Math, asked by raghavi09092005, 1 year ago

show that one and only one out of n,n+4,n+8,n+12,n+16 is divisible by 5 where my is any positive integer​

Answers

Answered by tharunstar85
0

Answer:

hey buddy here's your answer

let n be 1

n=1

n+4=4+1 = 5

n+8=1+8 = 9

n+12 = 1+12 = 13

n+16= 1+16=17

so if their sum is divisible by 5 then the formula is true........

1+5+9+13+17=45

as 45 is divisible by 5 then the formula holds true!

HOPE IT HELPED!!!PLZ BRAINLIEST AND I ALSO FOLLOW YOU IF YOU MARK AS THE BRAINLIEST^_^

Answered by llTheUnkownStarll
1

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.

 \fbox \blue{Hence Proved}

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