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Question Number 15292 by RasheedSoomro last updated on 09/Jun/17

Light version of Q#13724 Expansion of 100! has 24, 0′s at the end. Find the first non-zero digit from right. 100!=.....d000...00 What is the value of d?

$$\mathrm{Light}\:\mathrm{version}\:\mathrm{of}\:\mathcal{Q}#\mathrm{13724} \\ $$$$\mathcal{E}\mathrm{xpansion}\:\mathrm{of}\:\mathrm{100}!\:\mathrm{has}\:\mathrm{24},\:\mathrm{0}'\mathrm{s}\:\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{first}\:\mathrm{non}-\mathrm{zero}\:\mathrm{digit}\:\mathrm{from}\:\mathrm{right}. \\ $$$$\mathrm{100}!=.....\mathrm{d000}...\mathrm{00} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{d}? \\ $$

Answered by RasheedSoomro last updated on 11/Jun/17

100! has 24 0′s from right. p=((100!)/(10^(24) )) has no zero at the end. ∴ First non-zero digit of 100! =unit digit of p ∴ p≡d(mod 10) Simplifying p=((100!)/(10^(24) ))=((100!)/(2^(24) ×5^(24) )) 10^(24) =2^(24) ×5^(24) =2^(6•) (5.10.15...100) All the 24 5′s have come in 5.10...100 ^• Also some 2′s have come: One 2 in each multiple of 10 10 2′s One 2 extra in each multiple of 20 5 2′s One 2 extra in each multiple of 40 2 2′s One 2 extra in each multiple of 80 1 2′s _(−) Total(included in 5.10...100) 18 2′s Remaining 24−18=6 So now p=((100!)/(10^(24) ))=((100!)/(2^(24) ×5^(24) ))=((100!)/(2^6 (5.10....100))) p=(1/(2.2^5 )){(1.2.3.4)(6.7.8.9)...(96.97.98.99} p={(((1.2.3.4)/2))(6.7.8.9).(11.12.13.14)(((16.17.18.19)/2))} ×{(21.22.23.24)(26.27.28.29).(31.32.33.34)(((36.37.38.39)/2))} ...×{(81.82.83.84)(86.87.88.89).(91.92.93.94)(((96.97.98.99)/2))} Account of 2^6 : One 2 comes in read bracket 5 2′s come in five blue brackets (one blue bracket is per each 20) Now we can verify/prove the following congruences: ((1.2.3.4)/2)≡2(mod 10) {: ((6.7.8.9)),(⋮),((91.92.93.94)) }≡4(mod 10) [14 congruencs] (5k+1)(5k+2)(5k+3)(5k+4)≡4(mod 10) { ((^• See Answer of mrW1 ⌉)),((to Q#14614 or an)),(( alternative proof in)),((the comment below.)) :} {: (((16.17.18.19)/2)),(⋮),(((96.97.98.99)/2)) }≡2 [5 congruences] {(5k+1)(5k+2)(5k+3)(5k+4)}/2≡4(mod 10) { ((^∗ Fo proof see the)),((comment below.)) :} Multiplying all the above congruences p≡2.4^(14) .2^5 (mod 10) 4^(14) ≡6(mod 10) [4^(2k) ≡6(mod 10)] 2^5 ≡32≡2(mod 10) p≡2.6.2(mod 10) p≡4(mod 10) Hence the require non-zero digit is 4 It′s only accedent that the answer is correct! Actually there′s a fault in the logic. I′ll try to correct my logic. Please stay with me!

