Question

prove that one and one only out of n, n+4 , n+8, n+12, n+16 is divisible by 5 where n is postie integer

Answer 1

Answer:

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Explanation:

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.