# Question:- The sum of the squares of the digits of a two-digit number is 13.If we subtract 9 from that number , we get a num

By Ivy

Question:-
The sum of the squares of the digits of a two-digit number is 13.If we subtract 9 from that number , we get a number consisting of the same
digits written in the reverse order. Find the number.

### 2 thoughts on “Question:-<br /> The sum of the squares of the digits of a two-digit number is 13.If we subtract 9 from that number , we get a num”

1. Acçórding to the quéstion:-

Súbtract 9 fróm the oríginál númber , we get a numbér consísting of the sáme digits writtén in the revérse órder.

$$\sf{ \to \color{blue}óriginal\:number-9 = réverse\: fórm\:óf \: óriginàl\:númbér}$$

$$\sf{ \implies(10y+ a) – 9 = (10a +y)}$$

Tránspósíng the líke térms

$$\sf{ \implies10y -y -9 = 10a-a} \\ \\ \sf{ \implies9y -9 = 9a} \\ \\ \sf{ \implies9×(y-1 )= 9a} \\ \\ \bold{ \implies \red{y -1 = x}\: \: \: \: ….(2.)}$$

Nów súbsitúte the válúe of x = ( y-1) in a ²+y² = 13

$$\sf{ \implies a²+y² = 13} \\ \\ \sf{ \implies(y-1)² +y² =13 }\\ \\ \sf{ \implies y²+1 -2y +y² =13 }\\ \\ \sf{ \implies 2y² -2y +1 =13 }\\ \\ \sf{ \implies 2y² -2y = 12 }\\ \\ y²- y = 6$$

míddlé splít the quàdratiç éqúatión

$$\sf{ \implies y²+2y -3y -6 = 0 }\\ \\ \sf{ \implies y (y+2)-3(y+2) = 0} \\ \\ \sf{\implies(y+2)(y-3) = 0 }\\ \\ \sf{ \implies y = – 2 \: ór \: 3}$$

So the válue of y is 3

pút the valúe of y in eqúatíon (2)

$$\sf{ \implies x = y-1 }\\ \\ \sf{ \implies a= 3-1 }\\ \\ \sf{ \implies \red{ a= 2}}$$

Theréfóre the númbér fórméd is 32

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2. ## Question:-

• The sum of the squares of the digits of a two-digit number is 13.If we subtract 9 from that number , we get a number consisting of the same digits written in the reverse order. Find the number.

## To Find:–

• Fine the number.

## Solution:–

Here ,

Let the numbers be x and y

Given ,

Sun of the squares of number is 13:-

• $$\tt \: { x }^{ 2 } + { y }^{ 2 } = 13$$

According to the Question:-

$$\tt\implies \: 10x + y – 9 = 10y + x$$

$$\tt\implies \: 10x – x – 10y + y = 9$$

$$\tt\implies \: 9x – 9y = 9$$

$$\tt\implies \: 9( x – y ) = 9$$

$$\tt\implies \: x – y = \cancel\dfrac { 9 } { 9 }$$

$$\tt\implies \: x – y = 1 . . . . ( i )$$

Now ,

$$\tt\implies \: { ( x – y ) }^{ 2 } = { x }^{ 2 } + { y }^{ 2 } – 2xy$$

$$\tt\implies \: 1 = 13 – 2xy$$

$$\tt\implies \: 2xy = 12$$

$$\tt\implies \: xy = \cancel\dfrac { 12 } { 2 }$$

$$\tt\implies \: y = \dfrac { 6 } { x }$$

Now ,

Substitute y in equation ( i ):-

$$\tt\implies \: x – \dfrac { 6 } { x } = 1$$

$$\tt\implies \: { x }^{ 2 } – x – 6 = 0$$

$$\tt\implies \: { x }^{ 2 } + 2x – 3x – 6 = 0$$

$$\tt\implies \: x( x + 2 ) – 3( x + 2 ) = 0$$

$$\tt\implies \: ( x + 2 ) ( x – 3 ) = 0$$

$$\tt\implies \: x = – 2 , 3$$

Hence ,

• The number is 32