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Question Number 18884 by Tinkutara last updated on 31/Jul/17

$$\mathrm{Determine}\mathrm{the}\mathrm{smallest}\mathrm{positive}\mathrm{integer}$$$$\mathrm{x},\mathrm{whose}\mathrm{last}\mathrm{digit}\mathrm{is}6\mathrm{and}\mathrm{if}\mathrm{we}\mathrm{erase}$$$$\mathrm{this}6\mathrm{and}\mathrm{put}\mathrm{it}\mathrm{in}\mathrm{left}\mathrm{most}\mathrm{of}\mathrm{the}$$$$\mathrm{number}\mathrm{so}\mathrm{obtained},\mathrm{the}\mathrm{number}$$$$\mathrm{becomes}4\mathrm{x}.$$

Answered by mrW1 last updated on 01/Aug/17

$${\mathrm{let}}^{\prime }\mathrm{s}\mathrm{say}\mathrm{the}\mathrm{number}\mathrm{has}\mathrm{n}+1\mathrm{digits}$$$$\mathrm{let}\mathrm{x}=[\mathrm{abcd}...\mathrm{pqr}6]=10\mathrm{u}+6$$$$\mathrm{let}\mathrm{y}=[6\mathrm{abcd}...\mathrm{pqr}]=6\times {10}^{\mathrm{n}}+\mathrm{u}$$$$\mathrm{y}=4\mathrm{x}$$$$\Rightarrow 6\times {10}^{\mathrm{n}}+\mathrm{u}=4(10\mathrm{u}+6)$$$$\Rightarrow 6({10}^{\mathrm{n}}-4)=39\mathrm{u}$$$$\Rightarrow 2(2^{\mathrm{n}}\times 5^{\mathrm{n}}-2^2)=13\mathrm{u}$$$$\Rightarrow 2^3(2^{\mathrm{n}-2}\times 5^{\mathrm{n}}-1)=13\mathrm{u}$$$$\Rightarrow 2^3(25\times {10}^{\mathrm{n}-2}-1)=13\mathrm{u}$$$$\Rightarrow \mathrm{u}=\frac{2^3(25\times {10}^{\mathrm{n}-2}-1)}{13}$$$$\Rightarrow 25\times {10}^{\mathrm{n}-2}-1=13\mathrm{k}$$$$\mathrm{the}\mathrm{smallest}\mathrm{n}=5:$$$$25\times {10}^{5-2}-1=25000-1=24999=13\times 1923$$$$\Rightarrow \min .\mathrm{u}=2^3\times 1923=15384$$$$\Rightarrow \min .\mathrm{x}=153846$$$$$$$$\mathrm{the}\mathrm{further}\mathrm{numbers}\mathrm{are}:$$$$\mathrm{x}=153846153846$$$$\mathrm{x}=153846153846153846$$$$\mathrm{x}=153846153846153846153846$$$$......$$

Commented by Tinkutara last updated on 01/Aug/17

$$\mathrm{Please}\mathrm{also}\mathrm{explain}\mathrm{why}25\times {10}^{n-2}-1$$$$=13k\mathrm{because}\mathrm{it}\mathrm{should}\mathrm{be}\frac{13k}{8}.$$

Commented by mrW1 last updated on 01/Aug/17

$$\mathrm{yes},\mathrm{u}\mathrm{should}=\frac{2^3(25\times {10}^{\mathrm{n}-2}-1)}{13}.\mathrm{I}\mathrm{had}\mathrm{just}\mathrm{a}\mathrm{typo}.$$$$\mathrm{for}\mathrm{u}=\frac{2^3(25\times {10}^{\mathrm{n}-2}-1)}{13}\mathrm{and}13\mathrm{is}\mathrm{prime}$$$$25\times {10}^{\mathrm{n}-2}-1\mathrm{must}\mathrm{be}\mathrm{a}\mathrm{multiple}\mathrm{of}13,$$$$\mathrm{otherwise}\mathrm{u}\mathrm{woulde}\mathrm{be}\mathrm{no}\mathrm{integer}.$$$$\mathrm{i}.\mathrm{e}.25\times {10}^{\mathrm{n}-2}-1=13\mathrm{k}\mathrm{with}\mathrm{k}=\mathrm{integer}.$$$$\mathrm{then}\mathrm{u}=2^3\mathrm{k}.$$$$\mathrm{so}\mathrm{our}\mathrm{task}\mathrm{is}\mathrm{to}\mathrm{solve}\mathrm{the}\mathrm{equation}$$$$25\times {10}^{\mathrm{n}-2}-1=13\mathrm{k}\mathrm{with}\mathrm{n},\mathrm{k}\in \mathbb{N}.$$

Commented by mrW1 last updated on 01/Aug/17

$$\mathrm{certainly}\mathrm{you}\mathrm{can}\mathrm{solve}\mathrm{this}\mathrm{question}$$$$\mathrm{directly}\mathrm{like}\mathrm{this}:$$$$\mathrm{abcdef}6$$$$\times 4$$$$---------$$$$=6\mathrm{abcdef}$$$$\Rightarrow \mathrm{f}=4$$$$\Rightarrow \mathrm{e}=8$$$$\Rightarrow \mathrm{d}=3$$$$\Rightarrow \mathrm{c}=5$$$$\Rightarrow \mathrm{b}=1$$$$\Rightarrow \mathrm{a}=6!{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{right}.\Rightarrow \mathrm{our}\mathrm{first}\mathrm{number}\mathrm{is}153846$$$$\mathrm{just}\mathrm{go}\mathrm{on}\mathrm{with}\mathrm{these}\mathrm{steps}\mathrm{if}\mathrm{you}\mathrm{want}\mathrm{to}\mathrm{get}\mathrm{the}$$$$\mathrm{next}\mathrm{suitable}\mathrm{number}.$$

Commented by Tinkutara last updated on 01/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{mrW}1\mathrm{Sir}!$$

Commented by mrW1 last updated on 01/Aug/17

$$\mathrm{I}\mathrm{tried}\mathrm{to}\mathrm{use}\mathrm{an}\mathrm{equation}\mathrm{to}\mathrm{solve}\mathrm{the}$$$$\mathrm{question},\mathrm{because}\mathrm{the}\mathrm{question}\mathrm{can}$$$$\mathrm{be}\mathrm{generalised}:$$$$\mathrm{Find}\mathrm{such}\mathrm{numbers}\mathrm{when}\mathrm{we}\mathrm{erase}\mathrm{their}\mathrm{last}$$$$\mathrm{digit}\mathrm{and}\mathrm{put}\mathrm{it}\mathrm{before}\mathrm{the}\mathrm{first}\mathrm{digit},$$$$\mathrm{the}\mathrm{new}\mathrm{number}\mathrm{will}\mathrm{be}\mathrm{m}\mathrm{times}\mathrm{as}$$$$\mathrm{the}\mathrm{original}\mathrm{number}.$$$$\mathrm{m}\mathrm{can}\mathrm{be}2\mathrm{til}9.$$

Commented by Tinkutara last updated on 01/Aug/17

$$\mathrm{Which}\mathrm{equation}?\mathrm{Can}\mathrm{you}\mathrm{give}\mathrm{its}$$$$\mathrm{derivation}?$$

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