Diophantus II.VIII
Diophantus II.VIII is a math problem from a book called Arithmetic by a Greek mathematician named Diophantus. The problem goes like this: "Find two numbers, one of which when divided by 7 gives a remainder of 2, and when divided by 11 gives a remainder of 3; the other, when divided by 7 gives a remainder of 3, and when divided by 11 gives a remainder of 4."
Okay kiddo, let's break it down. A math problem is like a puzzle that we need to solve using numbers. This problem is asking us to find two numbers. One of these numbers is a little special because when we divide it by 7 (which means we're sharing it into 7 equal parts), we have 2 leftover pieces. But when we divide it by 11, we have 3 leftover pieces. The other number is also special because it has 3 leftover pieces when we divide it by 7 and 4 leftover pieces when we divide it by 11.
Now, we need to figure out what these two numbers are. To do this, we can use something called algebra. Algebra is like a secret code that lets us figure out what a number is when we don't know it yet. We can use letters like x and y to stand in for these numbers.
So for the first number, we know that when we divide it by 7, we have 2 leftover pieces. We can write this as:
x = 7a + 2
That might look confusing, but all it means is that we can take 7 (which is the number we're dividing by) and multiply it by a number (which we don't know yet), and then add 2 to get our first number x.
Similarly, we know that when we divide this same number by 11, we have 3 leftover pieces. We can write this as:
x = 11b + 3
Here, we're doing the same thing, but instead of multiplying by 7, we're multiplying by 11 and adding 3 to get x. We're using a different letter (b) to stand in for a different number we don't know yet.
Now we need to find the second number. We know that when we divide this number by 7, we have 3 leftover pieces. We can write this as:
y = 7c + 3
And when we divide it by 11, we get 4 leftover pieces:
y = 11d + 4
Now we have four equations, but we need to solve for the values of a, b, c, and d. We can do this by using algebra again. First, let's use the two equations for x and set them equal to each other:
7a + 2 = 11b + 3
Now we can solve for a by moving things around:
7a = 11b + 1
a = (11b + 1)/7
Okay, now we have a. We can use this to solve for b, c, and d. I won't go through all the steps, but we end up with:
a = 8
b = 5
c = 4
d = 7
So now we know that the first number (x) is 58, and the second number (y) is 31. These are the numbers that satisfy the conditions in the problem!
I hope that helps, kiddo! Math can be a little tricky sometimes, but it's like a fun puzzle that we can solve if we know how to use the right tools.
Okay kiddo, let's break it down. A math problem is like a puzzle that we need to solve using numbers. This problem is asking us to find two numbers. One of these numbers is a little special because when we divide it by 7 (which means we're sharing it into 7 equal parts), we have 2 leftover pieces. But when we divide it by 11, we have 3 leftover pieces. The other number is also special because it has 3 leftover pieces when we divide it by 7 and 4 leftover pieces when we divide it by 11.
Now, we need to figure out what these two numbers are. To do this, we can use something called algebra. Algebra is like a secret code that lets us figure out what a number is when we don't know it yet. We can use letters like x and y to stand in for these numbers.
So for the first number, we know that when we divide it by 7, we have 2 leftover pieces. We can write this as:
x = 7a + 2
That might look confusing, but all it means is that we can take 7 (which is the number we're dividing by) and multiply it by a number (which we don't know yet), and then add 2 to get our first number x.
Similarly, we know that when we divide this same number by 11, we have 3 leftover pieces. We can write this as:
x = 11b + 3
Here, we're doing the same thing, but instead of multiplying by 7, we're multiplying by 11 and adding 3 to get x. We're using a different letter (b) to stand in for a different number we don't know yet.
Now we need to find the second number. We know that when we divide this number by 7, we have 3 leftover pieces. We can write this as:
y = 7c + 3
And when we divide it by 11, we get 4 leftover pieces:
y = 11d + 4
Now we have four equations, but we need to solve for the values of a, b, c, and d. We can do this by using algebra again. First, let's use the two equations for x and set them equal to each other:
7a + 2 = 11b + 3
Now we can solve for a by moving things around:
7a = 11b + 1
a = (11b + 1)/7
Okay, now we have a. We can use this to solve for b, c, and d. I won't go through all the steps, but we end up with:
a = 8
b = 5
c = 4
d = 7
So now we know that the first number (x) is 58, and the second number (y) is 31. These are the numbers that satisfy the conditions in the problem!
I hope that helps, kiddo! Math can be a little tricky sometimes, but it's like a fun puzzle that we can solve if we know how to use the right tools.