This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 8: Math 3360 Applicable Algebra Chapter 12: Chinese Remainder Theorem 19. There is an unknown number of objects. When counted by threes the remainder is 2; when counted by fives the remainder is 3; and when counted by sevens the remainder is 2. How many objects are there? Solution: This gives us the system of congruence equations: x ≡ 2( mod 3) x ≡ 3( mod 5) x ≡ 2( mod 7) Thus x = 2 + 3 k , x = 3 + 5 i and x = 2 + 7 j for some integers i,j,k . Combining the first two equations we get that 3 k ≡ 1( mod 5), i.e. k ≡ 2( mod 5), so k = 2 + 5 m . Combining the first and last we get that 3 k ≡ 0( mod 7), i.e. k ≡ 0( mod 7), so k = 7 n . Taking our two equations find that 5 m ≡ 5( mod 7), so m ≡ 1( mod 7). Setting m = 1 we get that k = 7, and hence x = 23. Checking this modulo 3 , 5 , 7 we see that this is a solution. 44. Find the least nonnegative residue of 100 246 ( mod 247). Solution: We know that 247 = 13 · 19, so let us examine this modulo 13 and 19. We know that 100 ≡ 9( mod 13) and 100 ≡ 5( mod 19). Then using Euler’s theorem we get that 9 12 ≡ 1( mod 12) and 5 18 ≡ 1( mod 19). Applying this and multiplying out we see that 100 246 ≡ 9 246 ( mod 13) 100 246 ≡ 5 246 ( mod 19) ≡ (9 12 ) 20 9 6 ( mod 13) ≡ (5 12 ) 13 5 12 ( mod 19) ≡ 9 6 ( mod 13) ≡ 5 12 ( mod 19) ≡ 1( mod 13) ≡ 11( mod 19) Then we get that for x = 100 246 then x = 1+13 k and x = 11+19 i for integers i,k . Combining these we get that 13 k ≡ 10( mod 19), so then k ≡ 30( mod 19) ≡ 11( mod 19). Then substituting back in we get that x = 144. 49. Write down the elements of Z / 10 Z and of Z / 2 Z × Z / 5 Z and identify which elements correspond under the map ψ from Z / 10 Z to Z / 2 Z × Z / 5 Z . Solution: ψ ([0] 10 ) ≡ ([0] 2 , [0] 5 ) ψ ([5] 10 ) ≡ ([1] 2 , [0] 5 ) ψ ([1] 10 ) ≡ ([1] 2 , [1] 5 ) ψ ([6] 10 ) ≡ ([0] 2 , [1] 5 ) ψ ([2] 10 ) ≡ ([0] 2 , [2] 5 ) ψ ([7] 10 ) ≡ ([1] 2 , [2] 5 ) ψ ([3] 10 ) ≡ ([1] 2 , [3] 5 ) ψ ([8] 10 ) ≡ ([0] 2 , [3] 5 ) ψ ([4] 10 ) ≡ ([0] 2 , [4] 5 ) ψ ([9] 10 ) ≡ ([1] 2 , [4] 5 ) 52. Examine the map ψ : Z / 24 Z → Z / 6 Z × Z / 4 Z given by ψ ([ a ] 24 ) = ([ a ] 6 , [ a ] 4 ). What is the kernel of ψ ? Which elements of Z / 6 Z × Z / 4 Z are in the image of ψ ? Solution: Writing out the entire map we see ψ ([0] 24 ) ≡ ([0] 6 , [0] 4 ) ψ ([12] 24 ) ≡ ([0] 6 , [0] 4 ) ψ ([1] 24 ) ≡ ([1] 6 , [1] 4 ) ψ ([13] 24 ) ≡ ([1] 6 , [1] 4 ) ψ ([2] 24 ) ≡ ([2] 6 , [2] 4 ) ψ ([14] 24 ) ≡ ([2] 6 , [2] 4 ) ψ ([3] 24 ) ≡ ([3] 6 , [3] 4 ) ψ ([15] 24 ) ≡ ([3] 6 , [3] 4 ) ψ ([4] 24 ) ≡ ([4] 6 , [0] 4 ) ψ...
View
Full
Document
 Spring '08
 BILLERA
 Algebra, Remainder Theorem, Remainder, Prime number

Click to edit the document details