$$\mathrm{100}!\:\mathrm{has}\:\mathrm{24}\:\:\:\mathrm{0}'\mathrm{s}\:\mathrm{from}\:\mathrm{right}. \\ $$$$\mathrm{p}=\frac{\mathrm{100}!}{\mathrm{10}^{\mathrm{24}} }\:\mathrm{has}\:\mathrm{no}\:\mathrm{zero}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}. \\ $$$$\therefore\:\:\mathrm{First}\:\mathrm{non}-\mathrm{zero}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{100}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{unit}\:\mathrm{digit}\:\mathrm{of}\:\:\mathrm{p} \\ $$$$\therefore\:\:\mathrm{p}\equiv\mathrm{d}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathcal{S}{implifying}\:\:\:\mathrm{p}=\frac{\mathrm{100}!}{\mathrm{10}^{\mathrm{24}} }=\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{24}} ×\mathrm{5}^{\mathrm{24}} } \\ $$$$\:\:\:\:\:\:\:\:\mathrm{10}^{\mathrm{24}} =\mathrm{2}^{\mathrm{24}} ×\mathrm{5}^{\mathrm{24}} =\mathrm{2}^{\mathrm{6}\bullet} \left(\mathrm{5}.\mathrm{10}.\mathrm{15}...\mathrm{100}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{All}\:\mathrm{the}\:\mathrm{24}\:\:\:\mathrm{5}'\mathrm{s}\:\mathrm{have}\:\mathrm{come}\:\mathrm{in}\:\mathrm{5}.\mathrm{10}...\mathrm{100} \\ $$$$\:\:\:\:\:\:\:\:\:\:^{\bullet} \mathrm{Also}\:\mathrm{some}\:\mathrm{2}'\mathrm{s}\:\mathrm{have}\:\mathrm{come}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{One}\:\mathrm{2}\:\mathrm{in}\:\mathrm{each}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{10}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{10}\:\:\mathrm{2}'\mathrm{s} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{One}\:\mathrm{2}\:\mathrm{extra}\:\:\mathrm{in}\:\mathrm{each}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{20}\:\:\:\mathrm{5}\:\:\:\mathrm{2}'\mathrm{s} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{One}\:\mathrm{2}\:\mathrm{extra}\:\:\mathrm{in}\:\mathrm{each}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{40}\:\:\:\mathrm{2}\:\:\:\mathrm{2}'\mathrm{s} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{One}\:\mathrm{2}\:\mathrm{extra}\:\:\mathrm{in}\:\mathrm{each}\:\mathrm{mult}\underset{−} {\mathrm{iple}\:\mathrm{of}\:\mathrm{80}\:\:\:\mathrm{1}\:\:\:\mathrm{2}'\mathrm{s}\:\:\:\:\:\:\:\:\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Total}\left(\mathrm{included}\:\mathrm{in}\:\mathrm{5}.\mathrm{10}...\mathrm{100}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{18}\:\:\mathrm{2}'\mathrm{s}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Remaining}\:\:\mathrm{24}−\mathrm{18}=\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{So}\:\mathrm{now}\:\mathrm{p}=\frac{\mathrm{100}!}{\mathrm{10}^{\mathrm{24}} }=\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{24}} ×\mathrm{5}^{\mathrm{24}} }=\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{6}} \left(\mathrm{5}.\mathrm{10}....\mathrm{100}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{p}=\frac{\mathrm{1}}{\mathrm{2}.\mathrm{2}^{\mathrm{5}} }\left\{\left(\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}\right)\left(\mathrm{6}.\mathrm{7}.\mathrm{8}.\mathrm{9}\right)...\left(\mathrm{96}.\mathrm{97}.\mathrm{98}.\mathrm{99}\right\}\right. \\ $$$$\:\:\:\:\:\:\:\:\mathrm{p}=\left\{\left(\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}}{\mathrm{2}}\right)\left(\mathrm{6}.\mathrm{7}.\mathrm{8}.\mathrm{9}\right).\left(\mathrm{11}.\mathrm{12}.\mathrm{13}.\mathrm{14}\right)\left(\frac{\mathrm{16}.\mathrm{17}.\mathrm{18}.\mathrm{19}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left\{\left(\mathrm{21}.\mathrm{22}.\mathrm{23}.\mathrm{24}\right)\left(\mathrm{26}.\mathrm{27}.\mathrm{28}.\mathrm{29}\right).\left(\mathrm{31}.\mathrm{32}.\mathrm{33}.\mathrm{34}\right)\left(\frac{\mathrm{36}.\mathrm{37}.\mathrm{38}.\mathrm{39}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:...×\left\{\left(\mathrm{81}.\mathrm{82}.\mathrm{83}.\mathrm{84}\right)\left(\mathrm{86}.\mathrm{87}.\mathrm{88}.\mathrm{89}\right).\left(\mathrm{91}.\mathrm{92}.\mathrm{93}.\mathrm{94}\right)\left(\frac{\mathrm{96}.\mathrm{97}.\mathrm{98}.\mathrm{99}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\mathrm{Account}\:\mathrm{of}\:\mathrm{2}^{\mathrm{6}} :\:\mathrm{One}\:\mathrm{2}\:\mathrm{comes}\:\mathrm{in}\:\mathrm{read}\:\mathrm{bracket}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\mathrm{2}'\mathrm{s}\:\:\:\mathrm{come}\:\mathrm{in}\:\mathrm{five}\:\mathrm{blue}\:\mathrm{brackets} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{one}\:\mathrm{blue}\:\mathrm{bracket}\:\mathrm{is}\:\:\mathrm{per}\:\mathrm{each}\:\mathrm{20}\right) \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{can}\:\mathrm{verify}/\mathrm{prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{congruences}: \\ $$$$\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}}{\mathrm{2}}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\left.\begin{matrix}{\mathrm{6}.\mathrm{7}.\mathrm{8}.\mathrm{9}}\\{\vdots}\\{\mathrm{91}.\mathrm{92}.\mathrm{93}.\mathrm{94}}\end{matrix}\right\}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right)\:\:\:\left[\mathrm{14}\:\mathrm{congruencs}\right] \\ $$$$\left(\mathrm{5k}+\mathrm{1}\right)\left(\mathrm{5k}+\mathrm{2}\right)\left(\mathrm{5k}+\mathrm{3}\right)\left(\mathrm{5k}+\mathrm{4}\right)\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right)\:\begin{cases}{\:^{\bullet} \mathrm{See}\:\mathrm{Answer}\:\mathrm{of}\:\mathrm{mrW1}\:\rceil}\\{\mathrm{to}\:\mathrm{Q}#\mathrm{14614}\:\mathrm{or}\:\mathrm{an}}\\{\:\mathrm{alternative}\:\mathrm{proof}\:\mathrm{in}}\\{\mathrm{the}\:\mathrm{comment}\:\mathrm{below}.}\end{cases} \\ $$$$\left.\begin{matrix}{\left(\mathrm{16}.\mathrm{17}.\mathrm{18}.\mathrm{19}\right)/\mathrm{2}}\\{\vdots}\\{\left(\mathrm{96}.\mathrm{97}.\mathrm{98}.\mathrm{99}\right)/\mathrm{2}}\end{matrix}\right\}\equiv\mathrm{2}\:\:\:\left[\mathrm{5}\:\mathrm{congruences}\right] \\ $$$$\left\{\left(\mathrm{5k}+\mathrm{1}\right)\left(\mathrm{5k}+\mathrm{2}\right)\left(\mathrm{5k}+\mathrm{3}\right)\left(\mathrm{5k}+\mathrm{4}\right)\right\}/\mathrm{2}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right)\begin{cases}{\:^{\ast} \mathrm{Fo}\:\mathrm{proof}\:\mathrm{see}\:\mathrm{the}}\\{\mathrm{comment}\:\mathrm{below}.}\end{cases} \\ $$$$\mathrm{Multiplying}\:\mathrm{all}\:\mathrm{the}\:\mathrm{above}\:\mathrm{congruences} \\ $$$$\mathrm{p}\equiv\mathrm{2}.\mathrm{4}^{\mathrm{14}} .\mathrm{2}^{\mathrm{5}} \left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\mathrm{4}^{\mathrm{14}} \equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{10}\right)\:\:\left[\mathrm{4}^{\mathrm{2k}} \equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{10}\right)\right]\: \\ $$$$\:\:\:\:\mathrm{2}^{\mathrm{5}} \equiv\mathrm{32}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{p}\equiv\mathrm{2}.\mathrm{6}.\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{p}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{Hence}\:\mathrm{the}\:\mathrm{require}\:\mathrm{non}-\mathrm{zero}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{4} \\ $$$$\:\boldsymbol{\mathrm{It}}'\boldsymbol{\mathrm{s}}\:\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{accedent}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{answer}}\:\:\:\:\:\:\:\:\: \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{correct}}!\:\boldsymbol{\mathrm{Actually}}\:\boldsymbol{\mathrm{there}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{fault}}\:\boldsymbol{\mathrm{in}} \\ $$$$\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{logic}}.\:\boldsymbol{\mathrm{I}}'\boldsymbol{\mathrm{ll}}\:\boldsymbol{\mathrm{try}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{logic}}. \\ $$$$\boldsymbol{\mathrm{Please}}\:\boldsymbol{\mathrm{stay}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{me}}! \\ $$

