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Question Number 15543 by RasheedSoomro last updated on 11/Jun/17

$$\text{You can't use 'macro parameter character \#' in math mode}$$$${\mathcal{E}}_{}^{}\mathrm{xpansion}\mathrm{of}1000!$$$$\mathrm{has}240^{\prime }\mathrm{s}\mathrm{at}\mathrm{the}\mathrm{end}.\mathrm{Find}$$$$\mathrm{the}\mathrm{first}\mathrm{non}-\mathrm{zero}\mathrm{digit}$$$$\mathrm{from}\mathrm{right}.$$$$1000!=....\mathrm{d}0000...0$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{d}?$$$$$$

Answered by RasheedSoomro last updated on 11/Jun/17

$$\mathrm{Used}\mathrm{results}:$$$$(5\mathrm{k}+1)(5\mathrm{k}+2)(5\mathrm{k}+3)(5\mathrm{k}+4)\equiv 4\left(\mathrm{mod}10\right)$$$$\frac{(5\mathrm{k}+1)(5\mathrm{k}+2)(5\mathrm{k}+3)(5\mathrm{k}+4)}{2}\equiv 2\left(\mathrm{mod}10\right)$$$$-----------------------------$$$$\mathcal{E}\mathrm{xpansion}\mathrm{of}1000!\mathrm{has}2490^{\prime }\mathrm{s}\mathrm{at}\mathrm{the}\mathrm{end}.$$$$\mathrm{Let}\mathrm{p}=\frac{1000!}{{10}^{249}},\mathrm{p}\mathrm{has}\mathrm{no}0\mathrm{at}\mathrm{the}\mathrm{end}.$$$$\mathrm{p}=\frac{1000!}{2^{249}\times 5^{249}}$$$$\mathrm{We}\mathrm{can}\mathrm{write}$$$$5^{249}=5\times \frac{10}{2}\times \frac{15}{3}\times ...\frac{965}{193}\times ...\frac{985}{197}\times \frac{990}{2.3^2.11}\times .\frac{995}{199}\times \frac{1000}{2^3}$$$$=\frac{5.10.15....1000}{2^{{\alpha }_2}\times 3^{{\alpha }_3}\times 7^{{\alpha }_7}\times ...{193}^{{\alpha }_{193}}\times {197}^{{\alpha }_{197}}\times {199}^{{\alpha }_{199}}}$$$${\alpha }_2=\lfloor \frac{1000}{10}\rfloor +\lfloor \frac{1000}{20}\rfloor +\lfloor \frac{1000}{40}\rfloor +\lfloor \frac{1000}{80}\rfloor +\lfloor \frac{1000}{160}\rfloor +\lfloor \frac{1000}{320}\rfloor +\lfloor \frac{1000}{640}\rfloor$$$$=100+50+25+12+6+3+1=197$$$${\alpha }_3=\lfloor \frac{1000}{15}\rfloor +\lfloor \frac{1000}{45}\rfloor +\lfloor \frac{1000}{135}\rfloor +\lfloor \frac{1000}{405}\rfloor =66+22+7+2=97$$$$\mathrm{By}\mathrm{same}\mathrm{formula}$$$${\alpha }_7=32,{\alpha }_{11}=19,{\alpha }_{13}=16,{\alpha }_{17}=11,{\alpha }_{19}=10,{\alpha }_{23}=8,$$$${\alpha }_{29}={\alpha }_{31}=6,{\alpha }_{37}=5,{\alpha }_{41}={\alpha }_{43}={\alpha }_{47}=4,{\alpha }_{53}={\alpha }_{59}={\alpha }_{61}=3,$$$${\alpha }_{67}={\alpha }_{71}={\alpha }_{73}={\alpha }_{79}={\alpha }_{83}={\alpha }_{89}={\alpha }_{97}=2$$$${\alpha }_{101}={\alpha }_{103}=\dots ={\alpha }_{199}=1$$$$\mathrm{Hence}$$$$5^{249}=\frac{5.10.15....1000}{2^{197}.3^{97}.7^{32}...{199}^1}$$$$\mathrm{So}$$$$\mathrm{p}=\frac{1000!}{2^{249}\times 5^{249}}=\frac{1000!}{2^{249}\times \left(\frac{5.10.15....1000}{2^{197}.3^{97}.7^{32}...{199}^1}\right)}$$$$\mathrm{p}=\frac{1000!}{2^{249}\times 5^{249}}=\frac{1000!}{2^{52}\times \left(\frac{5.10.15....1000}{3^{97}.7^{32}...{199}^1}\right)}$$$$\mathrm{p}=(3^{97}.7^{32}...{199}^1)\left(\frac{1000!}{2^{52}(5.10...1000}\right)$$$$\mathrm{p}=(3^{97}.7^{32}...{199}^1)\left\{\left(\frac{1.2.3.4}{2}\right)(6...9)(11....14)\left(\frac{16...19}{2}\right)\right\}$$$$\times \left\{(21...24)(26....29)(31...34)\left(\frac{36...39}{2}\right)\right\}$$$$\times ...\times \left\{(981...984)(986...989)(991...994)\left(\frac{996...999}{2}\right)\right\}$$$$\mathrm{Now}\mathrm{we}\mathrm{can}\mathrm{see}\mathrm{that}$$$$3^{97}\equiv 3,7^{32}\equiv 1,{11}^{19}\equiv 1,{13}^{16}\equiv 1,{17}^{11}\equiv 3,{19}^{10}\equiv 1,{23}^8\equiv 1$$$$$$$$\mathrm{Continue}$$

