Question

How many two digit whole numbers are increased by 18 when their digits are reversed?

Answer 1

Hello ganeshie8. Long time since we have seen each other. :)

Answer 2

I don't really know math so I think my approach would be a simple and quick program.

Answer 3

:) lets write an equation in powers of 10

Answer 4

Ok

Answer 5

say the two digit number = \(xy = 10^1x+10^0y = 10x+y\)

then, its reverse wud be : \(yx = 10^1y+10^0x = 10y+x\)

Answer 6

\(10x + y = 10y-x+18\)

solve it in whole numbers :)

Answer 7

Wait. How did you get this equation?

Answer 8

typo :

\(10x + y = 10y \color{red}{+}x+18\)

Answer 9

How many two digit whole numbers are increased by 18 when their digits are reversed

so the equation is incorrect still :|

Answer 10

lets fix it :-

reversed number = original number + 18
\(10y + x = 10x +y + 18\)

Answer 11

see if that makes sense; we can go wid solving the equation...

Answer 12

So in short, you just reversed the position of x and y to be with 10 right? And then you just added 18. But why did you do it only to the right equation, not the left?

Answer 13

okay, lets see if i understood ur q correctly...

take some number :-
\(2013\)

can you write it in powers of 10 ? :)

Answer 14

I am not too sure with this. But I will use my prior knowledge for this.
2.013 x 10^3
20.13 x 10^2
201.3 x 10^1
2013 x 10^0

Answer 15

thats right, but i was asking even more simpler question; we can express \(2013\) in sum of powers as below :-

\(2013\)

\(2\) thousands = \(2\times 10^3\)
\(0\) hundreds= \(0\times 10^2\)
\(1\) tens = \(1\times 10^1\)
\(3\) ones = \(3\times 10^0\)
----------------------------------
\(2013 = 2 \times 10^3 + 0 \times 10^2 + 1 \times 10^1 + 3 \times 10^0\)

Answer 16

Yes :)

Answer 17

cool :) so a number \(xy\) can be expressed as \(x \times 10^1 + y \times 10^0 \)

Answer 18

Makes sense. What is the next step?

Answer 19

write its reverse also in powers of 10 and form an equation

Answer 20

Couldn't get you there. Can you explain quite more of what you meant?

Answer 21

\(xy = x \times 10^1 + y \times 10^0 = 10y + x\)

reverse of \(xy = yx = y\times 10^1 + x \times 10^0 = 10y+x\)

Answer 22

yahh, wid me so far ?

Answer 23

Why did u put 10y though and then plus x? (Sorry for asking u so many question :/)

Answer 24

whew ! typo again, it should be :-

\(xy = x \times 10^1 + y \times 10^0 = 10x + y\)
reverse of \(xy = yx = y\times 10^1 + x \times 10^0 = 10y+x\)

Answer 25

see if it looks okay now ?

Answer 26

Yes now makes sense :)

Answer 27

nice :) next we're asked to find numbers such that reverse is 18 more than the original :-

reversed number = original number + 18
\(10y+x=10x+y+18\)

Answer 28

rearrange and simplify a bit...

Answer 29

Ok

Answer 30

Um.. But can you help me with the rearranging and the simplification though?

Answer 31

let the no. at unit place=x
no.at ten's place=y
no=10y+x
according to the given condition
10x+y=10y+x+18
9x=9y+18
x=y+2
now y>0
\[0 \le x \le 9 also x>y as x=y+2 if we take 1,2,3,4,5,6,7 ,x=3,4,5,6,,7,8,9 respectively\]
so nos are 13,24,,35,46,57,68,79
Hence they are 7 in no.

Answer 32

What is no.?

Answer 33

you can see thee are seven two digit nos.

Answer 34

And surjithayer how do you get x=y+2

Answer 35

divide by 9,9x=9y+18

Answer 36

note y >0
if y=0
then it is not a two digit no.