Alright, here’s a fun little challenge for you. Imagine you’ve got a random assortment of digits, like the leftovers from a game of Scrabble, and your mission is to see if you can rearrange them to create a number that’s divisible by 60. Sounds simple enough, right? But there’s a catch; you have to keep a couple of math rules in mind.
First off, for a number to be divisible by 60, it has to tick off two main boxes: it must be divisible by 3 and by 20. Let’s break this down a bit.
To figure out if your number is divisible by 3, you need to add up all the digits you have. If that total is divisible by 3, congrats, you’ve conquered the first hurdle! So, if your digits include 2, 7, 1, and 8, you’d add those together to get 18, which is divisible by 3. Sweet!
Now, let’s talk about divisibility by 20. This one’s a little trickier. Your number has to end in a combination that allows it to be divisible by 20. This means it needs to end in at least one ‘0’ (because you can’t have a number that ends in 2 and be 20 without also having a 0, right?). Also, you’ve got to have at least one even digit (like 2, 4, 6, or 8) somewhere in there to make it all work.
So, let’s put this into action. Say you have the digits [3, 2, 0, 1, 5]. First, check the sum: 3 + 2 + 0 + 1 + 5 = 11. Oops, not divisible by 3, so we can stop there, right? But what if you had [6, 9, 0, 2, 4]? Add those up: 6 + 9 + 0 + 2 + 4 = 21, which is divisible by 3. Now, you’ve got a 0, and you also have even digits—2 and 4. Now the question is: can you rearrange these digits to make a number that fits the bill?
So, gather your digits, run the checks, and let’s see if you can whip up a number that skips happily in dance with 60! Can you do it?
Okay, so I’m trying to figure out if I can make a number from a bunch of digits that’s divisible by 60. Here’s what I got!
First, I need to check if my digits add up to something that can be divided by 3. That’s easy, right? Just add them all up! So for example, if I have these digits:
I’d do:
3 + 2 + 0 + 1 + 5 = 11Um, 11 isn’t divisible by 3…so, I guess that’s a no-go for that set of digits.
But what about this set:
Let’s add those up:
6 + 9 + 0 + 2 + 4 = 21Yay! 21 is divisible by 3! Whoohoo!
Now, I have a 0 in there, and I need at least one even digit. I see I have 2 and 4, so that’s cool.
Now comes the fun part! Can I rearrange them to create a number that ends with a 0 and still looks good? I could do something like:
240Or maybe:
460Both of these end in a 0 and use my digits. Plus, they are divisible by 60!
So, fingers crossed, I think I did it! Rearranging digits can be pretty fun, who knew math could be like a puzzle?
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To determine whether a random assortment of digits can be rearranged to create a number divisible by 60, we must ensure it meets two criteria: it must be divisible by 3 and by 20. To check for divisibility by 3, sum all the digits and see if the result is divisible by 3. For example, from the digits [6, 9, 0, 2, 4], we would calculate: 6 + 9 + 0 + 2 + 4 = 21, which is divisible by 3, satisfying the first condition. It’s essential to include a ‘0’ in the last digit to check for divisibility by 20, along with at least one even digit such as 2, 4, or 6 to ensure that the number is correctly structured to meet the requirement.
Next, let’s assemble the digits to confirm that they can indeed form a valid number. Given our example with the digits [6, 9, 0, 2, 4], we can end the number with ‘0’ and create various combinations with the remaining digits, like 240, 420, 640, etc. Each of these numbers checks out: they are all divisible by 20 because they end in ‘0’ and contain the even digits necessary for the assembly. Thus, after validating that the sum of the digits is divisible by 3 and properly closing off with ‘0’, it’s clear that it is feasible to rearrange the digits to create a number that is happily divisible by 60!
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