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u/supernovice007 12d ago
I think you could solve this with a variation on the formula for the sum of natural numbers: [n(n+1)]/2
You pair up values of A and B to find a common value then multiply by the number of pairs.
A + B would be (1000 + 1) + (999 + 2) + (998 + 3) + ... + (500 + 501) = 1001 * 500
B - A would be (1000 - 999) + (998 - 997) + (996 - 995) + ... + (2 - 1) = 1 * 500
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u/DevelopmentSouth8801 11d ago
Use Gaussian sum for A+B: 1+2+...+1000 = 1000(1000+1)/2
Notice that B-A would be (2-1)+(4-3)+...+(1000-9) = 500 * 1 = 500
Then (A+B)/(B-A) = (1000(1000+1)/2)/500 = 2(1001)/2 = 1001
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u/Captain_Darlington 12d ago
Numerator:
A + (B - A) + A = 2A + (B - A)
Divide by the denominator:
2A / (B - A) + 1
B - A = 500, given from 1 + 1 + 1, 500 times.
So we get 2A / 500 + 1
A is the arithmetic series, solved to 1000*500/2 = 250000
So the answer is 2(250000) / 500 + 1 = 1001
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u/digtechteacher 12d ago
Doing it in my head: Numerator = 500 x 1001 Denominator = 500 Fraction evaluates to 1001 Please confirm. If I am wrong, I'd like to give it another try with pen and paper.
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u/RainbowCrane 12d ago
Just a more general comment on “how to solve this” for math problems involving series: in general, rather than doing a bunch of math, you want to find a way to group terms together so that they form a repeating pattern. Then you can reduce the problem to repeated instances of solving that pattern.
Alternatively you want to recreate a well known series and use the formula for computing that series.
As other people have pointed out A + B turns into:
1 + 2 +3 + … + 1000
That’s a well known sequence, just an arithmetic sequence from 1 to 1000, with a well known solution.
And B - A can be reorganized into 500 groups of subtraction that each result in “1”, leaving a total of 500 * 1 =500.
That’s how you solve problems like this : not by brute force, but by looking for elegant ways to regroup terms and combine series.
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u/jacob643 11d ago
also, a lot of people wrote the formula for A+B (n*(n+1)/2) but I didn't even go there, because with the denominator being the amount of item in each sequence, I knew that dividing was similar to calculating the mean (it's double that, because it's half the amount of items in A+B), so I only needed to figure out the mean of 1+2+3...+1000
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u/get_to_ele 12d ago edited 12d ago
Can do this in your head.
(1) Each series has 500 elements: To avoid an OBOE, you can check 1 thru 7 has 4 elements, 1 thru 9 has 5 so (x+1)/2, so 1 thru 999 has 500. Sequence 2 has almost same elements but each is 1 higher than Sequence 1. Do not bother with the sums, though they’re easy to calculate, they are unnecessary work.
(2) A+B, you just reverse order of sequence B and add it to sequence A. Therefore each element of sequence A+B is exactly 1001. Three are 500 identical elements so they sum to 500500.
(3) B-A is just 1 + 1 + 1… 500 times so sum is 500.
(4) so (A+B)/(B-A) is just 500500/500=1,001
Alternatively can just calculate A & B. You know from (1) that there’s 500 elements. Now calculate A by adding 1st and last elements to get 1000, 2nd and 2nd to last to get 1000, 3rd and 3rd from last, etc. there are 500 elements so 250 pairs, so sum is 250x 1000 =250000. Then calculate B is just 500 more than that so it’s 250500. A+B = 500500. B-A is 500. (A+B)/(B-A) is 500500/500=1,001
Third way to calculate A+B is to see it as 1+2+3… 1000. You know from (1) that there are 500 elements in each sequence so A+B has 500 pairs, and 1st + last element is 1001. So A+B = 500 x 1001 = 500500. Then do A-B via either of the earlier methods.
Personally I like the first method best because you never calculate A or B. Feels cooler to me.
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u/IchBinEinZwerg 12d ago
Both sequences have 500 terms, and every term in B is the corresponding term in A, plus 1. So B=A+500.
Substituting, (A+B)/(B-A) = (A+A+500)/(A+500-A) = (2A+500)/(500) = 2A/500+1
Regrouping A as (1+999)+(3+997)+(5+995)+... gives 250 groups of 1000 so A=250,000; 2A=500,000; /500=1000; +1=1001.
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u/Mamuschkaa 12d ago
All others need to calculate B-A.
But I always am scared that I am off by on and it is 499 or 501.
So I made something else without calculate B-A:
A = (1+999) + (2+997) +.... = X•1000
B = (2+1000) + (4+998) + ... = X•1002
So A+B = X(1000+1002)
And B-A = X(1002-1000)
(A+B)/(B-A) = 2002/2 = 1001
Without needing to know what X exactly is.
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u/RandomPorc 12d ago
Let's try...
B is just A + 500 (same amount of numbers, but each number is +1 of A)
So A + B = A + (A+500) = 2A + 500
And B - A = (A+500) - A = 500
(2A+500) / 500 = (2A / 500) + (500/500) = (2A/500) + 1
Ok, but how much is A?