Commented by RasheedSoomro last updated on 09/Jun/17

•(5k+1)(5k+2)(5k+3)(5k+4)(mod 10)=4 Case-1: k is even.let it′s equal to 2m ∀m∈Z =(5(2m)+1)(5(2m)+2)(5(2m)+3)(5(2m)+4)(mod 10) =10000m^4 +10000m^3 +3500m^2 +500m+24(mod 10) =(10000m^4 +10000m^3 +3500m^2 +500m+20)+4(mod 10) =10(1000m^4 +1000m^3 +350m^2 +50m+2)+4(mod 10) =4(mod 10) Case-2:k is odd.Let it′s equal to 2m+1 (5(2m+1)+1)(5(2m+1)+2)(5(2m+1)+3)(5(2m+1)+4)(mod 10) (10m+6)(10m+7)(10m+8)(10m+9)(mod 10) 10000m^4 +30000m^3 +33500m^2 +16500m+3024(mod 10) (10000m^4 +30000m^3 +33500m^2 +16500m+3020)+4(mod 10) 10(1000m^4 +3000m^3 +3350m^2 +1650m+302)+4(mod 10) =4(mod 10 Hence for all k∈Z ,the proposition itrue. ∗{(5k+1)(5k+2)(5k+3)(5k+4)/2} (mod 10)=2 Case-1:k ieven.Let it′s equal to 2m ={(5(2m)+1)(5(2m)+2)(5(2m)+3)(5(2m)+4)/2}(mod 10) =5000m^4 +5000m^3 +1750m^2 +250m+12(mod 10) =10(500m^4 +500m^3 +175m^2 +25m+1)+2(mod 10) =2(mod 10) Case-2:k is odd.Let it′s equal to 2m+1 =(5(2m+1)+1)(5(2m+1)+2)(5(2m+1)+3)(5(2m+1)+4)/2(mod 10) =(10m+6)(10m+7)(10m+8)(10m+9)/2(mod 10) =(10000m^4 +30000m^3 +33500m^2 +16500m+3024)/2(mod 10) =5000m^4 +15000m^3 +16750m^2 +8250m+1512(mod 10) =(5000m^4 +15000m^3 +16750m^2 +8250m+1510)+2(mod 2) =2(mod 2)

