No need to show your work, just tell me the answer, it is due tomorrow.
How many integers $n$ satisfy the condition $100 < n < 200$ and the condition $n$ has the same remainder whether it is divided by $6$ or by $8$? I am stuck at the mid part, please help
2 is wrong, its 25
at least understand the question smh
The answer is 25 btw
cheater
Well, well, well, looks like you're stuck between a rock and a hard place! Don't worry, my friend, I'm here to clown around and help you out.
So, we need to find the integers $n$ that satisfy two conditions. Condition one is that $n$ is between 100 and 200, which means our possibilities are 101, 102, 103, and so on, all the way up to 199.
The second condition is a bit trickier. We need $n$ to have the same remainder when divided by 6 or by 8. Let's think about this - if a number has the same remainder when divided by both 6 and 8, what can we conclude?
Do you hear that? It's the sound of a clown horn honking! It's a hint! The key to this question is finding a common multiple of 6 and 8. Can you think of a number that satisfies this?
Alright, alright, I'll let the cat out of the bag! The common multiple of 6 and 8 is 24! So, any number that leaves the same remainder when divided by 6 and 8 must be a multiple of 24.
Now, we go back to our list of numbers between 100 and 200 and see how many of them are multiples of 24. Hmm, where are my juggling balls when I need them? Ah, there they are!
Let's see: 120, 144, 168, and 192 are all multiples of 24. That's a total of 4 numbers that satisfy both conditions.
So, to answer your question, there are 4 integers that satisfy the given conditions. Ta-da! I hope my clowning around helped you out, and remember, laughter is the best way to solve any problem!
I understand that you're looking for the answer to the problem, but I encourage you to try solving it on your own first. Solving math problems yourself not only helps to improve your skills, but it also allows you to better understand the concepts behind the problem. In this case, let's break down the problem step by step.
We are given that $100 < n < 200$ and that $n$ has the same remainder whether it is divided by $6$ or by $8$. To find the number of integers that satisfy both conditions, we need to find the common multiples of $6$ and $8$ within the given range.
To do this, we can find the least common multiple (LCM) of $6$ and $8$, which is the smallest positive integer that is divisible by both $6$ and $8$. The prime factorization of $6$ is $2 \times 3$ and the prime factorization of $8$ is $2^3$. The LCM is found by taking the highest power of each factor, so in this case, the LCM is $2^3 \times 3$, which equals $24$.
Next, we need to count the multiples of $24$ between $100$ and $200$. To do this, we divide the upper and lower limits by $24$ and find the whole number parts.
$\frac{100}{24} = 4$ (with a remainder of $4$)
$\frac{200}{24} = 8$ (with a remainder of $8$)
From $4$ to $8$, there are $8 - 4 - 1 = 3$ whole numbers (excluding the endpoints).
Therefore, there are $3$ integers $n$ that satisfy both conditions. So the answer is $\boxed{3}$.