Question Number 14614 by RasheedSoomro last updated on 03/Jun/17
$$\mathrm{If}\:\:\mathrm{5}\:\:\mathrm{doesn}'\mathrm{t}\:\mathrm{divide}\:\mathrm{any}\:\mathrm{of}\:\mathrm{n},\mathrm{n}+\mathrm{1}, \\ $$$$\mathrm{n}+\mathrm{2},\mathrm{n}+\mathrm{3}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)\equiv\mathrm{24}\left(\mathrm{mod100}\right) \\ $$
Commented by mrW1 last updated on 03/Jun/17
$$\mathrm{n}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{divide}\:\mathrm{n}? \\ $$$${how}\:{is}\:{this}\:{to}\:{understand}? \\ $$
Commented by RasheedSoomro last updated on 03/Jun/17
$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{error}\:\mathrm{sir}.\:\mathrm{Corrected}\:\mathrm{now}. \\ $$
Answered by mrW1 last updated on 03/Jun/17
![n=5k+r with 1≤r≤4 n+1=5k+r+1 n+2=5k+r+2 n+3=5k+r+3 since r+1≠5,r+2≠5,r+3≠5 ⇒r=1 n=5k+1 n+1=5k+2 n+2=5k+3 n+3=5k+4 P=n(n+1)(n+2)(n+3)= (5k+1)(5k+4)(5k+2)(5k+3) =(25k^2 +25k+4)(25k^2 +25k+6) =(25k^2 +25k)^2 +10(25k^2 +25k)+24 =2100[((k(k+1))/2)]^2 +1000[((k(k+1))/2)]+24 =2100m^2 +1000m+24 =100m(21m+10)+24 with m=((k(k+1))/2) since one of k and k+1 must be even, m is integer. ⇒P≡24 (mod 100)](https://www.tinkutara.com/question/Q14718.png)
$${n}=\mathrm{5}{k}+{r}\:{with}\:\mathrm{1}\leqslant{r}\leqslant\mathrm{4} \\ $$$${n}+\mathrm{1}=\mathrm{5}{k}+{r}+\mathrm{1} \\ $$$${n}+\mathrm{2}=\mathrm{5}{k}+{r}+\mathrm{2} \\ $$$${n}+\mathrm{3}=\mathrm{5}{k}+{r}+\mathrm{3} \\ $$$${since}\:{r}+\mathrm{1}\neq\mathrm{5},{r}+\mathrm{2}\neq\mathrm{5},{r}+\mathrm{3}\neq\mathrm{5} \\ $$$$\Rightarrow{r}=\mathrm{1} \\ $$$${n}=\mathrm{5}{k}+\mathrm{1} \\ $$$${n}+\mathrm{1}=\mathrm{5}{k}+\mathrm{2} \\ $$$${n}+\mathrm{2}=\mathrm{5}{k}+\mathrm{3} \\ $$$${n}+\mathrm{3}=\mathrm{5}{k}+\mathrm{4} \\ $$$${P}={n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)= \\ $$$$\left(\mathrm{5}{k}+\mathrm{1}\right)\left(\mathrm{5}{k}+\mathrm{4}\right)\left(\mathrm{5}{k}+\mathrm{2}\right)\left(\mathrm{5}{k}+\mathrm{3}\right) \\ $$$$=\left(\mathrm{25}{k}^{\mathrm{2}} +\mathrm{25}{k}+\mathrm{4}\right)\left(\mathrm{25}{k}^{\mathrm{2}} +\mathrm{25}{k}+\mathrm{6}\right) \\ $$$$=\left(\mathrm{25}{k}^{\mathrm{2}} +\mathrm{25}{k}\right)^{\mathrm{2}} +\mathrm{10}\left(\mathrm{25}{k}^{\mathrm{2}} +\mathrm{25}{k}\right)+\mathrm{24} \\ $$$$=\mathrm{2100}\left[\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} +\mathrm{1000}\left[\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}\right]+\mathrm{24} \\ $$$$=\mathrm{2100}{m}^{\mathrm{2}} +\mathrm{1000}{m}+\mathrm{24} \\ $$$$=\mathrm{100}{m}\left(\mathrm{21}{m}+\mathrm{10}\right)+\mathrm{24} \\ $$$${with}\:{m}=\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${since}\:{one}\:{of}\:{k}\:{and}\:{k}+\mathrm{1}\:{must}\:{be}\:{even}, \\ $$$${m}\:{is}\:{integer}. \\ $$$$\Rightarrow{P}\equiv\mathrm{24}\:\left({mod}\:\mathrm{100}\right) \\ $$
Commented by RasheedSoomro last updated on 04/Jun/17
$$\boldsymbol{\mathrm{T}}\mathrm{hanks}\:\boldsymbol{\mathrm{S}}\mathrm{ir}! \\ $$