# 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...

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?

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(1) Let

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.**

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.

7+4

9+2

6+5

8+3

So all of these give you the required answer.

**So the original number can be one of the following;**

**74**

**92**

**65**

**83**

**If take the reverse of the above numbers first you will get another set of answers.**

**47**

**56**

**38**

**29**