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):
@addax @ajayyy @athunderstruck @bwar @contrabourdon @golddeck @heraclio @hokkaido @iampolite @masoom @mmunited @mytechtrail @onecent @sidekickmatt @sparkesy43 @syalla @tonimontana @vote-transfer @zuerich
In case no one gets a result(which I doubt), I will give away the prize to anyone who comments.
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@contrabourdon sponsors this contest with 2 STEEM weekly.
You can support him by using a witness vote on untersatz, so he can further support this and other contests.
hey @quantum are you sure u wanna check divisibility by 909 in the third one..
let me know asap coz my answer is ready
Yes, I am totally sure.
B) Yes sum =42=12=3
C) No Sum is 70
B) No Sum = 108
C) Yes Sum = 165
B) Yes sum is divisible by 9 but not sure of trick to find out about 909. because common factors are 9 and 101
c) Yes Sum is divisible by 9 but not sure of trick to find out about 909
4 A) 13
B) 27, ?,?,?
I just discovered that I made a small mistake in 4 a):
You need to find 2 missing factors!
And by the way your solution for 4 a) is wrong anyway.
It is still about multi-digit sums. You just need to find out how many digits.
18 & 33
Stop guessing, start thinking, although that is only 50% wrong.
That's the last time I'll correct you!
My sample size wasn't large enough....
18137=2466=90/18=5
18159=2862=90/18=5
18189=3402=36/18=2
It wasn't till I did 1818=324=27 that I knew 18 was wrong... but I couldn't find my original comment to delete it.
123456789 is also divisible by 3.
4327384926 is also divisible by 3 .
7777777777 is also not divisible 3.
2.Checking for divisibility by 11
308 is also divisible 11.
453825 is not also divisible 11.
187869 is also divisible by 11.
3.Checking for divisibility by 909.
46385600 is not also divisible by 909.
36369999 is also divisible by 909.
4072320 is also divisible by 909.
4.a 3,9,11,15.
4.b don't now.
You are doing really strange things at 3 and 4a)
Yes. For 3 divided by 909 is impossible so I sumed up 909 and divided with the result 18. I did the same thing in 4.a.
That's an interesting idea. But it's just a coincidence that it works here.
Check for example "18" which is not divisible 909, but according to your rule it would be.
Actually you were supposed to do something else, much closer to task 1 and 2.
Also 4 a) is wrong. And also you need to find 2 numbers there
5.You can check if a number is divisible by a number d with a n-digit sum if there some m such that m*d=10^n-1. So for 11 it would be 9*11=99.
2.And this even works almost the same if you can find a multiple that is close to a power of 10. For example 11 is 10+1 so you can check divisibility by 11 by doing a alternating 1-digit sum:
3-0+8=11 so 308 is divisible by 11
4-5+3-8+2-5=-9 not divisible by 11
1-8+7-8+6-9=-11 divisible by 11
For 17 you can check divisibility using 2-digits because 17*6=100+2.
So if you wanted to check if 165616 is divisible by 17 you would divide 165616 into 2-digit pairs and starting from the left subtract 2 times the digit pair from the next pair. Also subtract multiples of 17 if possible.
16 56 16 -> 24 16 -> 7 16 -> 2
11 86 60 -> 64 60 ->13 60 -> 34 -> 0
That's another interesting concept you mention there.
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