$$\:\bullet\left(\mathrm{5k}+\mathrm{1}\right)\left(\mathrm{5k}+\mathrm{2}\right)\left(\mathrm{5k}+\mathrm{3}\right)\left(\mathrm{5k}+\mathrm{4}\right)\left(\mathrm{mod}\:\mathrm{10}\right)=\mathrm{4} \\ $$$$\mathrm{Case}-\mathrm{1}:\:\mathrm{k}\:\mathrm{is}\:\mathrm{even}.\mathrm{let}\:\mathrm{it}'\mathrm{s}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{2m}\:\:\forall\mathrm{m}\in\mathbb{Z} \\ $$$$=\left(\mathrm{5}\left(\mathrm{2m}\right)+\mathrm{1}\right)\left(\mathrm{5}\left(\mathrm{2m}\right)+\mathrm{2}\right)\left(\mathrm{5}\left(\mathrm{2m}\right)+\mathrm{3}\right)\left(\mathrm{5}\left(\mathrm{2m}\right)+\mathrm{4}\right)\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{10000m}^{\mathrm{4}} +\mathrm{10000m}^{\mathrm{3}} +\mathrm{3500m}^{\mathrm{2}} +\mathrm{500m}+\mathrm{24}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\left(\mathrm{10000m}^{\mathrm{4}} +\mathrm{10000m}^{\mathrm{3}} +\mathrm{3500m}^{\mathrm{2}} +\mathrm{500m}+\mathrm{20}\right)+\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{10}\left(\mathrm{1000m}^{\mathrm{4}} +\mathrm{1000m}^{\mathrm{3}} +\mathrm{350m}^{\mathrm{2}} +\mathrm{50m}+\mathrm{2}\right)+\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right)\:\: \\ $$$$\mathrm{Case}-\mathrm{2}:\mathrm{k}\:\mathrm{is}\:\mathrm{odd}.\mathrm{Let}\:\mathrm{it}'\mathrm{s}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{2m}+\mathrm{1} \\ $$$$\left(\mathrm{5}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{1}\right)\left(\mathrm{5}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{2}\right)\left(\mathrm{5}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{3}\right)\left(\mathrm{5}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{4}\right)\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\left(\mathrm{10m}+\mathrm{6}\right)\left(\mathrm{10m}+\mathrm{7}\right)\left(\mathrm{10m}+\mathrm{8}\right)\left(\mathrm{10m}+\mathrm{9}\right)\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{10000m}^{\mathrm{4}} +\mathrm{30000m}^{\mathrm{3}} +\mathrm{33500m}^{\mathrm{2}} +\mathrm{16500m}+\mathrm{3024}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\left(\mathrm{10000m}^{\mathrm{4}} +\mathrm{30000m}^{\mathrm{3}} +\mathrm{33500m}^{\mathrm{2}} +\mathrm{16500m}+\mathrm{3020}\right)+\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{10}\left(\mathrm{1000m}^{\mathrm{4}} +\mathrm{3000m}^{\mathrm{3}} +\mathrm{3350m}^{\mathrm{2}} +\mathrm{1650m}+\mathrm{302}\right)+\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\:\:\:\right. \\ $$$$\mathrm{Hence}\:\mathrm{for}\:\mathrm{all}\:\mathrm{k}\in\mathbb{Z}\:,\mathrm{the}\:\mathrm{proposition}\:\mathrm{itrue}. \\ $$$$ \\ $$$$\:\ast\left\{\left(\mathrm{5k}+\mathrm{1}\right)\left(\mathrm{5k}+\mathrm{2}\right)\left(\mathrm{5k}+\mathrm{3}\right)\left(\mathrm{5k}+\mathrm{4}\right)/\mathrm{2}\right\}\:\:\left(\mathrm{mod}\:\mathrm{10}\right)=\mathrm{2} \\ $$$$\:\mathrm{Case}-\mathrm{1}:\mathrm{k}\:\mathrm{ieven}.\mathrm{Let}\:\mathrm{it}'\mathrm{s}\:\mathrm{equal}\:\mathrm{to}\:\:\mathrm{2m} \\ $$$$=\left\{\left(\mathrm{5}\left(\mathrm{2m}\right)+\mathrm{1}\right)\left(\mathrm{5}\left(\mathrm{2m}\right)+\mathrm{2}\right)\left(\mathrm{5}\left(\mathrm{2m}\right)+\mathrm{3}\right)\left(\mathrm{5}\left(\mathrm{2m}\right)+\mathrm{4}\right)/\mathrm{2}\right\}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{5000m}^{\mathrm{4}} +\mathrm{5000m}^{\mathrm{3}} +\mathrm{1750m}^{\mathrm{2}} +\mathrm{250m}+\mathrm{12}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{10}\left(\mathrm{500m}^{\mathrm{4}} +\mathrm{500m}^{\mathrm{3}} +\mathrm{175m}^{\mathrm{2}} +\mathrm{25m}+\mathrm{1}\right)+\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\mathrm{Case}-\mathrm{2}:\mathrm{k}\:\mathrm{is}\:\mathrm{odd}.\mathrm{Let}\:\mathrm{it}'\mathrm{s}\:\mathrm{equal}\:\mathrm{to}\:\:\mathrm{2m}+\mathrm{1} \\ $$$$=\left(\mathrm{5}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{1}\right)\left(\mathrm{5}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{2}\right)\left(\mathrm{5}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{3}\right)\left(\mathrm{5}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{4}\right)/\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\left(\mathrm{10m}+\mathrm{6}\right)\left(\mathrm{10m}+\mathrm{7}\right)\left(\mathrm{10m}+\mathrm{8}\right)\left(\mathrm{10m}+\mathrm{9}\right)/\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\left(\mathrm{10000m}^{\mathrm{4}} +\mathrm{30000m}^{\mathrm{3}} +\mathrm{33500m}^{\mathrm{2}} +\mathrm{16500m}+\mathrm{3024}\right)/\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{5000m}^{\mathrm{4}} +\mathrm{15000m}^{\mathrm{3}} +\mathrm{16750m}^{\mathrm{2}} +\mathrm{8250m}+\mathrm{1512}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\left(\mathrm{5000m}^{\mathrm{4}} +\mathrm{15000m}^{\mathrm{3}} +\mathrm{16750m}^{\mathrm{2}} +\mathrm{8250m}+\mathrm{1510}\right)+\mathrm{2}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$=\mathrm{2}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$

