Question

What is the smallest number that when divided by 35,56,91 leaves remainder of 5 in each case

Answer 1

Given :-

Smallest number  when divided by 35 , 56 and 91 leaves remainder as 5 .

To Find :-

Tat smallest number .

Solution :-

Let's assume that number as x .

It's given that the number is leaving 5 as a remainder in each case . So it means x-5 will be completely divisible by the given numbers .

So we conclude that the number is multiple of all these given numbers .

By taking LCM of the given numbers .

$\begin{array}{c|c} 5 & 35 \\ \cline{1-2} 7 & 7 \\ \cline{1-2 } & 1 \end{array} \;\;\; \begin{array}{c|c} 2 & 56 \\ \cline{1-2} 2 & 28 \\ \cline{1-2 } 2 & 14 \\ \cline{1-2} 7 & 7 \\ \cline{1-2} & 1 \end{array}\;\;\;\; \begin{array}{c|c} 7 & 91 \\ \cline{1-2} 13 & 13 \\ \cline{1-2 } & 1 \end{array}$

Prime factors are :-

35 = 5 × 7

56 = 2×2×2×7

91 = 7×13 .

$\sf{\implies \; LCM = 2^{3} \times 5 \times 7 \times 13 }$

$\sf{\implies \; LCM = 3640 }$

So 3640 is the smallest number divisible by 35 , 56 and 91 .

$\sf{\implies \; x-5 = 3640 }$

$\sf{\implies \; x = 3640 + 5 }$

So the required number is 3645

Answer 2

$\large\underline\bold{Question:-}$

$\large\underline\bold{Answer:-}$

$\small\underline\bold{Given:-}$

$\sf\ The\: numbers\: given\: are\: 35, 56, 91.$

$\sf\ When\: the\: smallest\: number\: divided\:by\: the$

$\sf\ above\: numbers,\: they\: leave\: remainder\: of \:5$

$\sf\ in\: each \:case.$

$\small\underline\bold{To\:Find:-}$

$\sf\ The\: smallest \:number \:which\: when\: divided\: by \:the$

$\sf\ given \:numbers \:leaving\: remainder\: 5$

$\small\underline\bold{Required\: Solution:-}$

$\sf\ Any \:smallest\: number \:dividing\: the\: numbers\: is$

$\sf\ nothing\: but\: the\: Least\: Common\: Multiple$

$\sf\ of\: all\: the \:numbers,\: which\: is \:LCM.$

$\sf\ Let\: us \:find\: the\: LCM\: of\: all\: the$

$\sf\ given\: numbers \:35,56,91$

$\sf\ Prime\: factors\:of\:given\: numbers$

⟹$\sf\ 35=5×7$

⟹$\sf\ 56=2×2×7$

⟹$\sf\ 91=7×13$

$\sf\ So,\:LCM= 2^3×5×7×13$

⟹$\sf\ LCM=3640$

$\sf\ Now\: the\: smallest\: number\: when\: divided\: by$

$\sf\ 35,\: 56,\: 91\: leaves\: the\: remainder\: 5\: in\: each \:case$

⟹$\sf\ x+5=3640$

⟹$\sf\ x=3640+5$

$\large\underline\bold\purple{x=3645}$

$\large\bold{Hence\:the\:answer\:is\:3645.}$