Advertisements
Advertisements
प्रश्न
For any positive integer n , prove that n3 − n divisible by 6.
उत्तर
We have n3 − n = n(n2 − 1) = (n − 1) (n) (n + 1)
Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or, 6q + 2 or, 6q + 3 or, 6q + 4 or, 6q + 5.
If n = 6q, then
(n − 1)(n)(n + 1) = (6q − 1)(6q)(6q + 1)
= 6[(6q − 1)(q)(6q + 1)]
= 6m, which is divisible by 6
If n = 6q + 1, then
(n − 1)(n + 1) = (6q)(6q + 1)(6q + 2)
= 6[(q)(6q + 1)(6q + 2)]
= 6m, which is divisible by 6
If n = 6q + 2, then
(n − 1)(n)(n + 1) = (6q + 1)(6q + 2)(6q + 3)
= 6[(6q + 1)(3q + 1)(2q + 1)]
= 6m, which is divisible by 6
If n = 6q + 3, then
(n − 1)(n)(n + 1) = (6q + 3)(6q + 4)(6q + 5)
= 6[(3q + 1)(2q + 1)(6q + 4)]
= 6m, which is divisible by 6
If n = 6q + 4, then
(n − 1)(n)(n + 1) = (6q + 3)(6q + 4)(6q + 5)
= 6[(2q + 1)(3q + 2)(6q + 5)]
= 6m, which is divisible by 6
If n = 6q + 5, then
(n − 1)(n)(n + 1) = (6q + 4)(6q + 5)(6q + 6)
= 6[(6q + 4)(6q + 5)(q + 1)]
= 6m, which is divisible by 6
Hence, for any positive integer n, n3 – n is divisible by 6.
APPEARS IN
संबंधित प्रश्न
Show that every positive integer is of the form 2q and that every positive odd integer is of the from 2q + 1, where q is some integer.
Prove that the square of any positive integer of the form 5q + 1 is of the same form.
Define HOE of two positive integers and find the HCF of the following pair of numbers:
155 and 1385
Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
Without actual division show that each of the following rational numbers is a non-terminating repeating decimal.
(i)`77/210`
In Q.No. 7, HCF (a, b) is
The HCF of 95 and 152, is
If p and q are co-prime numbers, then p2 and q2 are
A positive integer, when divided by 88, gives the remainder 61. What will be the remainder when the same number is divided by 11?
Prove that n2 – n divisible by 2 for every positive integer n
