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If you take two-digit number, reverse its digit to make a second two digit number & add these two numbers together, their sum will be 121. What is the original number?
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Lets say the original number is xy.
Every decimal can be expressed in the following manner.
xy = x*10^1+y*10^0 = 10x+y
For example lets say 25.
25 = 2*10^1+1^10^0 = 20+5 = 25
So for first number we can say;
xy = 10x+y
For second number we can say;
yx = 10y+x
then sum of xy and yx is 121.
10x+y+10y+x = 121
(x+y)*10^1 +(x+y)*10^0 = 1*10^2+2*10^1+1*10^0
Now it is clear that (x+y)>10 that's why we get 10^2 term in the right side.
Then considering 10^0 coefficient (x+y)=1
If (x+y) = 1 then (x+y)>10 is violated.
we know 11= 1*10^1+1*10^0
So x+y = 11
consider how we get 11 by various numbers.
So all of these give you the required answer.
So the original number can be one of the following;
If take the reverse of the above numbers first you will get another set of answers.
y & z - represent the digits of the numbers
A - the original number
B - the number with reversed digit
Set up the equation for the original number. To do so, let y represent the ten's digit and z the unit's digit.
`A = y(10) + z`
`A = 10y + z`
Also, set up the equation for the number with reversed digit.
`B = z(10) + y`
`B = 10z + y`
Since the digits are reversed, the z here becomes the ten's digit and y the unit's digit.
Then, add A and B.
`A + B = 121`
Susbtitute the equations of A and B.
`10y + z + 10z + y = 121`
`11y + 11z = 121`
Divide both sides by 11 to simplify.
`y + z = 11`
Note that y and z represents digits of a number which indicates that their values vary from 0 to 9.
So, assign values of x and y that will give a sum of 11. Then, check if it satisfy the given condition.
y=2 & x=9 ==> A=29 & B=92 ==> `29+92=121`
y=3 & x=8 ==> A=38 & B=83 ==> `38+83=121`
y=4 & x=7 ==> A=47 & B=74 ==> `47+74=121`
y=5 & x=6 ==> A=56 & B=65 ==> `56+65=121`
Also, take note that in the problem. it does not indicate if the orginal number A is less than B. So, if we have A greater than B and add the two numbers, their sum is still 121.
Hence, there are 8 possible values of the original number. These are 29, 38, 47, 56, 65, 74, 83 and 92.
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