The number 2019 can be written as the sum of 2019 consecutive integers. What is the sum of the smallest and largest of these integers?
#Math comp
brazen lily
sacred coralBOT
brazen lily
@fast forge
fast forge
(1+2+...+n) = n(n+1)/2
brazen lily
how i would need to use this formula???
fast forge
n(n+1) is a lot like n^2
brazen lily
ok, i dont understand the application for the n(n+1)/2 to this problem tho
fast forge
believe in yourself
brazen lily
consecutive integers could also be negative as well
fast forge
ok, what does k+(k+1)+...+n = ?
brazen lily
alr thanks so much i got the answer 2
i checked possible factor pairs for the corresponding values of (n+1)(2k+n), also noting that 2k+n>0 which means k>-n/2
i found that 2k+n = 2 which means 2 is the sum. Is there a faster way than to check for factor pairs???
brazen lily
i see now i didnt read the problem correctly, i already know that n+1=2019
which makes things much faster
alr thanks so much for ur help
fast forge
(k+1)+(k+2) + ... + (k+n) = n(n+1)/2 + nk
n = 2019
brazen lily
Yes i did it this way:
(k+0)+(k+1)+(k+2)+...+(k+n)
==> k(n+1)+n(n+1)/2 = 2019
2k(n+1)+n(n+1) = 4038
(2k+n)(n+1) = 4038
Here i notice that the sum of the biggest and smallest is actually 2k+n because (k+n)+k => 2k+n
We also know that there is 2019 consecutive integers (which i overlooked at the beginning), which means that 2019 must be the factor of 4038 since n+1=2019.
Therefore the answer is 4038/2019 = 2 = 2k+n
.close
sacred coralBOT
Solved
brazen lily
i just realized i couldve solved it this way which was 10 times easier:
==> x = sum of the first and last
==> 2019/2 = number of pairs
(2019/2)(x) = 2019
2019x=2(2019)
x = 2
fast forge
2019 = 2019(2020)/2 + 2019(K) = 2019(1010 + k)
1 = 1010 + k
k = -1009
(k+1)+(k+n) = -1009 + 1 -1009 + 2019 = 2
brazen lily
there was really no need for the n(n+1)/2 formula, just say x = first+last and multiply by number of pairs which was 2019/2 and this equals to 2019. im really stupid
alr thank you anyways tho im going to sleep now 🙂
fast forge
the pairs is how n(n+1)/2 is proven
brazen lily
yes i am familiar with this, i was just super confused at the start because i didnt realize that the problem literally tells you how many numbers are in the sequence. the n(n+1)/2 is called the gauss formula or something i believe
tender mango
The number 2019 can be written as the sum of 2019 consecutive integers. What is ...
if you remember the summation formula for arithmetic series it's:
S(n) = (n/2)(T(1)+T(n))
where T(n) represents the n-th term
what we want is T(1)+T(n)
let's denote it as x
and here n=2019
therefore:
(2019/2)x = 2019
hence x = 2
brazen lily
if you remember the summation formula for arithmetic series it's:
S(n) = (n/2)(T...
yes that was the same solution i gave couple messages ago
brazen lily
thats okay thanks anyways tho