# 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.

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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.

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Solution:
By Euclid’s division lemma, we have
a = bq + r; 0 ≤ r < b
For a = n and b = 5, we have
n = 5q + r …(i)
Where q is an integer
and 0 ≤ r < 5, i.e. r = 0, 1, 2, 3, 4.
Putting r = 0 in (i), we get
n = 5q
=> n is divisible by 5.
n + 4 = 5q + 4
=> n + 4 is not divisible by 5.
n + 8 = 5q + 8
=> n + 8 is not divisible by 5.
n + 12 = 5q + 12
=> n + 12 is not divisible by 5.
n + 16 = 5q + 16
=> n + 16 is not divisible by 5.
Putting r = 1 in (i), we get
n = 5q + 1
=> n is not divisible by 5.
n + 4 = 5q + 5 = 5(q + 1)
=> n + 4 is divisible by 5.
n + 8 = 5q + 9
=> n + 8 is not divisible by 5.
n + 12 = 5q + 13
=> n + 12 is not divisible by 5.
n + 16 = 5q + 17
=> n + 16 is not divisible by 5.
Putting r = 2 in (i), we get
n = 5q + 2
=> n is not divisible by 5.
n + 4 = 5q + 9
=> n + 4 is not divisible by 5.
n + 8 = 5q + 10 = 5(q + 2)
=> n + 8 is divisible by 5.
n + 12 = 5q + 14
=> n + 12 is not divisible by 5.
n + 16 = 5q + 18
=> n + 16 is not divisible by 5.
Putting r = 3 in (i), we get
n = 5q + 3
=> n is not divisible by 5.
n + 4 = 5q + 7
=> n + 4 is not divisible by 5.

n + 8 = 5q + 11
=> n + 8 is not divisible by 5.
n + 12 = 5q + 15 = 5(q + 3)
=> n + 12 is divisible by 5.
n + 16 = 5q + 19
=> n + 16 is not divisible by 5.
Putting r = 4 in (i), we get
n = 5q + 4
=> n is not divisible by 5.
n + 4 = 5q + 8
=> n + 4 is not divisible by 5.
n + 8 = 5q + 12
=> n + 8 is not divisible by 5.
n + 12 = 5q + 16
=> n + 12 is not divisible by 5.
n + 16 = 5q + 20 = 5(q + 4)
=> n + 16 is divisible by 5.
Thus for each value of r such that 0 ≤ r < 5 only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.