Question

In a 3 digit number , unit's digit is one more than the hundred's digit and ten's digit is one less than the hundred's digit . If the sum of the original 3 digit number and numbers obtained by changing the order of the digits cyclically is 2664 , find the number​

Answer 1

Answer:

Hey mate here is ur ans

Step-by-step explanation:

Let hundred digit =x

ten digit =x−1

unit digit =x−1

First form of Number =100×x+100(x−1)+x+1

=111x−9

Second form of Number =(x+1)100+x×10+x−1

=111x+99

third form of number =(x−1)100+(x+1)10+x

=111x−90

So 111x−9+111x+99+111x−90=2664

111x=2664

x=8

So the number 8×100+7×10+9

=879

Answer 2

Let us consider the hundreds digit be ‘x’

Unit digit be ‘x + 1’

and ten’s digit be ‘x – 1’

So the number = (x + 1) + 10(x – 1) + 100 × x

= x + 1 + 10x – 10 + 100x

= 111x – 9

By reversing the digits,

Unit digit be ‘x – 1’

Tens digit be ‘x’

Hundred digit be ‘x + 1’

So the number = x – 1 + 10x + 100x + 100

= 111x + 99

and sum of original 3-digit number = x + 10(x + 1) + 100(x – 1)

= x + 10x + 10 + 100x – 100

= 111x – 90

Now according to the condition,

111x – 9 + 111x + 99 + 111x – 90 = 2664

333x + 99 – 99 = 2664

333x = 2664

x = 2664/333

= 8

∴ The number = 111x – 9

= 111(8) – 9

= 888 – 9

= 879

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