Use Euclid’s division lemma to show that the cube. of any positive integer is of the form 9m,9m+1 or 9m+8 About the author Rylee
Let us consider a and b where a be any positive number and b is equal to 3. According to Euclid’s Division Lemma a = bq + r where r is greater than or equal to zero and less than b (0 ≤ r < b) a = 3q + r so r is an integer greater than or equal to 0 and less than 3. Hence r can be either 0, 1 or 2. Case 1: When r = 0, the equation becomes a = 3q Cubing both the sides a3 = (3q)3 a3 = 27 q3 a3 = 9 (3q3) a3 = 9m where m = 3q3 Case 2: When r = 1, the equation becomes a = 3q + 1 Cubing both the sides a3 = (3q + 1)3 a3 = (3q)3 + 13 + 3 × 3q × 1(3q + 1) a3 = 27q3 + 1 + 9q × (3q + 1) a3 = 27q3 + 1 + 27q2 + 9q a3 = 27q3 + 27q2 + 9q + 1 a3 = 9 ( 3q3 + 3q2 + q) + 1 a3 = 9m + 1 Where m = ( 3q3 + 3q2 + q) Case 3: When r = 2, the equation becomes a = 3q + 2 Cubing both the sides a3 = (3q + 2)3 a3 = (3q)3 + 23 + 3 × 3q × 2 (3q + 1) a3 = 27q3 + 8 + 54q2 + 36q a3 = 27q3 + 54q2 + 36q + 8 a3 = 9 (3q3 + 6q2 + 4q) + 8 a3 = 9m + 8 Where m = (3q3 + 6q2 + 4q)therefore a can be any of the form 9m or 9m + 1 or, 9m + 8. [tex]\huge{\tt{\red{S}\green{H}\purple{I}\pink{N}\blue{C}\orange{H}\red{A} \green{N} \: \pink{✿༉}}}[/tex] Reply
❥คɴᎦᴡєя Let us consider a and b where a be any positive number and b is equal to 3. According to Euclid’s Division Lemma a = bq + r So r is an integer greater than or equal to 0 and less than 3. Hence r can be either 0, 1 or 2. Case 1: When r = 0, the equation becomes a = 3q Cubing both the sides a3 = (3q)3 a3 = 27 q3 a3 = 9 (3q3) a3 = 9m where m = 3q3 Case 2: When r = 1, the equation becomes a = 3q + 1 Cubing both the sides a3 = (3q + 1)3 a3 = (3q)3 + 13 + 3 × 3q × 1(3q + 1) a3 = 27q3 + 1 + 9q × (3q + 1) a3 = 27q3 + 1 + 27q2 + 9q a3 = 27q3 + 27q2 + 9q + 1 a3 = 9 ( 3q3 + 3q2 + q) + 1 a3 = 9m + 1 Where m = ( 3q3 + 3q2 + q) Case 3: When r = 2, the equation becomes a = 3q + 2 Cubing both the sides a3 = (3q + 2)3 a3 = (3q)3 + 23 + 3 × 3q × 2 (3q + 1) a3 = 27q3 + 8 + 54q2 + 36q a3 = 27q3 + 54q2 + 36q + 8 a3 = 9 (3q3 + 6q2 + 4q) + 8 a3 = 9m + 8 Where m = (3q3 + 6q2 + 4q)therefore a can be any of the form 9m or 9m + 1 or, 9m + 8. Reply
Let us consider a and b where a be any positive number and b is equal to 3.
According to Euclid’s Division Lemma
a = bq + r
where r is greater than or equal to zero and less than b (0 ≤ r < b)
a = 3q + r
so r is an integer greater than or equal to 0 and less than 3.
Hence r can be either 0, 1 or 2.
Case 1: When r = 0, the equation becomes
a = 3q
Cubing both the sides
a3 = (3q)3
a3 = 27 q3
a3 = 9 (3q3)
a3 = 9m
where m = 3q3
Case 2: When r = 1, the equation becomes
a = 3q + 1
Cubing both the sides
a3 = (3q + 1)3
a3 = (3q)3 + 13 + 3 × 3q × 1(3q + 1)
a3 = 27q3 + 1 + 9q × (3q + 1)
a3 = 27q3 + 1 + 27q2 + 9q
a3 = 27q3 + 27q2 + 9q + 1
a3 = 9 ( 3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m = ( 3q3 + 3q2 + q)
Case 3: When r = 2, the equation becomes
a = 3q + 2
Cubing both the sides
a3 = (3q + 2)3
a3 = (3q)3 + 23 + 3 × 3q × 2 (3q + 1)
a3 = 27q3 + 8 + 54q2 + 36q
a3 = 27q3 + 54q2 + 36q + 8
a3 = 9 (3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m = (3q3 + 6q2 + 4q)therefore a can be any of the form 9m or 9m + 1 or, 9m + 8.
[tex]\huge{\tt{\red{S}\green{H}\purple{I}\pink{N}\blue{C}\orange{H}\red{A} \green{N} \: \pink{✿༉}}}[/tex]
❥คɴᎦᴡєя
Let us consider a and b where a be any positive number and b is equal to 3.
According to Euclid’s Division Lemma
a = bq + r
So r is an integer greater than or equal to 0 and less than 3.
Hence r can be either 0, 1 or 2.
Case 1: When r = 0, the equation becomes
a = 3q
Cubing both the sides
a3 = (3q)3
a3 = 27 q3
a3 = 9 (3q3)
a3 = 9m
where m = 3q3
Case 2: When r = 1, the equation becomes
a = 3q + 1
Cubing both the sides
a3 = (3q + 1)3
a3 = (3q)3 + 13 + 3 × 3q × 1(3q + 1)
a3 = 27q3 + 1 + 9q × (3q + 1)
a3 = 27q3 + 1 + 27q2 + 9q
a3 = 27q3 + 27q2 + 9q + 1
a3 = 9 ( 3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m = ( 3q3 + 3q2 + q)
Case 3: When r = 2, the equation becomes
a = 3q + 2
Cubing both the sides
a3 = (3q + 2)3
a3 = (3q)3 + 23 + 3 × 3q × 2 (3q + 1)
a3 = 27q3 + 8 + 54q2 + 36q
a3 = 27q3 + 54q2 + 36q + 8
a3 = 9 (3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m = (3q3 + 6q2 + 4q)therefore a can be any of the form 9m or 9m + 1 or, 9m + 8.