Question

show that only one out of n,n+4,n+8,n+12,n+16 is divisible by 5 , where n is a positive integer.

Answer 1

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

Answer 2

Answer:

By Euclid's division lemma, every positive integer 'a' is of the form bq + r, 0<= r < b

Thus, for b = 5, every positive integer is of the form 5q + r, 0 <= r < 5

i.e. Every positive integer is of the form 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4

Here, our positive integer is 'n'

CASE - 1: When n = 5q

Then, only n is divisible by 5 and n+4, n+8, n+12, n+16 are not.

CASE - 2: When n = 5q + 1

Then, only n+4 = 5q+5 = 5(q+2) is divisible by 5 and rest are not.

CASE - 3: When n = 5q + 2

Then, only n+8 = 5q+10 = 5(q+2) is divisible by 5

CASE - 4: When n = 5q + 3

Then, only n+12 = 5q+15 = 5(q+3) is divisible by 5

CASE - 5: When n = 5q + 4

Then, only n+16 = 5q+20 = 5(q+4) is divisible by 5

From the above 5 cases we can conclude that one and only one out of n,n+4,n+8,n+12,n+16 is divisible by 5