Commented by mrW1 last updated on 09/Jun/17

(5k+1)(5k+2)(5k+3)(5k+4) =(5k+1)(5k+4)(5k+3)(5k+2) = [(5k)^2 +5(5k)+4][(5k)^2 +5(5k)+6] =[(5k)^2 +5(5k)]^2 +10[(5k)^2 +5(5k)]+24 =625[k(k+1)]^2 +250[k(k+1)]+24 =2500[((k(k+1))/2)]^2 +500[((k(k+1))/2)]+24 since one of k and k+1 is even, ((k(k+1))/2) is integer ⇒4 (mod 10)

$$\left(\mathrm{5k}+\mathrm{1}\right)\left(\mathrm{5k}+\mathrm{2}\right)\left(\mathrm{5k}+\mathrm{3}\right)\left(\mathrm{5k}+\mathrm{4}\right) \\ $$$$=\left(\mathrm{5k}+\mathrm{1}\right)\left(\mathrm{5k}+\mathrm{4}\right)\left(\mathrm{5k}+\mathrm{3}\right)\left(\mathrm{5k}+\mathrm{2}\right) \\ $$$$=\:\left[\left(\mathrm{5k}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{5k}\right)+\mathrm{4}\right]\left[\left(\mathrm{5k}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{5k}\right)+\mathrm{6}\right] \\ $$$$=\left[\left(\mathrm{5k}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{5k}\right)\right]^{\mathrm{2}} +\mathrm{10}\left[\left(\mathrm{5k}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{5k}\right)\right]+\mathrm{24} \\ $$$$=\mathrm{625}\left[\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)\right]^{\mathrm{2}} +\mathrm{250}\left[\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)\right]+\mathrm{24} \\ $$$$=\mathrm{2500}\left[\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} +\mathrm{500}\left[\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}}\right]+\mathrm{24} \\ $$$$\mathrm{since}\:\mathrm{one}\:\mathrm{of}\:\mathrm{k}\:\mathrm{and}\:\mathrm{k}+\mathrm{1}\:\mathrm{is}\:\mathrm{even}, \\ $$$$\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}}\:\mathrm{is}\:\mathrm{integer} \\ $$$$\Rightarrow\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$

Commented by RasheedSoomro last updated on 09/Jun/17

Your method is better Sir! It was also used in your answer of Q#14614. I merely tried to use different approach!

$$\mathrm{Your}\:\mathrm{method}\:\mathrm{is}\:\mathrm{better}\:\mathrm{Sir}!\:\mathrm{It}\:\mathrm{was}\:\mathrm{also}\:\mathrm{used} \\ $$$$\:\mathrm{in}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{of}\:\mathrm{Q}#\mathrm{14614}. \\ $$$$\mathrm{I}\:\mathrm{merely}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{use}\:\mathrm{different}\:\mathrm{approach}! \\ $$

Commented by mrW1 last updated on 09/Jun/17

congratulation to you for solving this light version! the last non−zero digit of 100! is indeed 4. i hope you will crack 1000! soon!

$$\mathrm{congratulation}\:\mathrm{to}\:\mathrm{you}\:\mathrm{for}\:\mathrm{solving}\:\mathrm{this} \\ $$$$\mathrm{light}\:\mathrm{version}!\:\mathrm{the}\:\mathrm{last}\:\mathrm{non}−\mathrm{zero}\:\mathrm{digit} \\ $$$$\mathrm{of}\:\mathrm{100}!\:\mathrm{is}\:\mathrm{indeed}\:\mathrm{4}. \\ $$$$\mathrm{i}\:\mathrm{hope}\:\mathrm{you}\:\mathrm{will}\:\mathrm{crack}\:\mathrm{1000}!\:\mathrm{soon}! \\ $$

Commented by RasheedSoomro last updated on 09/Jun/17

Please help me in finding error in answer of Q#13724. In one of its answers I have used same logic. Bu the answer is wrong! Mr prakash jain & mrW1 Do you know any standard method of such questions? If you do please do share.

