Question

Find the least number which when divided by 18 24 30 and 42 will leave in each case same remainder​

Answer 1

AnswEr :

2521.

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

When divided by 18, 24, 30 and 42 will leave in each case same remainder 1.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The least number.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Taking L.C.M. of 18,24,30 and 42 we get;

$\begin{array}{l|r}\cline{2-2} 2& 18,24,30,42 \\ \cline{2-2}2&9,12,15,21 \\ \cline{2-2} 2& 9,6,15,21 \\ \cline{2-2} 3& 9,3,15,21\\ \cline{2-2} 3& 3,1,5,7\\ \cline{2-2} 5& 1,1,5,7\\ \cline{2-2} 7& 1,1,1,7\\ \cline{2-2} & 1,1,1,1\end{array}}$

∴ L.C.M = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520.

Thus,

The least required number is 2520 + 1 = 2521.

$\boxed{\begin{minipage}{6 cm} \underline{Full-form of L.C.M and H.C.F\::} \\ \\ {\sf L.C.M. = Lowest \:Common\:Multiple}\\ \\ \sf{H.C.F=Highest\:Common\:Factor} \end{minipage}}$

Answer 2

Step-by-step explanation:

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