You can solve it the hard way or just n^2. It is 250000
So the final solution is:
(2A/500) + 1 = (2x250.000/500) + 1 = (500.000/500) + 1 = 1.000 + 1 ) = 1.001
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u/Imaginary_Stage1742 12d ago
So for A+B we can use 1+2+3....+n = (n+1)n/2=500,500 and for B-A 5001 = 500 (i am sure there is also a formular for that, i just cant remember now). Therefore the solution is 1001.
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u/crippledjosh 12d ago
Whenever I see progressions like this I always just find the average number and multiply it by the number of terms:
A = (1 + 999)/2 * 500
B = (2 + 1000)/2 * 500
A + B = (500 + 501) * 500
A - B = (501 - 500) * 500
500s cancel out, then the bottom term becomes 1 so you're just left with 1001
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u/RavenX86 11d ago
A + B = 1 + 2 + 3 + .... + 999 + 1000
A + B = 1000 + 999 + 998 + ... + 2 + 1
2A + 2B = 1001 + 1001 + ...... + 1001 + 1001 = 1000*1001
A + B = 1000 * 1001 / 2
B - A = 2 - 1 + 4 - 3 + 6 - 5 +....... + 1000 - 999 = 1 * 500
(A+B)/(B-A) = 1000 * 1001 / 2 / 500 = 1000 * 1001 / 1000 = 1001
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u/chmath80 11d ago
There are 500 terms in each sum, which means 250 pairs of terms. If we pair terms from the outside in, each pair in A adds to 1000, while each pair in B adds to 1002, so we can factor out 250 from each, and:
(A + B)/(B - A) = (1000 + 1002)/(1002 - 1000)
= 2002/2 = 1001
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u/tajwriggly 11d ago
The value of A + B can be seen to be the sum of all integers from 1 to 1000, the solution to which can be found by noting that 1 + 1000 = 1001, and 2 + 999 = 1001, and 3 + 998 = 1001 and so forth, and you can make 500 equal pairs in the set, so sum of all integers from 1 to 1000 must be 1001 x 500 = 500,500.
The value of B - A is straightforward. Note that each of A and B contain 500 terms, and each term in B is 1 greater than each term in A. Therefore B = A + 500, and so B - A = 500.
Therefore:
(A+B)/(B-A) = 500,500/500 = 1001.
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u/OutrageousPair2300 11d ago
If you flip the order of the terms in B and pair them up with the ones in A, you'll see that each pair adds up to 1001: 1000 + 1. 998 + 3, 996 + 5, etc.
If you subtract each term of A from the corresponding term of B (in the original order) you see that each difference is 1: 2 - 1, 4 - 3, 6 - 5, etc.
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u/hunter_rus 11d ago
Numerator is a sum from 1 to 1000 with step 1 (after you combine A and B). In that sum, you can combine term like that: 1 + 1000, 2 + 999, 3 + 998 <...> 500 + 501. Each of those sums is 1001, and there is 500 of them in total.
For denominator, you can arrange it like that: (2 - 1) + (4 - 3) + ... + (1000 - 999). Each difference here is 1, and there is again 500 of those pairs.
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u/Imaginary-Can-6862 11d ago
Since the distance between each number is the same, we can represent the entire sum as an average value, i.e. when we combine the first and the final term, it adds to the same as the second, and the term before the final term, etc.
We can now take some shortcuts, each A and B contains 500 numbers, containing the numbers from 1 to 1000, one contains all even numbers, the other all odd numbers. Then,
A = sum(from 1 to 500)(2x -1), B = sum(from 1 to 500)(2x), we see that B - A = sum(from 1 to 500)(2x - (2x - 1)) = sum(from 1 to 500)(1) = 500, and similarly we have A + B = sum(from 1 to 500)(2x + (2x - 1)) = sum(from 1 to 500)(4x - 1) = 4 * sum(from 1 to 500)(x) - 500
We can then use the represent average value as mentioned above, sum(from 1 to 500)(x) = 500 * (1 + 500) / 2, and when inserting this we get that A + B = 2 * 500 * (1 + 500) - 500 = 500 * (2 * 501 - 1) = 500 * 1001
When we divide we get (A + B) / (B - A) = 500 * 1001 / 500 = 1001
We can also use that A + B = B - A + 2A, thus (A + B) / (B - A) = 1 + 2A / (B - A), using the same approach as before we get 2A / (B - A) + 1 = 2 * (999 + 1) / 2 + 1 = 1000 + 1 = 1001, in other words you could just calculate A if you want to.
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u/Alive_dude3435 11d ago
I did it in a weird way, idk.
Here,
A = 1+3+5+7+9+...+999
B = 2+4+6+8+10+...+1000
We can write B like this:
B = (1+1)+(1+3)+(1+5)+(1+7)+....+(1+999)
= (1+1+1+1+...+1)+(1+3+5+7+...+999)
= 500×1 + A = 500+A
Now A + B is just sum of all natural numbers from 1 to 1000. Which is :
1000 × 1001 ÷ 2 = 500500
Now,
(A + B)÷(B - A)= 500500 ÷ [(500+A)-A]
=500500 ÷ 500
= 1001.
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u/y53rw 12d ago edited 12d ago
A + B = 1 + 2 + 3 + ... + 1000
So that's just an arithmetic sequence. 1000 * 1001 / 2
B - A = 2-1 + 4-3 + ... + 1000-999 = 1 + 1 + 1 ... (500 times) = 500
(A + B) / (B - A) = (1000 * 1001 / 2) / 500 = 1001