# 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. .

Suppose the digits at ten's place is 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 100y + 10x  + 0 = 100y + 10x
when the left hand and middle digits be interchanged
New number = 100x + 10y
According to question, when the left hand and middle digits be interchanged the no. is diminished by 180, so we have;

Now when the left hand digit be halved and middle and right hand digits be interchanged;
The number is i.e. 50y +
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;

So from (i) and (ii) we get;
x = 6 and y = 8
so the number is (100y + 10x) = $100×8+10×6$ = 860

• 8

lets find thethe original number

let x be in hundred place

y be in tens place

third digit is 0 so dont mind it

original number =100x+10y

as per first condition new number

2nd digit becomes first digit

so new number =100y + 10x + 0

EQUATION 1

100x + 10y-180=100y+10x

simplifying we get 90x - 90y=180

90(x - y)=180

x-y=180/90=2

as per second condition new number

fiest digit gets halfed

100x/2=50x

next 2nd digit becomes 1st and viceversa

so 0+1*y

EQUATION 2

100x + 10y -454=50x + y

simplifying we get 50x + 9y=454

third step = multiply 50 in equation 1

so50(x - y)=50*2

50x-50y=100

next subtract thirs step from equation 2

50x + 9y=454

--------

50x-50y=100

___________

0+ 59y =354

we get 59y=354

so y = 6

putting this value to equation 1 we get x as 8

so original number

100x+10y = 800 + 60=860

thumbs up if its correct because I worked hard on it

• 15
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