Question

A series of positive consective integers add upto 2000. Find the least value of m where m is the first term

Answer 1

Well if m is the first term, and these are consecutive numbers, then doesn't that mean the next term is (m+1)?

Answer 2

yes

Answer 3

Wait, I think I misinterpreted this as being a really easy question when it's actually a little more involved. Hmm.

Answer 4

that should work right..

m + (m+1) + ... = 2000

Answer 5

Actually that's completely right. Just need to find a general summation formula.

Answer 6

yes

Answer 7

say there are "n" terms in the series

Answer 8

m + (m+1) + ... + (m + n-1) = 2000

Answer 9

well okay

Answer 10

nm + n(n-1)/2 = 2000

Answer 11

hmm guess we need to mess with this equation a bit..

Answer 12

@ganeshie8 Technically I think your equation should be: \[m(n+1)+\frac{n(n+1)}{2}=2000\]

Answer 13

yes if there are n+1 terms...

Answer 14

Wait I wrote it wrong and it happened to agree with what you wrote. What I mean is that it should agree with both.\[m(n+1)+\frac{n(n-1)}{2}=2000\]

Answer 15

Wait maybe I'm confusing myself, I thought one of your summations in n was off by 1 but now I'm not even sure. I think I need to go to bed now. Good luck, this is a very interesting problem. @No.name

Answer 16

n terms :

(m+0) + (m+1) + ... + (m + n-1) = 2000

(m+m+... n times) + (0+1+ ... n-1)= 2000

(nm) + (n(n-1)/2) = 2000

but we still need to interpret this equation in integers and find the minimum value for m

Answer 17

there must be a trick about pascal number right ?

Answer 18

Is it correct to say that small \(m\) means large \(n\) ?

Answer 19

smallest first term => largest number of terms in the series

Answer 20

like
0+1+2+3+...+m+(m+1)+(m+2)+...+n = 2000 + m(m+1)/2

Answer 21

i agree with that.
smallest 1st term = largest no. of terms.
also i am getting m=47 (n=32) by trial and error

Answer 22

m=47 is correct hartnn, i still don't have a solution though
http://www.wolframalpha.com/input/?i=solve+n*m+%2B+n%28n-1%29%2F2+%3D+2000%2Cn%3E0%2Cm%3E0+over+integers

Answer 23

(2m-1)^2 +16000 shold be a perfect square

Answer 24

Oh yes the quadratic gives that condition - but thats a necessary condition - i feel the solution is going to be complex than it appeared initially !

Answer 25

mmm isnt there a a condition to how sumation from 0 should be like or some form ?

Answer 26

1+2+3+... +n is always a triangular number, so ?

Answer 27

so , whats the nearest triangular number to 2000 ?

Answer 28

62(62+1)/2 = 1953

Answer 29

63(63+1)/2 = 2016

Answer 30

mmm whats the nearest triangular number that devide 3

Answer 31

mm forget it ...its trial method

Answer 32

Alright, lets work it using basic number theory

Answer 33

n*m + n(n-1)/2 = 2000

2nm + n(n-1) = 4000

n(2m + n - 1) = 4000

Answer 34

=>

n | 4000
2m+n-1 | 4000

Answer 35

i was like trying to use this
[0+1+2+3+...+(m-1) ]+[m+(m+1)+(m+2)+...+n] = m(m-1)/2+2000
and find triangular number (s) which is above 2000
such that s-2000 is also triangle number

Answer 36

oh , diophantine equation :O

Answer 37

yep

Answer 38

but to this point we have or
n | 4000
2m+n-1 | 4000
it's not necessary that both divides 4000

Answer 39

ab = c

=>

a|c
b|c

Answer 40

hehe yeah xD
sorry im just hungry hehehe
ok continue plz

Answer 41

ok gtg nw , ill think of it ltr

Answer 42

I don't have any such method but getting m= 47

Answer 43

how ?

Answer 44

I actually wanted a algebraic sollution mine is a more lengthier method , but i doon't know diaphante... or anything similar

Answer 45

Just apply the Sum of terms in A.P formula , then you will get a quadratic when solve that
then figure out some combinations

Answer 46

oh, so trial and error...

Answer 47

yeah you can say it trial and error , but is there any method

Answer 48

not in a high school level math, if there is, even i haven't studied it

Answer 49

so @ganeshie8 lets continue from here
n(2m + n - 1) = 4000

Answer 50

\[\large n(2m + n - 1) = 2^5\times 5^3\]

Answer 51

total # of factors = (5+1)(3+1) = 24

Answer 52

so n can take these values
\(n=5\)
\(n=5^2\)
\(n=5^3\)
\(n=2×5\)
\(n=2×5^2\)
\(n=2×5^3\)
\(n=2^2×5\)
\(n=2^2×5^2\)
\(n=2^2×5^3\)
\(n=2^3×5\)
\(n=2^3×5^2\)
\(n=2^3×5^3\)
\(n=2^4×5\)
\(n=2^4×5^2\)
\(n=2^4×5^3\)
\(n=2^5×5\)
\(n=2^5×5^2\)
\(n=2^5×5^3\)

hehehe xD

Answer 53

yeah lol !
got it , continue

Answer 54

you forgot one value, n can take \(1\) too

Answer 55

u can do minimizing function right

Answer 56

callcooolus

Answer 57

will no since if n=1
then m=2000

Answer 58

callcoolus is not so cool when it comes to solving in integers danny

Answer 59

ohh right :)

Answer 60

im kind of stupid with calcules xD

Answer 61

ikram, did u get how we narrowed down to 2 cases ?

Answer 62

not realy

Answer 63

\[\large n(2m + n - 1) = 2^5\times 5^3\]

n = 2^5 or n = 5^3
we can ignore all other cases cuz n and (2m+n-1) can not be both even / both odd at the same time.

Answer 64

why ?

Answer 65

if n is even, then (2m+n-1) has to be odd, right ?

Answer 66

right ?

Answer 67

if n is odd, then (2m+n-1) has to be even, right ?

Answer 68

ok ?

Answer 69

that means, n and (2m+n-1) cannot share the factor "2", right ?

Answer 70

aha ok

Answer 71

Maybe since there are two unknowns, we just need to create one more equation. Perhaps since there's a kind of symmetry we can come up with an equation for the average of the numbers.

Answer 72

Why aren't there nonuniform operators in math? It's always every term is exactly 1 term greater than the next or division is always into equal parts instead of unequal parts.

If we could just divide this into just slightly less equal parts we could get our answer real easily lol.

Answer 73

but then mmm more cases could be ,
n =5,5^2,5^3 why u only chosed 5^3
also
n=2,2^2....

Answer 74

Well the average distance between points always has the average equal to the middle distance if there's an odd number since the one on the left is one less than the center and the one on the right has one more so they cancel out. Does this make sense or am I not really doing anything useful here.