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Unformatted text preview: M ath 210 Section O il: Test #3
NAME: _  __
This test has 6 pages. The points per page are 4, 4, 5, 5, 5, 5. [2 points] Question la.
Find all .solutions of the congruence 2.x = 18 ( mod 5 0), w here 0 < x < 50. = 15 a f
Iv ~ Q* fi/i^o^ 25 )
—/ [2 points] Question I b. Find all solutions of the congruence 5x = 1 (mod 11), where 0 < x < 11. [4 points] Question 2.
Find all solutions of t he system of congruences
',' ) 2x 4 3?y = 0
3.x + 5(/ = 6 (mod 7)
(mod 7) whore 0 < x < 1 and 0 < y < 7. =0 3 = 12, sS I = 0 ^w 7 ) S 3 [5 p oints] Question 3.
Fiiid all solutions of the; system of congruences
(j) x=3 (mod 8) / ', } x=7 (mod 9) where 0 < x < 72. oc =
=7 Sul ^i
^ 10
=5 5o oc [5 points] Question 4. Prove t hat n'4 + In is divisible by 3 for all n atural numbers n.
(You may give either an inductive or noninductive proof, as long as your
proof is c orrect.) (n+l t 2 + /; = 3n  / ISO
or n  1 ^ = 23i =<S^ =
fl,  2  S "  J = 1 0*  tfs = 2<?  i  tti  n
Questions. ** = 2'?7  / = 3f  i =33 ^ = 2  3 3  Jf = 66/ = 69 Consider the following recursively defined sequence. Let 0,1 = 3, and let
r?.n = 2 a n _i — 1 for all n > 2, [1 point] (i) Give the values of a 2l a;$, ^,4, a5, and a,6[2 points] ( ii) Guess a nonrecursive formula for a n .
[2 points] ( iii) Use induction to prove that your formula from (ii) is correct. (i) O'O
se case. a  2' +1 " 3 T O
^ ± [5 points] Question 6.
Use i nduction to prove that n\ ?r! for all n > 6.
(Recall t hat n\s 1 x 2 x 3 x • • • x 71.)
To save you t ime with computations. I include sonic values of n\d nA.
Note t hat, t here's a very good reason I told you to start w ith n — G.
5
n 12 3 4
G
n\3 1 2 6 24 120 720
1 8 27 64 125 216 720 > n = 2 Xf ase sf: A SSUME i I f.e . (n i ]
n h ! > ^ 3 fir , > > (t\ i )
// V\ + . .3 Seme ...
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This note was uploaded on 12/07/2011 for the course MATH 210 taught by Professor Staff during the Spring '08 term at University of Delaware.
 Spring '08
 Staff
 Congruence

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