Question

A number when divided by a divisor leaves a remainder of 5, and when divided by twice the divisor leaves a remainder of 45. Find the divisor?

Answer 1

Dear student
$$\begin{gathered} \mathrm{Let}\mathrm{d}\mathrm{is}\mathrm{divisor},{\mathrm{Q}}_1\mathrm{and}{\mathrm{Q}}_2\mathrm{are}\mathrm{quotient}\mathrm{of}\mathrm{first}\mathrm{and}\mathrm{second}\mathrm{cases}\mathrm{respectively}. \\ \mathrm{First}\mathrm{case}: \\ \mathrm{Number}=\mathrm{d}\times {\mathrm{Q}}_1+5 \\ \mathrm{Second}\mathrm{case}: \\ \mathrm{Number}=2\mathrm{d}\times {\mathrm{Q}}_2+45 \\ \mathrm{From}\mathrm{above}\mathrm{cases} \\ \mathrm{d}\times {\mathrm{Q}}_1+5=2\mathrm{d}\times {\mathrm{Q}}_2+45 \\ \mathrm{d}\times {\mathrm{Q}}_1-2\mathrm{d}\times {\mathrm{Q}}_2=45-5 \\ \mathrm{d}\left({\mathrm{Q}}_1-2{\mathrm{Q}}_2\right)=40 \\ \mathrm{d}=\frac{40}{\left({\mathrm{Q}}_1-2{\mathrm{Q}}_2\right)} \\ \mathrm{So},\mathrm{divisors}\mathrm{may}\mathrm{be}40\mathrm{or}\mathrm{factors}\mathrm{of}40\left(\mathrm{i}.\mathrm{e}2,4,5,8,10,20\right) \\ \mathrm{Now},\mathrm{acc}\mathrm{to}\mathrm{Euclid}\mathrm{division}\mathrm{lemma} \\ \mathrm{We}\mathrm{know} \\ \mathrm{a}=\mathrm{bq}+\mathrm{r},\mathrm{where}0\le \mathrm{r}<\mathrm{b} \\ \mathrm{Case}\mathrm{I}:\mathrm{When}\mathrm{remainder}\left(\mathrm{r}\right)\mathrm{is}5\mathrm{and}\mathrm{b}=\mathrm{d} \\ \mathrm{So},0\le 5<\mathrm{d} \\ \mathrm{Case}\mathrm{II}:\mathrm{When}\mathrm{remainder}\left(\mathrm{r}\right)\mathrm{is}45\mathrm{and}\mathrm{b}=2\mathrm{d} \\ \mathrm{So},0\le 45<2\mathrm{d} \\ \Rightarrow 0\le 22.5<\mathrm{d} \\ \mathrm{Hence}\mathrm{d}>22.5 \\ \mathrm{So},\mathrm{it}\mathrm{is}\mathrm{possible}\mathrm{only}\mathrm{when}\mathrm{we}\mathrm{take}\mathrm{d}=40 \end{gathered}$$
Regards