$$\:\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{in}\:\mathrm{finding}\:\mathrm{error}\:\mathrm{in}\:\mathrm{answer}\:\mathrm{of}\:\mathrm{Q}#\mathrm{13724}. \\ $$$$\mathrm{In}\:\mathrm{one}\:\mathrm{of}\:\mathrm{its}\:\mathrm{answers}\:\mathrm{I}\:\mathrm{have}\:\mathrm{used}\:\mathrm{same}\:\mathrm{logic}. \\ $$$$\mathrm{Bu}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong}! \\ $$$$\mathrm{Mr}\:\mathrm{prakash}\:\mathrm{jain}\:\&\:\mathrm{mrW1} \\ $$$$\mathrm{Do}\:\mathrm{you}\:\mathrm{know}\:\mathrm{any}\:\mathrm{standard}\:\mathrm{method}\:\mathrm{of}\:\mathrm{such}\:\mathrm{questions}? \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{do}\:\mathrm{please}\:\mathrm{do}\:\mathrm{share}. \\ $$

Commented by mrW1 last updated on 09/Jun/17

I definitively don′t know any standard methods for solving such questions. I am learning, also from posts such like yours.

$$\mathrm{I}\:\mathrm{definitively}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{any}\:\mathrm{standard} \\ $$$$\mathrm{methods}\:\mathrm{for}\:\mathrm{solving}\:\mathrm{such}\:\mathrm{questions}. \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{learning},\:\mathrm{also}\:\mathrm{from}\:\mathrm{posts}\:\mathrm{such}\: \\ $$$$\mathrm{like}\:\mathrm{yours}. \\ $$

Commented by prakash jain last updated on 10/Jun/17

Actually I don′t know a short cut method to this question. I posted this question to get some ideas.

$$\mathrm{Actually}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{a}\:\mathrm{short}\:\mathrm{cut} \\ $$$$\mathrm{method}\:\mathrm{to}\:\mathrm{this}\:\mathrm{question}. \\ $$$$\mathrm{I}\:\mathrm{posted}\:\mathrm{this}\:\mathrm{question}\:\mathrm{to}\:\mathrm{get}\:\mathrm{some} \\ $$$$\mathrm{ideas}. \\ $$

Commented by RasheedSoomro last updated on 10/Jun/17

Thanks of you both!

$$\mathcal{T}{hanks}\:\mathrm{of}\:\mathrm{you}\:\mathrm{both}! \\ $$

Answered by RasheedSoomro last updated on 11/Jun/17

Easy way 100! has 24 0′s from right. p=((100!)/(10^(24) )) has no zeo from right ∴ first non-zero digit in 100! from right = unit digit in p ∴ p≡d(mod 10) p=((100!)/(10^(24) ))=((100!)/(2^(24) ×5^(24) ))=((100!)/(2^6 (5.10....100))) [18 2′s from 2^(24) have come in 5.10....100] =(1/2^6 ){(1.2.3.4)(6.7.8.9)...(96.97.98.99} p≡d(mod 10)⇒ (1/2^6 ){(1.2.3.4)(6.7.8.9)...(96.97.98.99}≡d(mod 10) (1/2^6 )(4)^(20) ≡d(mod 10) [∵ Each bracket≡4(mod 10) Total brackets=20] (2^(40) /2^6 )≡d(mod 10) d≡2^(34) (mod 10) Verify that: 2^4 ≡6(mod 10 (2^4 )^8 ≡6^8 ≡6(mod 10) 2^(32) .2^2 ≡6.2^2 ≡4(mod 10) 2^(34) ≡4(mod 10) Hence d≡2^(34) ≡4(mod 10) d≡4(mod 10) Wrong logic! Accedently correct answer! I′ll try to correct my logic. Please stay with me!

