• Feb 8th 2008, 04:40 AM
blertta
1)Prove that if n is an odd positive integer, then
N = 2269n + 1779n + 1730n ¡ 1776n
is an integer multiple of 2001... 2)Factor the expression
30(a2 + b2 + c2 + d2) + 68ab ¡ 75ac ¡ 156ad ¡ 61bc ¡ 100bd + 87cd: 3) Prove that for every x∈R is true this equation: x^8+x^6-x^3-x+1>0,,, 4) Make the add:1^2+2^2-3^2+… +〖( n-1)〗^2+n^2 which n∈ N ,,,,5) Find the last digit of 777^777
• Feb 8th 2008, 04:52 AM
a tutor
Quote:

Originally Posted by blertta

1)Prove that if n is an odd positive integer, then
N = 2269n + 1779n + 1730n ¡ 1776n
is an integer multiple of 2001...

2)Factor the expression
30(a2 + b2 + c2 + d2) + 68ab ¡ 75ac ¡ 156ad ¡ 61bc ¡ 100bd + 87cd:

3) Prove that for every x∈R is true this equation: x^8+x^6-x^3-x+1>0,,,

4) find this sum :1^2+2^2-3^2+… +〖( n-1)〗^2+n^2 which n∈ N ,,,,

5) Find the last digit of 777^777

2) What does i represent here?
4) What's happening with the signs? Are they + + + + or + - + - ?

5) It's useful to know something (interesting?) about integers of the form $a^{4n+1}$
This is all you need if you don't have to prove it.
• Feb 8th 2008, 06:00 AM
Soroban
Hello, blertta!

Some of the problems are impossible to read.

Quote:

4) Find this sum: . $1^2+2^2-3^2+ \hdots + n^2$ where $n \in N$
There is a formula for this series: . $S_n \;=\;\frac{n(n+1)(2n+1)}{6}$

But I suppose they want to see its derivation . . .

Quote:

5) Find the last digit of $777^{777}$
We are concerned with the last digit (only) of any number.

Note the endings of consecutive powers of 7.

. . $\begin{array}{ccc} 7^1 & \to & 7 \\ 7^2 & \to & 9 \\ 7^3 & \to & 3 \\ 7^4 & \to & 1 \\ 7^5 & \to & 7 \\ \vdots & & \vdots \end{array}$

We see that $7^4$ ends in 1 ... and the sequence repeats.

We have: . $7^{777} \;=\;7^{776}\cdot 7 \;=\;(7^4)^{194}\cdot 7 \;\to \;1^{194}\cdot 7 \;=\;7$

Therefore, the last digit of $777^{777}$ is $7.$