Commented by tawa tawa last updated on 11/Jun/17

$$\mathrm{Weldone}\mathrm{sir}.$$

Commented by RasheedSoomro last updated on 12/Jun/17

$$\mathbb{T}\mathrm{han}\mathbb{k}\mathrm{s}\mathrm{miss}\mathrm{tawa}.\mathrm{pl}\mathrm{see}\mathrm{also}\mathrm{my}\mathrm{other}\mathrm{answer}.$$

Answered by RasheedSoomro last updated on 12/Jun/17

$$\underset{-}{\mathcal{NEW}\mathcal{APPROACH}}\left(\mathrm{The}\mathrm{simplest}\mathrm{approach}\right)$$$$\mathrm{Used}\mathrm{results}$$$$(5\mathrm{k}+1)(5\mathrm{k}+2)(5\mathrm{k}+3)(5\mathrm{k}+4)\equiv 4\left(\mathrm{mod}10\right)$$$$\frac{(5\mathrm{k}+1)(5\mathrm{k}+2)(5\mathrm{k}+3)(5\mathrm{k}+4)}{2}\equiv 2\left(\mathrm{mod}10\right)$$$$2^{4\mathrm{k}+1}\equiv 2\left(\mathrm{mod}10\right)$$$$4^{2\mathrm{k}}\equiv 6\left(\mathrm{mod}10\right)$$$$------------------------$$$$\mathcal{E}\mathrm{xpansion}\mathrm{of}1000!\mathrm{has}2490^{\prime }\mathrm{s}\mathrm{at}\mathrm{the}\mathrm{end}$$$$\mathrm{p}=\frac{1000!}{{10}^{249}}\mathrm{has}\mathrm{no}\mathrm{zero}\mathrm{at}\mathrm{the}\mathrm{end}.$$$$\mathrm{Let}\mathrm{f}\left(\mathrm{n}\right)=\mathrm{n}!$$$$\mathrm{f}\left(1000\right)=1000!=(1...4)(5...9)...(996...999)(5.10.15...1000)$$$$=(1...4)(6...9)...(996...999).5^{200}.(1.2.3...200)$$$$=5^{200}(1...4)(6...9)...(996...999)\times \mathrm{f}\left(200\right)$$$$\mathrm{f}\left(200\right)=(1...4)(6...9)...(196...\mathrm{l}99)(5.10...200)$$$$=5^{40}(1...4)(6...9)...(196...\mathrm{l}99)(1.2...40)$$$$=5^{40}(1...4)(6...9)...(196...\mathrm{l}99)\times \mathrm{f}\left(40\right)$$$$\mathrm{f}\left(40\right)=(1...4)(6...9)...(36...39)(5.10...40)$$$$=5^8(1...4)(6...9)...(36...39)(1.2...8)$$$$=5^8(1...4)(6...9)...(36...39)\times \mathrm{f}\left(8\right)$$$$\mathrm{f}\left(8\right)=(1.2.3.4)(5.6.7.8)\left(5\right)$$$$=5(1.2.3.4)(6.7.8)$$$$$$$$\mathrm{f}\left(1000\right)=5^{200}(1...4)(6...9)...(996...999)$$$$\times 5^{40}(1...4)(6...9)...(196...\mathrm{l}99)$$$$\times 5^8(1...4)(6...9)...(36...39)$$$$\times 5(1...4)(6.7.8)$$$$=5^{249}{(1...4)}^4{(6...9)}^3(6.7.8)...{(36...39)}^3$$$$\times {(41...44)}^2...{(196...199)}^2$$$$\times (201...204)...(996...999)$$$$\mathrm{p}=\frac{\mathrm{f}\left(1000\right)}{2^{249}.5^{249}}$$$$=\left(\frac{1}{2^{249}}\right)(6.7.8){(1...4)}^4{(6...9)}^3{(11...14)}^3...