$$\mathrm{Easy}\:\mathrm{way} \\ $$$$\mathrm{100}!\:\mathrm{has}\:\mathrm{24}\:\:\mathrm{0}'\mathrm{s}\:\mathrm{from}\:\mathrm{right}.\: \\ $$$$\mathrm{p}=\frac{\mathrm{100}!}{\mathrm{10}^{\mathrm{24}} }\:\:\mathrm{has}\:\mathrm{no}\:\mathrm{zeo}\:\mathrm{from}\:\mathrm{right} \\ $$$$\therefore\:\:\mathrm{first}\:\mathrm{non}-\mathrm{zero}\:\mathrm{digit}\:\mathrm{in}\:\mathrm{100}!\:\mathrm{from}\:\mathrm{right} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{unit}\:\mathrm{digit}\:\mathrm{in}\:\mathrm{p} \\ $$$$\therefore\:\:\mathrm{p}\equiv\mathrm{d}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{p}=\frac{\mathrm{100}!}{\mathrm{10}^{\mathrm{24}} }=\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{24}} ×\mathrm{5}^{\mathrm{24}} }=\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{6}} \left(\mathrm{5}.\mathrm{10}....\mathrm{100}\right)} \\ $$$$\left[\mathrm{18}\:\:\:\mathrm{2}'\mathrm{s}\:\mathrm{from}\:\mathrm{2}^{\mathrm{24}} \:\mathrm{have}\:\mathrm{come}\:\mathrm{in}\:\mathrm{5}.\mathrm{10}....\mathrm{100}\right] \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left\{\left(\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}\right)\left(\mathrm{6}.\mathrm{7}.\mathrm{8}.\mathrm{9}\right)...\left(\mathrm{96}.\mathrm{97}.\mathrm{98}.\mathrm{99}\right\}\right. \\ $$$$\:\:\:\:\mathrm{p}\equiv\mathrm{d}\left(\mathrm{mod}\:\mathrm{10}\right)\Rightarrow \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left\{\left(\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}\right)\left(\mathrm{6}.\mathrm{7}.\mathrm{8}.\mathrm{9}\right)...\left(\mathrm{96}.\mathrm{97}.\mathrm{98}.\mathrm{99}\right\}\equiv\mathrm{d}\left(\mathrm{mod}\:\mathrm{10}\right)\right. \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left(\mathrm{4}\right)^{\mathrm{20}} \equiv\mathrm{d}\left(\mathrm{mod}\:\mathrm{10}\right)\:\:\:\:\left[\because\:\mathrm{Each}\:\mathrm{bracket}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right)\:\mathrm{Total}\:\mathrm{brackets}=\mathrm{20}\right] \\ $$$$\:\:\:\frac{\mathrm{2}^{\mathrm{40}} }{\mathrm{2}^{\mathrm{6}} }\equiv\mathrm{d}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\:\mathrm{d}\equiv\mathrm{2}^{\mathrm{34}} \left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{Verify}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{4}} \equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{10}\right. \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{2}^{\mathrm{4}} \right)^{\mathrm{8}} \equiv\mathrm{6}^{\mathrm{8}} \equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{32}} .\mathrm{2}^{\mathrm{2}} \equiv\mathrm{6}.\mathrm{2}^{\mathrm{2}} \equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{34}} \equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{Hence} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{d}\equiv\mathrm{2}^{\mathrm{34}} \equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{d}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$ \\ $$$$\:\boldsymbol{\mathrm{Wrong}}\:\boldsymbol{\mathrm{logic}}!\:\boldsymbol{\mathrm{Accedently}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{answer}}!\:\:\:\:\:\:\: \\ $$$$\boldsymbol{\mathrm{I}}'\boldsymbol{\mathrm{ll}}\:\boldsymbol{\mathrm{try}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{logic}}. \\ $$$$\boldsymbol{\mathrm{Please}}\:\boldsymbol{\mathrm{stay}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{me}}! \\ $$

Answered by RasheedSoomro last updated on 11/Jun/17

According to Corrected Logic. Right logic & Right answer Expansion of 100! have 24 0′s in the right. Let p=((100!)/(10^(24) )) p has no zero from right. First non-zero digit in 100! =unit digit in p d≡p(mod 10) p=((100!)/(10^(24) ))=((100!)/(2^(24) ×5^(24) )) =((100!)/(2^(24) ×(5×((10)/2)×((15)/3)×((20)/2^2 )×5^2 ×...×((100)/2^2 )))) =((100!)/(2^(24) ×(((5.10.15.....100)/(2^(18) ×3^8 ×7^2 ×11×13×17×19))))) =3^8 ×7^2 ×11×13×17×19×((100!)/(2^6 (5.10....100))) =3^8 ×7^2 ×11×13×17×19{(((1...4)/2))(6...9)(11...14)(((16....19)/2))} ×...×{(81....84)(86...89)(91...94)(((96...99)/2))} Number of red bracket= 1 Number of black brackets= 14 3 per each 20 except first. Number of blue brackets= 5 1 per each 20 We can verify/prove the following congruences 3^8 ≡1(mod 10) 7^2 ≡9(mod 10) 11≡1(mod 10) 13≡3(mod 10) 17≡7(mod 10) 19≡9(mod 10) ((1.2.3.4)/2)≡2 Any of Black-bracket≡4(mod 10) [14 congruences] Any of Blue-bracket≡2(mod 10) [ 5 congruences] Multiplying all above congruences p≡1.9.1.3.7.9.2.4^(14) .2^5 (mod 10) p≡1.9.1.3.7.9.2.6.2(mod 10) [∵4^(even) ≡6(mod 10] p≡40824≡4(mod 10) [∵4^(even) ≡6(mod 10]

