If-5-doesn-t-divide-any-of-n-n-1-n-2-n-3-then-prove-that-n-n-1-n-2-n-3-24-mod100-

Question Number 14614 by RasheedSoomro last updated on 03/Jun/17

If 5 {doesn}^{'} t divide any of n, n + 1,

n + 2, n + 3 then prove that

n (n + 1) (n + 2) (n + 3) \equiv 24 (\mod 100)

Commented by mrW1 last updated on 03/Jun/17

n {doesn}^{'} t divide n ?

h o w i s t h i s t o u n d e r s t a n d ?

Commented by RasheedSoomro last updated on 03/Jun/17

Sorry for error sir . Corrected now .

Answered by mrW1 last updated on 03/Jun/17

n = 5 k + r w i t h 1 ⩽ r ⩽ 4

n + 1 = 5 k + r + 1

n + 2 = 5 k + r + 2

n + 3 = 5 k + r + 3

s i n c e r + 1 \neq 5, r + 2 \neq 5, r + 3 \neq 5

\Rightarrow r = 1

n = 5 k + 1

n + 1 = 5 k + 2

n + 2 = 5 k + 3

n + 3 = 5 k + 4

P = n (n + 1) (n + 2) (n + 3) =

(5 k + 1) (5 k + 4) (5 k + 2) (5 k + 3)

= (25 k^{2} + 25 k + 4) (25 k^{2} + 25 k + 6)

= {(25 k^{2} + 25 k)}^{2} + 10 (25 k^{2} + 25 k) + 24

= 2100 {[\frac{k (k + 1)}{2}]}^{2} + 1000 [\frac{k (k + 1)}{2}] + 24

= 2100 m^{2} + 1000 m + 24

= 100 m (21 m + 10) + 24

w i t h m = \frac{k (k + 1)}{2}

s i n c e o n e o f k a n d k + 1 m u s t b e e v e n,

m i s i n t e g e r .

\Rightarrow P \equiv 24 (m o d 100)

Commented by RasheedSoomro last updated on 04/Jun/17

T hanks S ir!

\boldsymbol T hanks \boldsymbol S ir!