Question

A two digit number is such that the product of its digits us 8 ,if 18is added to the number the digits interchange. Their places.find the number?​

Answer 1

Step-by-step explanation:

Let the no. be 10x+y

Here x × y = 8

x= 8/y

So, acc. to question

10x+y+18= 10y+x

9x-9y+18=0

9(x-y+2)=0

x-y+2=0/9

x-y+2=0

(8/y)-y+2=0

8-y²+2y=0×y

-(y²-2y-8) =0

y²-4y+2y-8=0

y(y-4) +2(y-4) =0

(y+2) (y-4) =0

y=-2 y=4

If y= -2

Then x= 8/-2 = -4

If y = 4

Then x= 8/4= 2

So, the no. is -42 or 24

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Answer 2

Step-by-step explanation:

Let unit's digit be x

$\therefore \: \tt ten's \: digit = \frac{8}{x}$

$\sf \: Original \: number = 10 \times \frac{8}{x} + x \\ \implies \: \tt \: \frac{80}{x} + x \\ \implies \tt \: \frac{ 80 + \tt{ x}^{2} } {x} \ \ \\ \\\ \tt \N umber \: after \: interchanging \: its \: digits \\ \implies \: \tt10x + \frac{8}{x} \\ \tt \implies \: \frac{ 10 {x }^{2} + 8 }{x}\\ \\ \tt \: ACQ \\ \therefore \tt\: \frac{ 10 {x}^{2} + 8}{x} = \frac{80 + {x}^{2}}{x} + 18 \\ \tt \longrightarrow \: 10 {x}^{2} + 8 = 80 + {x}^{2} + 18x \\ \tt \longrightarrow \: 9 {x }^{2} - 18x - 72 = 0 \\ \tt \longrightarrow {x }^{2} - 2x - 8 = 0 \\ \tt \longrightarrow {x }^{2} - 4x + 2x - 8 = 0 \\ \tt \longrightarrow \: x(x - 4) + 2(x - 4) = 0 \\ \tt \longrightarrow(x - 4)(x - 2) = 0 \\ \\ \tt \: If \: x - 4 = 0 \\ \tt x = 4 \\ \\ \tt \: Unit's \: digit = 4 \\ \tt \: Ten's \: digit = \frac{8}{4} = 2 \\ \tt \: \red{Number = 24}$