PI
Show that one and only one out of n, n +4, n +8, n +12 and n + 16 is divisible by 5 where
n is any positive integer.
Answers
Answer:
Consider the positive integer is of the form
5q,5q+1,5q+2,5q+3......
Here
b=5
r=0,1,2,3,4,
Where r=0,thenn=5q
Now, n=5q is divisible by 5
n+4=5q+4[not divisible by 5]
n+8=5q+8[not divisible by 5]
n+6=5q+6[not divisible by 5]
n+12=5q+12[not divisible by 5]
Where r=1,n=5q+1
n=5q+1
n+4=5q+5[divisible by 5]
n+8=5q+9[not divisible by 5]
n+6=5q+7[not divisible by 5]
n+12=5q+13[not divisible by 5]
Where r=2,n=5q+2
n=5q+2
n+4=5q+6[not divisible by 5]
n+8=5q+10[divisible by 5]
n+6=5q+8[not divisible by 5]
n+12=5q+14[not divisible by 5]
Where r=3,n=5q+3
n=5q+3
n+4=5q+7[not divisible by 5]
n+8=5q+11[not divisible by 5]
n+6=5q+9[not divisible by 5]
n+12=5q+15[ divisible by 5]
When,
r=4,n=5q+4
n=5q+4
n+4=5q+8[not divisible by 5]
n+8=5q+12[not divisible by 5]
n+6=5q+10[ divisible by 5]
n+12=5q+16[not divisible by 5]
From 1, 2, 3, 4, 5 Its clear that, one and only one out of n,n+4,n+12,n+6 is divisible by 5
Hence, this is the answer.
Solution:
According to Euclid's division Lemma,
Let the positive integer = n, b=5
n = 5q+r, where q is the quotient and r is the remainder
0 < r < 5 implies remainders may be 0, 1, 2, 3, 4 and 5
Therefore, n may be in the form of 5q, 5q+1, 5q+2, 5q+3, 5q+4
So, this gives us the following cases:
CASE 1:
When, n = 5q
n+4 = 5q+4
n+8 = 5q+8
n+12 = 5q+12
n+16 = 5q+16
Here, n is only divisible by 5
CASE 2:
When, n = 5q+1
n+4 = 5q+5 = 5(q+1)
n+8 = 5q+9
n+12 = 5q+13
n+16 = 5q+17
Here, n + 4 is only divisible by 5
CASE 3:
When, n = 5q+2
n+4 = 5q+6
n+8 = 5q+10 = 5(q+2)
n+12 = 5q+14
n+16 = 5q+18
Here, n + 8 is only divisible by 5
CASE 4:
When, n = 5q+3
n+4 = 5q+7
n+8 = 5q+11
n+12 = 5q+15 = 5(q+3)
n+16 = 5q+19
Here, n + 12 is only divisible by 5
CASE 5:
When, n = 5q+4
n+4 = 5q+8
n+8 = 5q+12
n+12 = 5q+16
n+16 = 5q+20 = 5(q+4)
Here, n + 16 is only divisible by 5
So, we can conclude that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.