What is the largest number which divides 60 and 75 leaving remainders 8 and 10?
Have you ever encountered a math problem that sounds a bit like a riddle? This is one of those! We're looking for a specific number that, when you use it to divide 60 and 75, doesn't divide them perfectly. Instead, it leaves behind specific remainders: 8 when dividing 60, and 10 when dividing 75. Let's break down how to find this "largest number."
Understanding the Problem
When we say a number "divides" another number and leaves a remainder, it means that after you've taken away as many full groups of the divisor as possible, there's a little bit left over. The remainder is always smaller than the divisor.
In our case:
- When we divide 60 by our mystery number, we get a remainder of 8.
- When we divide 75 by our mystery number, we get a remainder of 10.
This means that:
- 60 - 8 is perfectly divisible by our mystery number.
- 75 - 10 is perfectly divisible by our mystery number.
The Key Insight: Working with the Differences
Let's calculate those differences. These are the numbers that our mystery number must divide evenly.
Step 1: Subtract the remainders.
- For 60, the remainder is 8. So, 60 - 8 = 52.
- For 75, the remainder is 10. So, 75 - 10 = 65.
Now, our problem has transformed. We are looking for the largest number that divides both 52 and 65 perfectly. In mathematical terms, we are looking for the Greatest Common Divisor (GCD) of 52 and 65.
Finding the Greatest Common Divisor (GCD)
There are a couple of common ways to find the GCD of two numbers. Let's explore the method of listing factors.
Step 2: Find the factors of each number.
- Factors of 52: These are numbers that divide 52 evenly.
- 1 x 52 = 52
- 2 x 26 = 52
- 4 x 13 = 52
- Factors of 65: These are numbers that divide 65 evenly.
- 1 x 65 = 65
- 5 x 13 = 65
Step 3: Identify the common factors.
Now, let's see which numbers appear in both lists of factors:
- The common factors of 52 and 65 are 1 and 13.
Step 4: Select the greatest common factor.
We are looking for the *largest* number that divides both 52 and 65. Between 1 and 13, the largest is 13.
The Solution and Verification
Therefore, the largest number which divides 60 and 75 leaving remainders 8 and 10 is 13.
Let's quickly check this to make sure it works:
- Divide 60 by 13:
60 ÷ 13 = 4 with a remainder of 8. (13 x 4 = 52; 60 - 52 = 8)
- Divide 75 by 13:
75 ÷ 13 = 5 with a remainder of 10. (13 x 5 = 65; 75 - 65 = 10)
Our calculations are correct! The number 13 fits the conditions perfectly.
It's also important to remember that the divisor must be larger than the remainder. In this case, our divisor is 13, which is larger than both remainders (8 and 10). This confirms our answer is valid.
The problem asks for the largest number that satisfies specific division conditions. By subtracting the remainders from the original numbers, we transform the problem into finding the Greatest Common Divisor (GCD) of the resulting numbers. The GCD is the largest number that divides both of these new numbers without leaving a remainder.
Frequently Asked Questions (FAQ)
How do I know to subtract the remainders?
Subtracting the remainders is the key step because it removes the "leftover" part from the original numbers. When you subtract the remainder, what's left is a number that is perfectly divisible by the original divisor. So, if a number divides 60 and leaves a remainder of 8, it means that number divides (60 - 8) perfectly. The same logic applies to the second division.
Why is it called the "Greatest Common Divisor"?
It's called the "Greatest Common Divisor" (GCD) because it's the largest (greatest) number that can be found in the list of all numbers that divide (divisor) two or more numbers evenly (common). In our case, we found all the numbers that divide 52 and 65, and then we picked the biggest one that they both shared.
What if the remainders were the same?
If the remainders were the same, you would still subtract that common remainder from both original numbers. Then, you would find the GCD of the two resulting numbers, just like we did in this problem. The principle remains the same: the divisor must perfectly divide the number after the remainder is accounted for.
