happy21 said:
skipjack said:
It's simpler to use the original equation. For x = 2 or x > 3, the LHS ends in 3, so it cannot be N².
Okay, this is fine. Also, can you please point me to the link which proves that if a natural number ends in 2, 3, 7 or 8 it can never be a perfect square, so that I can give that too to the students?
As a matter of interest, can the following be clarified..., the same thing:
(1+3!/2 + 4!/2 + .... + x!/2)=2M, for x>1 which how serves the same result (like previously depicted), which I was not able to get...
But anyways, it (the reason of '3' in the end!) works.
Thanks.
Let N be some integer, form of N=10A+n, A is an integer and n is last digit of N. (n can go from 0 to 9)
Then N² is :
N² =(10A+n)²= 100A²+20An+n²=10(10A²+2Aa)+n²
We can see that last digit of N² depends on last digit of n², where n?<0,9> and n is natural.
for n=1, n²=1 last digit of N is 1
for n=2 n²=4 last digit of N is 4
for n=3, n²=9 last digit of N is 9
for n=4, n²=16 last digit of N is 6
for n=5, n²=25 last digit of N is 5
for n=6, n²=36 last digit of N is 6
for n=7, n²=49 last digit of N is 9
for n=8, n²=64 last digit of N is 4
for n=9, n²=81 last digit of N is 1
for n=0, n²=0 last digit of N is 0
possible last digits OF N² are : 0,1,4,9,6 and 5.