Question

find the greatest number that will divide 37,74 and 183 leaving 1 ,2 and 3 AS remainders respectively

Answer 1

$\underline{\underline{\Huge{\textbf{Question:}}}}$
Find the $\textsf{greatest number that will divide}$ 37, 74 and 183 leaving 1 ,2 and 3 as remainders respectively

$\underline{\underline{\LARGE{\textsf{Concept\: Behind:}}}}$

$\textsf{No. that will divide \implies Divisor}$

$\blacksquare \:$ When a No. Q divided by P leaves $\textit{zero as a remainder}$, then it is said that $\textbf{P divides Q}$. And hence P is a $\textsf{factor}$ of Q.

$\blacksquare \:$ If in the division, $\textsf{Remainder \neq 0}$
Subtracting the remainder from the Dividend gives the $\textsf{New - dividend}$ which when divided by divisor leaves Zero as the remainder

$\blacksquare \:$ When $\textit{more dividends}$ are considered,
Greatest No. that will divide(which means divisor) is the $\textsf{Highest Common Factor (H.C.F) of the New dividends}$

$\therefore$
$\bigstar \textsf{Greatest No. that will divide x, y and z leaves a, b and c is \textbf{H.C.F[(x-a), (y-b), (z-c)]}}$

H.C.F is also called as $\textbf{Greatest\: Common\: Divisor(G.C.D)}$

$\underline{\underline{\huge{\textbf{Solution:}}}}$

$\star$ Note the keyterms in the question given and analyse the concept behind.

Given,
Dividends are 37, 74, 183
Remainders are 1, 2, 3

Say
$\begin{array}{a\: b\: c}x= 37&y=74&z=183\\a=1&b=2&c=3\end{array}$

$\underline{\large{\textit{Step-1}}}$
Find the new dividends which will leave zero as remainder when divided by divisor.

Let ($d_{1},d_{2},d_{3}$) be the new dividends

$\begin{array}{a\: b\: c} d_{1}= x-a&d_{2} = y-b&d_{3}= z-c\\ d_{1} = 37-1=36& d_{2}=74-2=72&d_{3}=183-3=180 \end{array}$

$\underline{\large{\textit{Step-2}}}$
Find HCF of new dividends [i.e., ($d_{1},d_{2},d_{2})$]

$\diamondsuit{}\: \textsf{Using Prime Factorisation, Express the Numbers as Product of Prime factors}$

$\begin{array}{r|l}2 & 36 \\ \cline{1-2} 2&18\\\cline{1-2}3&9\\\cline{1-2}3&3\\\cline{1-2}&1\end{array} \quad\quad \begin{array}{r|l}2 & 72 \\ \cline{1-2} 2&36\\ \cline{1-2}2&18\\\cline{1-2}3&9\\\cline{1-2}3&3\\\cline{1-2}&1\end{array} \quad\quad \begin{array}{r|l}2& 180 \\ \cline{1-2} 2&90\\ \cline{1-2}5&45\\\cline{1-2}3&9\\\cline{1-2}3&3\\\cline{1-2}&1 \end{array}$

$\diamondsuit{}\: \textsf{Note the common Prime factors}$

$36 = \quad \: \: \: 2 \times 2 \times 3 \times 3\\72 = 2 \times 2 \times 2 \times 3 \times 3\\180 = 5 \times 2 \times 2 \times 3 \times 3$

Common Factors = 2, 2, 3, 3

$\diamondsuit{}\: \textsf{Product of common prime factors is the required HCF of the Numbers}$

$\implies$ Required HCF = $2\times 2\times 3\times 3 = 36$

$\therefore$

$\bigstar\: \textsf{The greatest number that will divide 37, 74 and 183 leaving 1 ,2 and 3 as remainders respectively is \underline{\large{\textbf{36}}}}$


$\underline{\underline{\Large{\textsf{Verification:}}}}$

$37 = \: \: 1(36) +1 \\ 74 = \: \: 2(36)+2 \\ 183 =5(36) +3$