratch your head? Today, we're tackling one such challenge: finding the smallest number that behaves in a very specific way when divided by other numbers. Specifically, we're looking for a number that, when you divide it by 5, 6, 7, or 8, always leaves you with a remainder of 3. But here's the twist: when you divide this same number by 9, it divides perfectly, with no remainder at all!
This might sound complicated, but by breaking it down step-by-step, we can uncover the solution. This type of problem is a classic example of using the concepts of the Least Common Multiple (LCM) and remainders in number theory.
Understanding the Problem's Conditions
Let's rephrase the conditions to make them crystal clear:
- When the number is divided by 5, the remainder is 3.
- When the number is divided by 6, the remainder is 3.
- When the number is divided by 7, the remainder is 3.
- When the number is divided by 8, the remainder is 3.
- When the number is divided by 9, the remainder is 0 (meaning it's perfectly divisible by 9).
Step 1: Dealing with the Common Remainder
The first four conditions all have the same remainder: 3. This is a crucial clue! If a number leaves a remainder of 3 when divided by 5, 6, 7, and 8, it means that if we were to subtract 3 from that number, the result would be perfectly divisible by 5, 6, 7, and 8.
So, we are essentially looking for a number (let's call it 'X') such that:
- X - 3 is divisible by 5
- X - 3 is divisible by 6
- X - 3 is divisible by 7
- X - 3 is divisible by 8
This means that X - 3 must be a common multiple of 5, 6, 7, and 8. Since we are looking for the *least* such number, we need to find the Least Common Multiple (LCM) of 5, 6, 7, and 8.
Step 2: Calculating the Least Common Multiple (LCM)
To find the LCM of 5, 6, 7, and 8, we can use prime factorization:
- Prime factorization of 5: 5
- Prime factorization of 6: 2 × 3
- Prime factorization of 7: 7
- Prime factorization of 8: 2 × 2 × 2 = 2³
The LCM is found by taking the highest power of each prime factor that appears in any of the numbers:
LCM(5, 6, 7, 8) = 2³ × 3 × 5 × 7
LCM(5, 6, 7, 8) = 8 × 3 × 5 × 7
LCM(5, 6, 7, 8) = 24 × 35
LCM(5, 6, 7, 8) = 840
So, any number that leaves a remainder of 3 when divided by 5, 6, 7, and 8 must be of the form 840k + 3, where 'k' is a non-negative integer (0, 1, 2, ...).
Step 3: Incorporating the Divisibility by 9
Now we need to consider the last condition: the number must be perfectly divisible by 9.
We are looking for a number of the form 840k + 3 that is also a multiple of 9.
Let's test values of 'k' starting from 0:
- If k = 0: The number is 840(0) + 3 = 3. Is 3 divisible by 9? No.
- If k = 1: The number is 840(1) + 3 = 843. Let's check if 843 is divisible by 9. A quick way to check divisibility by 9 is to sum the digits: 8 + 4 + 3 = 15. Since 15 is not divisible by 9, 843 is not divisible by 9.
- If k = 2: The number is 840(2) + 3 = 1680 + 3 = 1683. Sum of digits: 1 + 6 + 8 + 3 = 18. Since 18 is divisible by 9 (18 / 9 = 2), 1683 is divisible by 9.
We found a number that satisfies both sets of conditions! Since we started testing with the smallest possible values of 'k', 1683 is indeed the least number that meets all the requirements.
The Solution
The least number which when divided by 5, 6, 7, and 8 leaves a remainder of 3, but when divided by 9 leaves no remainder, is 1683.
Let's verify:
- 1683 ÷ 5 = 336 remainder 3
- 1683 ÷ 6 = 280 remainder 3
- 1683 ÷ 7 = 240 remainder 3
- 1683 ÷ 8 = 210 remainder 3
- 1683 ÷ 9 = 187 remainder 0
It all checks out!
Frequently Asked Questions (FAQ)
Q: How do I know if a number is divisible by 9?
A: A simple trick is to add up all the digits of the number. If the sum of the digits is divisible by 9, then the original number is also divisible by 9. For example, in 1683, 1 + 6 + 8 + 3 = 18, and 18 is divisible by 9.
Q: Why do I need to find the LCM first?
A: The LCM of 5, 6, 7, and 8 is the smallest number that is a multiple of all of them. Since our number, minus 3, has to be a multiple of 5, 6, 7, and 8, the smallest possible value for (Our Number - 3) is their LCM. This helps us establish a pattern for all numbers that fit the first four conditions.
Q: Can there be other numbers that fit these conditions?
A: Yes! Any number of the form 840k + 3 that is also divisible by 9 will satisfy these conditions. We found the *least* number by starting with the smallest possible value for 'k'. The next such number would be found by continuing to test larger values of 'k' until we find another number divisible by 9.
Q: What if the remainders were different?
A: If the remainders were different (e.g., remainder 1 for 5, remainder 2 for 6), the problem would be significantly more complex and typically solved using the Chinese Remainder Theorem. However, when the remainders are the same, the LCM approach works beautifully.
