Question

The integer N is greater than 120 When N is divided by 28 the remainder is 3 When N is divided by 120 the remainder is 3 Find the least value of N. You must show your working

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

☆ When integer N is divided by 28, its leaves the remainder 3.

and

☆ When integer N is divided by 120, it also leaves the remainder 3.

☆ It implies, the integer N should be

$\bf :\longmapsto\:N \: = \: LCM \: (28, \: 120) \: + \: 3$

So,

☆ Let find the LCM of 28, 120 using prime factorization.

$\green{\bf :\longmapsto\:Prime \: factorization \: of \: 28}$

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:28 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:14 \:\:}} \\\underline{\sf{7}}&\underline{\sf{\:\:7\:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\green{\bf :\longmapsto\:Prime \: factorization \: of \: 28 = {2}^{2} \times 7}$

$\red{\bf :\longmapsto\:Prime \: factorization \: of \: 120}$

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:120 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:60 \:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:30\:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:15 \:\:}} \\ {\underline{\sf{5}}}& \underline{\sf{\:\:5\:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\red{\bf :\longmapsto\:Prime \: factorization \: of \: 120 = {2}^{3} \times 5 \times 3}$

So,

$\blue{\bf:\longmapsto\:LCM(28, 120)}$

$\rm \:= \: \: {2}^{3} \times 3 \times 5 \times 7$

$\rm \:= \: \: 8 \times 15 \times 7$

$\rm \:= \: \: 120 \times 7$

$\rm \:= \: \:840$

$\blue{\bf:\longmapsto\:LCM(28, 120) = 840}$

Hence,

$\bf :\longmapsto\:N \: = \: LCM \: (28, \: 120) \: + \: 3$

$\bf :\longmapsto\:N \: = \: 840 \: + \: 3$

$\bf :\longmapsto\:N \: = \: 843 \: \:$