{(36...39)}^3$$$$\times {(41...44)}^2{(46...49)}^2...{(196...199)}^2$$$$\times (201...204)...(996...999)$$$$$$$$=\{(1...4)(6...9)...(996...999)\}$$$$=(6.7.8)($$$$=\left(\frac{1}{2^{249}}\right){\left\{\stackrel{8\mathrm{brackets}}{(1...4)...(36....39)}\right\}}^3$$$$\times {\left\{\stackrel{32\mathrm{brackets}}{(41...44)...(196...199)}\right\}}^2$$$$\times \left\{\stackrel{180\mathrm{brackets}}{(201...204)...(996...999)}\right\}$$$$\times \{(1...4)(6.7.8\}$$$$\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{brackets}=8\times 3+32\times 2+180+2=270$$$$\mathrm{Any}\mathrm{of}249\mathrm{brackets}\mathrm{could}\mathrm{be}\mathrm{given}2\mathrm{as}\mathrm{denominator}$$$$\left(\frac{\ast .\ast .\ast .\ast }{2}\right)\equiv 2\left(\mathrm{mod}10\right)\left[249\mathrm{congruences}\right]$$$$(\ast .\ast .\ast .\ast )\equiv 4\left(\mathrm{mod}10\right)\left[20\mathrm{congruences}\right]$$$$6.7.8\equiv 6\left(\mathrm{mod}10\right)\left[1\mathrm{congruence}\right]$$$$\mathrm{Multiplying}\mathrm{above}\mathrm{congruences}$$$$\mathrm{p}\equiv 2^{249\bullet }.4^{20\ast }.6\left(\mathrm{mod}2\right)$$$$\equiv 2.6.6\left(\mathrm{mod}10\right)$$$$\equiv 2(\mathrm{mod}10$$$$$$

Commented by mrW1 last updated on 12/Jun/17

$$\mathrm{I}\mathrm{appreciate}\mathrm{very}\mathrm{that}\mathrm{you}\mathrm{never}\mathrm{stop}$$$$\mathrm{before}\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{found}!$$

Commented by tawa tawa last updated on 12/Jun/17

$$\mathrm{Wow},\mathrm{kudox}\mathrm{sir}.$$

Commented by mrW1 last updated on 12/Jun/17

$$\mathcal{C}\mathcal{ONGRATULATION}\mathcal{SIR}!$$

Commented by RasheedSoomro last updated on 12/Jun/17

$$\mathcal{TH}\alpha nks\mathcal{SIR}!$$$$\mathcal{A}partofyourcongratulation$$$$returnstoyou,becauseitis$$$$basedonaresultprovedbyyou!$$

Commented by mrW1 last updated on 12/Jun/17

$$\mathrm{Can}\mathrm{you}\mathrm{find}\mathrm{the}\mathrm{last}\mathrm{non}-\mathrm{zero}\mathrm{digit}$$$$\mathrm{of}2000!\mathrm{using}\mathrm{your}\mathrm{method}?$$

Commented by RasheedSoomro last updated on 12/Jun/17

$$\mathrm{I}\mathrm{think}\mathrm{I}\mathrm{can}\mathrm{sir}.$$

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