# summation of consecutive natural numbers does not end in 7,4,2,9

I calculated sum of n consecutive natural numbers where n = 1 to 100 .What I mean is $$\sum_{n=1}^{1}n = 1$$ $$\sum_{n=1}^{2}n = 3$$ $$\sum_{n=1}^{3}n = 6$$

And I got answers and noticed that none of the summation answers ended in digits $7,4,2,9$

I verified with the code in python for the first 100 numbers


for n in range(1,100):
a =0
for i in range(1,n+1):
a = a + i
print a%10


Why is it so ? I just checked for 100 numbers . Does this hold good for any number n ? If so, can we prove this ?

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Yes, this is correct. Note that $8\cdot 1+1=9$, $8\cdot 3+1=25$ and $8\cdot 6+1=49$. In general, if $S$ is your sum, then $8S+1$ is a perfect square. So no $S$ ends in those digits because no square ends in $3$ or $7$. – Thomas Andrews Jul 24 '13 at 22:26
Thanks Thomas Andrews . Your comment was very useful to solve this problem. – Harish Kayarohanam Jul 24 '13 at 23:04

The numbers you're computing are called triangular numbers, and they are of the form:

$$\frac{n(n+1)}{2}$$

Suppose we have:

$$\frac{n(n+1)}{2}=10m+k$$

This tells us that the quantity $n(n+1)$ is congruent to $2k$ modulo $20$. So our task is to show that $n(n+1)$ is never congruent to $4,8,14,$ or $18$ modulo $20$. This you can check on a case by case basis, by letting $n$ range from $0$ to $19$.

Or, using the clever hint given by Thomas Andrews in the comments, notice that

$$8\frac{n(n+1)}{2}+1=4n^2+4n+1=(2n+1)^2=80m+8k+1$$

With $k=2,4,7,9$, it would follows that $(2n+1)^2$ ends in a $7$ or $3$, which is impossible.

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Amazing answer .Thanks . Your proof along with Thomas Andrews comment is really helpful . – Harish Kayarohanam Jul 24 '13 at 22:43
Tiny bit of cleanup: you can in fact just work mod 10 instead of mod 20, since '4, 8, 14, or 18 mod 20' is exactly equivalent to '4 or 8 mod 10'. – Steven Stadnicki Jul 24 '13 at 23:21

HINT: $$\sum_{n=1}^{r}n = n(n-1)/2$$

Now you are asking that why $n(n-1)/2$ does not end with digits like 2 ,4 ,7,9

Product of two consecutive Natural numbers always ends with 0,2,6 ( Intuitive Prove!)
So $n(n-1)/2$ will have digits ending with 0,1,3,5 . (where 5 comes in the case where the last two digits are 1 and 0 )

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Thanks Simar .I am getting your point. Your answer was helpful. can you please edit it and make it clear .Point is as 0,2,6 are only ending digits , possible ending digits after division by 2 ,are 0 or 5(for 0 as we can have 0 or 10) , 1 or 6 (for 2 as we can have 2 or 12) , 3 or 8 (for 6 as we can have 6 or 16) . so possible ending digits are 1,3,5,6,8 . so left out digits are 2,4,7,9. – Harish Kayarohanam Jul 24 '13 at 22:58