SOLUTION: Prove that the value of the expression is not divisible by 6 for any whole n: (2n+1)(n+5)–2(n+3)–(5n+13).

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Prove that the value of the expression is not divisible by 6 for any whole n: (2n+1)(n+5)–2(n+3)–(5n+13).      Log On


   



Question 1129675: Prove that the value of the expression is not divisible by 6 for any whole n:
(2n+1)(n+5)–2(n+3)–(5n+13).

Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

%282n%2B1%29%28n%2B5%29+-2%28n%2B3%29+-+%285n%2B13%29
Hint: an integer is divisible by 6+if and only if it is divisible by both 2 and 3.
Can you show that your expression is even,divisible by both 2 and 3 ?
first simplify it:
%282n%2B1%29%28n%2B5%29+-2%28n%2B3%29+-+%285n%2B13%29
2n%5E2%2B10n%2Bn%2B5+-2n+-6+-+5n-13
2n%5E2%2B11n+%2B5+-7n+-6+-13
2n%5E2+%2B4n%2B5-19
2+n%5E2+%2B+4+n+-+14
2+%28n%5E2+%2B+2+n+-+7%29 => GFC is 2;
so,it is divisible by 2,
but it is not divisible by 3
=>consequently, it is not divisible by 6 either

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The proof by the other tutor is not complete. She shows that the given expression in simplified form is equivalent to

2%28n%5E2%2B2n-7%29

and then states without proof that the expression is divisible by 2 but not by 3.

But to complete the proof that the expression is not divisible by 6 for ANY whole number, we have to prove that the factor n%5E2%2B2n-7 is NEVER divisible by 3.

We do that with modular arithmetic.

For any whole number n, there are 3 possible values for n, mod 3. We need to show that in all three cases the factor n%5E2%2B2n-7 is not divisible by 3 -- that is, that it is not equal to 0 mod 3.

(1) n = 0 mod 3: n^2+2n-7 mod 3 = 0+0-7 = -7 mod 3 (not = 0 mod 3)
(2) n = 1 mod 3: n^2+2n-7 mod 3 = 1+2-7 = -4 mod 3 (not = 0 mod 3)
(3) n = 2 mod 3: n^2+2n-7 mod 3 = 1+1-7 = -5 mod 3 (not = 0 mod 3)

The proof is now complete, because we have shown that, although the expression is always divisible by 2, it is never divisible by 3; and therefore it is never divisible by 6.