A no. consists of 3 digits , the right hand being zero. If the left hand and middle digits be interchanged the no. is diminished by 180.If the left hand digit be halved and middle and right hand digits be interchanged the no. is diminished by 454. Find the no. .

*x*

and the digit at hundred's place is

*y*.

And the digit at unit's place is 0 as per question.

So the number is 100

*y*+ 10

*x*+ 0 = 100

*y*+ 10

*x*

when the left hand and middle digits be interchanged

New number = 100

*x*+ 10

*y*

According to question, when the left hand and middle digits be interchanged the no. is diminished by 180, so we have;

$100y+10x-180=100x+10y\phantom{\rule{0ex}{0ex}}\Rightarrow 100y-10y+10x-100x=180\phantom{\rule{0ex}{0ex}}\Rightarrow 90y-90x=180\phantom{\rule{0ex}{0ex}}\Rightarrow y-x=2...\left(\mathrm{i}\right)$

Now when the left hand digit be halved and middle and right hand digits be interchanged;

The number is $\frac{100y}{2}+x$ i.e. 50

*y*+

*x*

And according to the question,when the left hand digit be halved and middle and right hand digits be interchanged the no. is diminished by 454, so we have;

$100y+10x-454=50y+x\phantom{\rule{0ex}{0ex}}\Rightarrow 100y-50y+10x-x=454\phantom{\rule{0ex}{0ex}}\Rightarrow 50y+9x=454...\left(\mathrm{ii}\right)$

So from (i) and (ii) we get;

*x*= 6 and

*y*= 8

so the number is (100

*y*+ 10

*x*) = $100\times 8+10\times 6$ = 860

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