210test3key - M ath 210 Section O il: Test #3 NAME: _ -- __...

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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 non-inductive proof, as long as your proof is c orrect.) (n+l t 2 + /; = 3n - / ISO or n - 1 ^ = 2-3-i =<S-^ = fl, -- 2 - S " - J = 1 0-* - tfs = 2-<? - i - tt-i -- 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 non-recursive 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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210test3key - M ath 210 Section O il: Test #3 NAME: _ -- __...

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