Help needed for number problems.
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?
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.