MODULE_09
Let three consecutive positive integer’s be
n – 1, n and n + 1.
By Euclid’s Division Algorithm,
Prove that product of three consecutive positive integers is divisible by 6.
a
6q
=
+
r
where
0
<
r
<
6
∴
The possible remainders are
0,
1,
2,
3,
4,
5
∴
a
6q
=
or
6q
+
1
or
6q
+
2
or
6q
+
3
or
6q
+
4
or
6q
+
5
, where q is some integer
Case 1 :
If,
n = 6q
n
–
1
=
6q
–
1
n
+
1
=
6q
+
1
So,
(n – 1)
(n + 1)
(n)
=
(6q – 1)
(6q)
(6q + 1)
=
6
[(6q – 1)(q)(6q + 1)]
=
6
m
[ where
=
(6q – 1)
(q)
(6q + 1) ]
Here, the above result is multiple of 6.
Hence, it is divisible by 6.
m
Q.
Sol.
Case 2 :
If,
n = 6q + 1
n
–
1
=
6q + 1
–
1
n
+
1
=
6q + 1
+
1
=
=
6q
6q + 2
So,
(n – 1)
(n + 1)
(n)
=
(6q + 1)
=
6
[q(6q + 1)(6q + 2)]
=
6
m
where
=
Here, the above result is multiple of 6.
Hence, it is divisible by 6.
(6q)
(6q + 2)
m
[q(6q + 1)(6q + 2)]
Let three consecutive positive integer’s be
n – 1, n and n + 1.
By Euclid’s Division Algorithm,
a
6q
=
+
r
where
0
<
r
<
6
∴
The possible remainders are
∴
a
6q
=
or
6q
+
1
or
6q
+
2
or
6q
+
3
or
6q
+
4
or
6q
+
5
, where q is some integer
Prove that product of three consecutive positive integers is divisible by 6.
Q.
Sol.
0,
1,
2,
3,
4,
5
Case 3 :
If,
n = 6q + 2
n
–
1
=
6q + 2
–
1
n
+
1
=
6q + 2
+
1
=
=
6q + 1
6q + 3
So,
(n – 1)
(n + 1)
(n)
=
(6q + 2)
=
6
[(6q + 1)(3q + 1)(2q + 1)]
=
6
m
where
=
Here, the above result is multiple of 6.
Hence, it is divisible by 6.
(6q + 1)
(6q + 3)
m
[(6q + 1)(3q + 1)(2q + 1)]
=
(3q + 1)
(6q + 1)
(2q + 1)
(2)
(3)
Let three consecutive positive integer’s be
n – 1, n and n + 1.
By Euclid’s Division Algorithm,
a
6q
=
+
r
where
0
<
r
<
6
∴
The possible remainders are
∴
a
6q
=
or
6q
+
1
or
6q
+
2
or
6q
+
3
or
6q
+
4
or
, where q is some integer
Prove that product of three consecutive positive integers is divisible by 6.
Q.
Sol.
6q
+
5
0,
1,
2,
3,
4,
5
Case 4 :
If,
n = 6q + 3
n
–
1
=
6q + 3
–
1
n
+
1
=
6q + 3
+
1
=
=
6q + 2
6q + 4
So,
(n – 1)
(n + 1)
(n)
=
(6q + 3)
=
6
[(3q + 1)(2q + 1)(6q + 4)]
=
6
m
where
=
Here, the above result is multiple of 6.
Hence, it is divisible by 6.
(6q + 2)
(6q + 4)
m
[(3q + 1)(2q + 1)(6q + 4)]
=
(3q + 1)
(6q + 4)
(2q + 1)
(2)
(3)
Prove that product of three consecutive positive integers is divisible by 6.
Q.
Sol.
Let three consecutive positive integer’s be
n – 1, n and n + 1.
By Euclid’s Division Algorithm,
a
6q
=
+
r
where
0
<
r
<
6
∴
The possible remainders are
∴
a
6q
=
or
6q
+
1
or
6q
+
2
or
6q
+
3
or
6q
+
4
or
6q
+
5
, where q is some integer
0,
1,
2,
3,
4,
5
Case 5 :
If,
n = 6q + 4
n
–
1
=
6q + 4
–
1
n
+
1
=
6q + 4
+
1
=
=
6q + 3
6q + 5
So,
(n – 1)
(n + 1)
(n)
=
(6q + 4)
=
6
[(2q + 1)(3q + 2)(6q + 5)]
=
6
m
where
=
Here, the above result is multiple of 6.
Hence, it is divisible by 6.
(6q + 3)
(6q + 5)
m
[(2q + 1)(3q + 2)(6q + 5)]
=
(2q + 1)
(6q + 5)
(3q + 2)
(3)
(2)
Prove that product of three consecutive positive integers is divisible by 6.
Q.
Sol.
Let three consecutive positive integer’s be
n – 1, n and n + 1.
By Euclid’s Division Algorithm,
a
6q
=
+
r
where
0
<
r
<
6
∴
The possible remainders are
∴
a
6q
=
or
6q
+
1
or
6q
+
2
or
6q
+
3
or
6q
+
4
or
6q
+
5
, where q is some integer
0,
1,
2,
3,
4,
5
Case 6 :
If,
n = 6q + 5
n
–
1
=
6q + 5
–
1
n
+
1
=
6q + 5
+
1
=
=
6q + 4
6q + 6
So,
(n – 1)
(n + 1)
(n)
=
(6q + 5)
=
6
[(6q + 4)(6q + 5)(q + 1)]
=
6
m
where
=
Here, the above result is multiple of 6.
Hence, it is divisible by 6.
(6q + 4)
(6q + 6)
m
[(6q + 4)(6q + 5)(q + 1)]
=
(6q + 4)
(q + 1)
(6q + 5)
(6)
Prove that product of three consecutive positive integers is divisible by 6.
Q.
Sol.
Let three consecutive positive integer’s be
n – 1, n and n + 1.
By Euclid’s Division Algorithm,
a
6q
=
+
r
where
0
<
r
<
6
∴
The possible remainders are
∴
a
6q
=
or
6q
+
1
or
6q
+
2
or
6q
+
3
or
6q
+
4
or
6q
+
5
, where q is some integer
0,
1,
2,
3,
4,
5