We write a program that finds a positive n number of integers. So, if all the integers from 1 to n in the form 1 + 2 + 3 + ⋯ + n are summed up, the result is between 100 and 1000. At the same time, the three numbers of this sum are the same (like 333
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We write a program that finds a positive n number of integers. So, if all the integers from 1 to n in the form 1 + 2 + 3 + ⋯ + n are summed up, the result is between 100 and 1000. At the same time, the three numbers of this sum are the same (like 333). Print the integer and sum of the program as output.
1 Comment
James Tursa
on 15 May 2017
What have you done so far? What specific problems are you having with your code?
Answers (2)
Cam Salzberger
on 15 May 2017
So you could solve this with a complete brute-force method, but let's not because that's no fun. If I were to approach this problem, I would first check out what the mathematical expression is for 1+2+...+n. There's a convenient Wikipedia article on this showing that it's n(n+1)/2.
0 = (n^2 + n -2*total)
or
n = roots([1 1 -2*total])
And you know your desired possible total values (111, 222, ..., 999). So if I were doing this just in the Command Window, I would run that through a loop, solving for n for each value:
s = [111 222 333 444 555 666 777 888 999];
for k = 1:length(s)
roots([1 1 -2*s(k)])
end
And then just see which of the (positive) n values was an integer. Double-check this with sum(1:n).
Now to automate it, you'd just have to automate the finding of which value is an integer. This is pretty simple to do with mod or floor, so I'll leave it to you.
Hope this helps!
-Cam
0 Comments
ramazan araz
on 17 May 2017
2 Comments
Cam Salzberger
on 17 May 2017
I didn't use 'tal' anywhere, so I assume you are talking about "total". This is representing the "all summed up" value in the question. For this case, we know we want the total to be 111, 222, 333, ..., or 999.
John D'Errico
on 17 May 2017
Please don't add an answer just to make a comment. As you see, it is very simple to make a comment. Use that instead.
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