Math Contest #13 [1 SBI + 20 or more STEM] Guaranteed prize for participation!
Here you can keep your brain fit by solving math related problems and also earn SBI by doing so.
The problems usually contain a mathematical equation that in my opinion is fun to solve or has an interesting solution.
I will also only choose problems that can be solved without additional tools(at least not if you can calculate basic stuff in your head), so don't grab your calculator, you won't need it.
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Every friday and tuesday!
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Rules
No upvote, No resteem, No follow required!
I will give 1 SBI to 1 random person who put a correct solution of the problem in a comment.
Although I can't control it, I would appreciate if you would stake the STEM to support the STEM(Science-Technology-Engineering-Math) community.
You have 72 hours to solve it.
Some additional rules for the case that the problem is an equation or a collection of equations.
In case there are multiple solutions, you have a higher chance of winning if you get all of them correctly.
Solutions can sometimes also be general like x = 20, y > -1. If there are multiple solutions submitted, that are part of the same general solution, they will be counted as one solution.
And one more thing: If there are infinitely many solutions, I only want integer solutions!
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Problem
Today I want you to do something different. This problem is not about solving some equation, but about understanding a new concept.
You probably already know the concept of digit sums, but I'll explain them again just to be sure.
The digit sum is the sum of all digits of a given number.
Example:
57825 → 5+7+8+2+5 = 27
But what's so special about digit sums?
You can use them to check if the number can be divided by 3 or 9. If the digit sum is divisible by 3, then the number is also divisible by 3. Similar with 9:
43267 → 4+3+2+6+7 = 22
22 is not divisible by 3 and 9 →43267 is also not divisible by 3 and 9.
3210 → 3+2+1+0 = 6
6 is divisible by 3 → 3210 is also divisible by 3.
I found a way to advance this concept, to be able to check for more factors.
If you add pairs of 2 digits instead of single digits you can check for divisibility by 3, 9, 11, ?, ?:
1126409108 → 11 + 26 + 40 +91 + 8 = 176 = 16*11 → 1126409108 is divisible by 11 and not divisible by 3, 9, ? and ?.
If you don't straight see that 176 is divisible by 11, you can repeat the 2-digit sum, until the case is trivial:
176 → 1 + 76 = 77 → 176 is divisible by 11.
The concept can be advanced to any number of digits thus being able to check for more and more factors.
But be careful with the factors. Although you can check for 3 and 9 with any number of digits, you can only check for 11 using an even-digit sum
And now your tasks:
In 4 a) there are TWO unknown factors!
Everyone who enters within time will get a certain amount of STEM when he correctly solves one of the task:
| Taks | Reward |
|---|---|
| 1 | 1 STEM |
| 2 | 2 STEM |
| 3 | 4 STEM |
| 4 a) | 1 STEM |
| 4 b) | 4 STEM |
| 5 | 8 STEM, only when you explain in your own words! |
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To everyone who already participated in a past contest, come back today and try a new problem(tell me if you don't want to be tagged):
In case no one gets a result(which I doubt), I will give away the prize to anyone who comments.
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You can support him by using a witness vote on untersatz, so he can further support this and other contests.