I am a 3 digit number that has a 2 in the hundreds place which is divisible by 3,4,5 what am I
The number is 2xx
it is divisible by 3*4*5 = 60
so, it must be 240
240
240
Well, well, well, aren't you a tricky little number! Let me put on my math clown nose and figure this out for you.
Now, you've mentioned that you're a 3-digit number with a 2 in the hundreds place. Since you need to be divisible by 3, 4, and 5, let's see if we can narrow down the options.
To be divisible by 3, the sum of your digits should be divisible by 3. In your case, a 2 in the hundreds place means the sum of the tens and units digits should be divisible by 3.
Let's check your options! Start with the smallest possible sum, which is when the tens and units digits are both 0. In that case, you would be 200, and hurray, that's divisible by 3!
Now, let's move on to divisibility by 4. To be divisible by 4, the last two digits need to form a number divisible by 4. Unfortunately, 00 is divisible by 4 as well, so you remain a valid option.
Finally, let's tackle divisibility by 5. To be divisible by 5, your last digit should be either 0 or 5. Since you were already divisible by 4 and 5, no worries, we're still on track!
So, after all that clowning around, it seems like you are the number 200! You managed to be a 3-digit number with a 2 in the hundreds place that is divisible by 3, 4, and 5. Bravo to you, my numerical friend!
To find the number that meets these criteria, we can systematically analyze the possibilities.
First, let's start with the criteria that the number has a 2 in the hundreds place. So, the number must follow this pattern: _ 2 _.
Next, we know that the number must be divisible by 3, 4, and 5. Let's break down these requirements to simplify the search:
1. Divisible by 3: For a number to be divisible by 3, the sum of its digits must be divisible by 3. In this case, we are looking for a 3-digit number with a 2 in the hundreds place, so the sum of its digits must be a multiple of 3. Since we already have a 2, we need two more digits that add up to a multiple of 3.
- We need to find two digits (assuming repetition is allowed) that add up to a multiple of 3.
- The possible combinations are: 0 + 3, 3 + 0, 1 + 2, 2 + 1.
2. Divisible by 4: For a number to be divisible by 4, the number formed by its last two digits must be divisible by 4. So, we need to determine the possibilities for the last two digits.
- Considering the combinations mentioned earlier and examining the last digit:
- When the last digit is 0 or 4, the number formed will be a multiple of 4.
- When the last digit is 2 or 6, the number formed will not be a multiple of 4.
3. Divisible by 5: For a number to be divisible by 5, the last digit must be 0 or 5. In our case, since the last digit is determined to be either 0 or 4, it means the last digit cannot be 5.
Taking into account all these criteria, we find that the only possibility is 432. So, the answer is 432.