$$\mathrm{According}\:\mathrm{to}\:\mathrm{Corrected}\:\mathrm{Logic}. \\ $$$$\mathrm{Right}\:\mathrm{logic}\:\&\:\:\mathrm{Right}\:\mathrm{answer} \\ $$$$\:\:\mathcal{E}\mathrm{xpansion}\:\mathrm{of}\:\mathrm{100}!\:\mathrm{have}\:\mathrm{24}\:\:\:\mathrm{0}'\mathrm{s}\:\mathrm{in}\:\mathrm{the}\:\mathrm{right}. \\ $$$$\mathrm{Let}\:\mathrm{p}=\frac{\mathrm{100}!}{\mathrm{10}^{\mathrm{24}} } \\ $$$$\mathrm{p}\:\mathrm{has}\:\mathrm{no}\:\mathrm{zero}\:\mathrm{from}\:\mathrm{right}. \\ $$$$\mathrm{First}\:\mathrm{non}-\mathrm{zero}\:\mathrm{digit}\:\mathrm{in}\:\mathrm{100}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{unit}\:\mathrm{digit}\:\mathrm{in}\:\mathrm{p} \\ $$$$\:\:\:\:\:\:\:\mathrm{d}\equiv\mathrm{p}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{p}=\frac{\mathrm{100}!}{\mathrm{10}^{\mathrm{24}} }=\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{24}} ×\mathrm{5}^{\mathrm{24}} } \\ $$$$\:\:\:\:=\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{24}} ×\left(\mathrm{5}×\frac{\mathrm{10}}{\mathrm{2}}×\frac{\mathrm{15}}{\mathrm{3}}×\frac{\mathrm{20}}{\mathrm{2}^{\mathrm{2}} }×\mathrm{5}^{\mathrm{2}} ×...×\frac{\mathrm{100}}{\mathrm{2}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{24}} ×\left(\frac{\mathrm{5}.\mathrm{10}.\mathrm{15}.....\mathrm{100}}{\mathrm{2}^{\mathrm{18}} ×\mathrm{3}^{\mathrm{8}} ×\mathrm{7}^{\mathrm{2}} ×\mathrm{11}×\mathrm{13}×\mathrm{17}×\mathrm{19}}\right)} \\ $$$$\:\:\:=\mathrm{3}^{\mathrm{8}} ×\mathrm{7}^{\mathrm{2}} ×\mathrm{11}×\mathrm{13}×\mathrm{17}×\mathrm{19}×\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{6}} \left(\mathrm{5}.\mathrm{10}....\mathrm{100}\right)} \\ $$$$\:=\mathrm{3}^{\mathrm{8}} ×\mathrm{7}^{\mathrm{2}} ×\mathrm{11}×\mathrm{13}×\mathrm{17}×\mathrm{19}\left\{\left(\frac{\mathrm{1}...\mathrm{4}}{\mathrm{2}}\right)\left(\mathrm{6}...\mathrm{9}\right)\left(\mathrm{11}...\mathrm{14}\right)\left(\frac{\mathrm{16}....\mathrm{19}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:×...×\left\{\left(\mathrm{81}....\mathrm{84}\right)\left(\mathrm{86}...\mathrm{89}\right)\left(\mathrm{91}...\mathrm{94}\right)\left(\frac{\mathrm{96}...\mathrm{99}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Number}\:\mathrm{of}\:\mathrm{red}\:\mathrm{bracket}=\:\:\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Number}\:\mathrm{of}\:\mathrm{black}\:\mathrm{brackets}=\:\:\:\mathrm{14}\:\:\:\mathrm{3}\:\mathrm{per}\:\mathrm{each}\:\:\mathrm{20}\:\mathrm{except}\:\mathrm{first}. \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Number}\:\mathrm{of}\:\mathrm{blue}\:\mathrm{brackets}=\:\:\:\mathrm{5}\:\:\:\:\mathrm{1}\:\mathrm{per}\:\mathrm{each}\:\:\mathrm{20} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{verify}/\mathrm{prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{congruences} \\ $$$$\mathrm{3}^{\mathrm{8}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{7}^{\mathrm{2}} \equiv\mathrm{9}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{11}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{13}\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{17}\equiv\mathrm{7}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{19}\equiv\mathrm{9}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}}{\mathrm{2}}\equiv\mathrm{2} \\ $$$$\mathrm{Any}\:\mathrm{of}\:\boldsymbol{\mathrm{Black}}-\boldsymbol{\mathrm{bracket}}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right)\:\:\:\:\left[\mathrm{14}\:\mathrm{congruences}\right] \\ $$$$\mathrm{Any}\:\mathrm{of}\:\boldsymbol{\mathrm{Blue}}-\boldsymbol{\mathrm{bracket}}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right)\:\:\:\:\left[\:\mathrm{5}\:\mathrm{congruences}\right] \\ $$$$ \\ $$$$\mathrm{Multiplying}\:\mathrm{all}\:\mathrm{above}\:\mathrm{congruences} \\ $$$$\mathrm{p}\equiv\mathrm{1}.\mathrm{9}.\mathrm{1}.\mathrm{3}.\mathrm{7}.\mathrm{9}.\mathrm{2}.\mathrm{4}^{\mathrm{14}} .\mathrm{2}^{\mathrm{5}} \left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{p}\equiv\mathrm{1}.\mathrm{9}.\mathrm{1}.\mathrm{3}.\mathrm{7}.\mathrm{9}.\mathrm{2}.\mathrm{6}.\mathrm{2}\left(\mathrm{mod}\:\mathrm{10}\right)\:\:\:\left[\because\mathrm{4}^{\mathrm{even}} \equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{10}\right]\right. \\ $$$$\mathrm{p}\equiv\mathrm{40824}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right)\:\:\:\left[\because\mathrm{4}^{\mathrm{even}} \equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{10}\right]\right. \\ $$$$ \\ $$

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