Brainsteem Compute #1 Prize Computational Maths Puzzle [Win 40% and 10% in SBD]

happychau123 (53) 7 years ago

I feel u.. 30% is really low lol

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beriberi (47) 7 years ago

So, the sum of all the multiples of 5 below 2000. I made a table and figured out that for any x, the sum of the multiples of 5 = 5(1+2+3...(x/5 -1)). Therefore 5 * (1+2+3....399) gives us the sum of all the values of 5 under 2000. That equals 399000.

For the 7, it works the same 7(1+2+3...(x/7-1)). So 7 *( 1+2+3....284). So, (284/2) * 285 * 7 = 283290

So, 399000+283290=682990.

But shit, I've double counted all the multiples of 35.

So, sum of multiples of 35 under 2000 is
35 * (1+2+3....56). AKA 35 * 56 * 57/2 = 55890

So, subtract that out and 682990 - 55890 = 627100

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3 votes

beriberi (47) 7 years ago

I bet I forgot a zero or added a 7 somewhere. But re-steeming because I had a lot of fun trying.

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beriberi (47) 7 years ago

So, I realized the x/5, x/7, x/35 had to be a whole number, but was wrong about having to round down. When rounded up, I get to the right answer. Close. But not playing horseshoes or hand grenades. Till next time.

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happychau123 (53) 7 years ago (edited)

td; dr: the answer is 626430

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happychau123 (53) 7 years ago

oops there is a /2 in the right side out of the picture lol

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doughtaker (56) 7 years ago

I'd like to give this a shot. beriberi's methodology is accurate but as they mentioned something was missing.

The sum of all positive integers from 1 to x is computed as x*(x+1)/2. This is the key component.

The multiples of 5 under 2000 go from 5 to 1995. Factoring out 5 yields 5*(1+2+...+399).
5 * [399 * 400 / 2] = 5 * 79800 = 399,000

The multiples of 7 under 2000 go from 7 to 1995. Factoring out 7 yields 7*(1+2+...+285).
7 * [285 * 286 / 2] = 7 * 40755 = 285,285

Multiples of 35 have thus been double-counted and need to be subtracted. Those go from 35 to 1995. Factoring out 35 yields 35*(1+2+...+57).
35 * [57 * 58 / 2] = 35 * 1653 = 57,855

The sum then is 399,000 + 285,285 - 57,855 = 626,430

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mage00000 (61) 7 years ago

same answer here with one small difference in the calculation:
To find for example the multiples of 5 below 2000, we notice the first is 5 and the last is 1995 and there are 399 such numbers. The first + the last = 5 + 1995 = 2000, so is the second + the second last = 10 + 1990 = 2000. This can be repeated 199 times and then we are left with 1000. So the sum is 199 x 2000 + 1000 = 399000.
The same can be done for 7 and 35, and the answer is 626430 as @doughtaker wrote.

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dkmathstats (61) 7 years ago (edited)

I've coded mine in the programming language R. (Can't upload images in post)

Here is my code (and output):

> ## Rycharde's Challenge:
> 
> # Find the sum of all the multiples of 5 or 7 below 2000:
> 
> # Limiting number:
> 
> limit_number <- 2000
> 
> # Empty Vector:
> candidates <- c()
> 
> for (i in 1:(limit_number - 1)){
+   if (i %% 5 == 0 | i %% 7 == 0){
+     candidates <- c(candidates, i) # Append.
+   }
+ }
> 
> sum(candidates) # Print Total; 626430
[1] 626430

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rycharde (63) 7 years ago

Many thanks!

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rycharde (63) 7 years ago

Starting to come to a consensus answer.
Worth re-checking calculations.
More upvotes and more answers will increase the overall payout.
But this is also about learning a style of mathematics that rarely appears in textbooks, but is at the heart of maths challenges and competitions!

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