Consecutive sums

Just use sum of n terms in an A.P with d=1 $\sum _{ i=a }^{ n\quad terms }{ i } =\quad (2a+(n-1))\quad \times \quad n/2\\ k=(2a+38)\times 39/2\quad \quad -\quad 38\quad consecutive\quad terms\\ k=(2b+39)\times 40/2\quad \quad \quad -\quad 40\quad consecutive\\ k=(2c+40)\times 41/2\quad ..\\ So\quad \\ 2k\quad =\quad 41\times 2(c-20)\quad =>\quad 41\times Integer\\ 2k\quad =\quad 40\times (b-39)\quad \quad =>\quad 40\times Integer\\ 2k\quad =\quad 39\times (a-19)\quad \quad =>\quad 39\times Integer\\ Smallest\quad of\quad 2k\quad =\quad G.C.D(39,40,41)\quad =\quad 63960\\ So\quad Final\quad k\quad =\quad 31980\quad [Pure\quad math\quad no\quad theory\quad ;)\quad ]$

let the number be x.

If a number can be expressed as the sum of n consecutive integers, then the number is divisible by n if n is odd, and gives remainder n/2 on dividing by n if n is even.

thus, we first need to find the LCM of 39 and 41, which is 1599.

now we need to find the smallest multiple of 1599 which gives remainder 20 on dividing by 40

.'. 1599/40 gives remainder -1

we need remainder 20 or -20

thus the number required must be 1599 x 20

.'. 31980

Let be $\begin{aligned}n~~~&=a+(a+1)+\cdots+(a+38)&=39a+38·39\,/\,2\\&=b+(b+1)+\cdots+(b+38)+(b+39)&=40b+39·40\,/\,2\\&=c+(c+1)+\cdots+(c+38)+(c+39)+(c+40)\!\!\!\!\!\!\!\!\!\!&=41c+40·41\,/\,2\end{aligned}$

This gives the modular system $\begin{cases}n\equiv0&\text{(mod 39)}\\n\equiv20&\text{(mod 40)}\\n\equiv0&\text{(mod 41)}\end{cases}$

Using chinese reminder theorem, we find $n\equiv31,980 \text{ (mod (}39\cdot40\cdot41\text{))}$ . When $\boxed{n=31,980}$ , we have $a=801,~b=780,~c=760$ , which are positive values.

Verification:

$\begin{aligned}801+802+\cdots+839&=31,980\\780+781+782+\cdots+819&=31,980\\760+761+762+763+\cdots+800&=31,980\end{aligned}$

For an odd number of consecutive integers, their sum would just be the product of that odd quantity with the middlemost integer in that sequence.

For an even number of consecutive integers, their sum would just be the product of HALF of that even quantity with the sum of the two middlemost integers in that sequence.

N = 39x = 41y = 20(a + b)

Since the above are all equal to N, than this number N must have factors of 39, 41 and 20 in it.

N = LCM (20, 39 & 41)
= 20 x 39 x 41
= 31980
because all three factors are co-prime to each other.

Note
Actually, the general formula for my first two statements is
S(n) = n(A + L) / 2
n : quantity of numbers in a particular sequence
A : the first number in that sequence
L : the last number in that sequence

Say the number is x. x = a + (a+1) + (a+2) + .... + (a+38) = 39a + 38 39/2 = 39(a + 19). Similarly, x = 40b + 39 40/2 = 20(2b + 39) and x = 41c + 40*41/2 = 41(c + 20).

Hence, 39, 41 and 20 are all factors of x. The smallest such number is 39x41x20 = 31980. a = 801 b = 780 c = 760