Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.4k views
in Number System by (58.8k points)

For any positive integer n, prove that n3 – n is divisible by 6.

1 Answer

+1 vote
by (44.3k points)
selected by
 
Best answer

Let s be any positive integer.

On dividing s by 6, let m be the quotient and r be the remainder.

By Euclid’s division lemma,

s = 6m + r, where 0 ≤ r ˂ 6

We can say that,

Any positive integer is of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 for some positive integer n.

Case 1: When n = 6m

n3 – n = (6m)3 – 6m = 216 m3 – 6m = 6m(36m2 – 1) = 6q, where q = m(36m2 -1)

n3 – n is divisible by 6

Case 2: When n = 6m + 1

n3 – n = n(n2 – 1) = n (n – 1) (n + 1) = (6m + 1) (6m) (6m + 2) = 6m(6m + 1) (6m + 2)

= 6q,

where q = m(6m + 1) (6m + 2)

n3 – n is divisible by 6

Case 3: When n = 6m + 2

n– n = n (n – 1) (n + 1) = (6m + 2) (6m + 1) (6m + 3) = (6m + 1) (36 m2 + 30m + 6)

= 6[m (36m2 + 30m + 6)] + 6 (6m2 + 5m + 1)

= 6p + 6q, where p = m (36m2 + 30m + 6) and q = 6m2 + 5m + 1

n3 – n is divisible by 6

Case 4: When n = 6m + 3

n3 – n = (6m + 3)3 – (6m + 3) = (6m + 3) [(6m + 3)2 – 1] = 6m [6m + 3)2 – 1] + 3 [(6m + 3)2 – 1]

= 6 [m [(6m + 3)2 – 1] + 6 [18m2 + 18m + 4]

= 6p + 3q,

where p = m[(6m + 3)2 – 1]

q = 18m2 + 18m + 4

n3 – n is divisible by 6

Case 5: When n = 6m + 4

n3 – n = (6m + 4)3 – (6m + 4) = (6m + 4) [(6m + 4)2 – 1]

= 6m [(6m + 4)2 – 1] + 4 [(6m + 4)2 – 1]

= 6m [(6m + 4)2 – 1] + 4 [36m2 + 48m + 16 – 1]

= 6m [(6m + 4)2 – 1] + 12 [12m2 + 16m + 5]

= 6p + 6q,

where p = m [(6m + 4)2 – 1]

q = 2 (12 m2 + 16m + 5)

n3 – n is divisible by 6

Case 6: When n = 6m + 5

n3 – n = (6m + 5) [(6m + 5)2 – 1] = 6m [(6m + 5)2 – 1] + 5 [(6m + 5)2 – 1]

= 6p + 30q

= 6 (p + 5q),

where p = m [(6m + 5)2 – 1]

q = 6m2 + 10m + 4

n3 – n is divisible by 6

Therefore, n3 – n is divisible by 6, for any positive integer n.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...