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Question Number 99058 by Learner-123 last updated on 18/Jun/20

Find a number that becomes N times smaller if the first digit is removed in the following cases: 1) N=17 2) N=27 3)N=37 4)N=47.

$${Find}\:{a}\:{number}\:{that}\:{becomes}\:{N}\:{times} \\ $$$${smaller}\:{if}\:{the}\:{first}\:{digit}\:{is}\:{removed} \\ $$$${in}\:{the}\:{following}\:{cases}: \\ $$$$\left.\mathrm{1}\right)\:{N}=\mathrm{17} \\ $$$$\left.\mathrm{2}\right)\:{N}=\mathrm{27} \\ $$$$\left.\mathrm{3}\right){N}=\mathrm{37} \\ $$$$\left.\mathrm{4}\right){N}=\mathrm{47}. \\ $$

Commented by Rasheed.Sindhi last updated on 18/Jun/20

1) N=17

$$\left.\mathrm{1}\right)\:{N}=\mathrm{17} \\ $$

Commented by mr W last updated on 18/Jun/20

i don′t understand what is meant. say the number is XYYYYY. it should be YYYYY=((XYYYYY)/N) with N=17?

$${i}\:{don}'{t}\:{understand}\:{what}\:{is}\:{meant}. \\ $$$${say}\:{the}\:{number}\:{is}\:{XYYYYY}. \\ $$$${it}\:{should}\:{be} \\ $$$${YYYYY}=\frac{{XYYYYY}}{{N}}\:{with}\:{N}=\mathrm{17}? \\ $$

Commented by Rasheed.Sindhi last updated on 18/Jun/20

Sir I understand by first digit ′unit-digit′.So if remove 7 from 17 it becomes 1which is 17 times smaller than 17.

$${Sir}\:{I}\:{understand}\:{by}\:{first}\:{digit} \\ $$$$'{unit}-{digit}'.{So}\:{if}\:{remove}\:\mathrm{7}\:{from} \\ $$$$\mathrm{17}\:{it}\:{becomes}\:\mathrm{1}{which}\:{is}\:\mathrm{17}\:{times} \\ $$$${smaller}\:{than}\:\mathrm{17}. \\ $$

Commented by mr W last updated on 18/Jun/20

thanks sir! (Xy)=10X+y=17X ⇒y=7X≤9 ⇒X≤(9/7) ⇒X=1 ⇒y=7 ⇒the number is 17. (Xy)=10X+y=27X ⇒y=17X≤9 ⇒X≤(9/(17)) ⇒no solution

$${thanks}\:{sir}! \\ $$$$\left({Xy}\right)=\mathrm{10}{X}+{y}=\mathrm{17}{X} \\ $$$$\Rightarrow{y}=\mathrm{7}{X}\leqslant\mathrm{9} \\ $$$$\Rightarrow{X}\leqslant\frac{\mathrm{9}}{\mathrm{7}}\:\Rightarrow{X}=\mathrm{1}\:\Rightarrow{y}=\mathrm{7} \\ $$$$\Rightarrow{the}\:{number}\:{is}\:\mathrm{17}. \\ $$$$ \\ $$$$\left({Xy}\right)=\mathrm{10}{X}+{y}=\mathrm{27}{X} \\ $$$$\Rightarrow{y}=\mathrm{17}{X}\leqslant\mathrm{9}\:\Rightarrow{X}\leqslant\frac{\mathrm{9}}{\mathrm{17}}\:\Rightarrow{no}\:{solution} \\ $$

Commented by Learner-123 last updated on 18/Jun/20

thanks both sirs!

$${thanks}\:{both}\:{sirs}! \\ $$

Commented by Rasheed.Sindhi last updated on 18/Jun/20

First-digit(unit-digit)removing: tu=10t+u→(((10t+u)−u)/(10))=u (((10t+u)−u)/(10))=((10t+u)/(10t+u))=1 t=1 The required number is 1u 11,12,13,...,19

$${First}-{digit}\left({unit}-{digit}\right){removing}: \\ $$$${tu}=\mathrm{10}{t}+{u}\rightarrow\frac{\left(\mathrm{10}{t}+{u}\right)−{u}}{\mathrm{10}}={u} \\ $$$$\frac{\left(\mathrm{10}{t}+{u}\right)−{u}}{\mathrm{10}}=\frac{\mathrm{10}{t}+{u}}{\mathrm{10}{t}+{u}}=\mathrm{1} \\ $$$${t}=\mathrm{1} \\ $$$${The}\:{required}\:{number}\:{is}\:\mathrm{1}{u} \\ $$$$\mathrm{11},\mathrm{12},\mathrm{13},...,\mathrm{19} \\ $$

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