Express that one and only one …

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Sia ? 6 years, 6 months ago
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.
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