Question

In two digit number the digit at units place is equal to square of digit at ten's place . If 54 is added to the number the digits gets interchanged . Find the numbers

Answer 1

Given :-

→ In two digit number the digit at units place is equal to square of digit at ten's place .

→ If 54 is added to the number then the digits gets interchanged .

To find :-

→ The two digit number

Solution :-

Let the digit at tens place of a two digit number be X

The place value of X = 1×X = X

The digit at ones place of the two digit number

= The square of the digit at tens place

= X²

The place value of the X² = 10×X² = 10X²

The two digit number = 10X²+X

The number is obtained by reversing the digits = 10X+X²

According to the given problem

If 54 is added to the number the digits gets interchanged.

=> 10X²+X +54 = 10X+X²

=> 10X²+X+54-10X-X² = 0

=> (10X²-X²)+(X-10X)+54 = 0

=> 9X²+(-9X)+54 = 0

=> 9X²-9X+54 = 0

=> 9(X²-X +6) = 0

=> X²-X+6 = 0/9

=> X²-X+6 = 0

=> X²-3X+2X+6 = 0

=> X(X-3)+2(X-3) = 0

=> (X-3)(X+2) = 0

=> X-3 = 0 or X+2 = 0

=> X = 3 or X = -2

Therefore, X = 3 and -2

X can't be negative,

So , X = 3

The digit at tens place = 3

The digit at ones place

= Square of the digit at tens place

= 3²

= 9

The two digit number =39

Answer :-

The required two digit number is 39

Check :-

Case -1:-

X = 3

The two digit number = 39

The number is obtained by reversing the digits

= 93

= 39+54

We get the relations in the given problem

Note :-

If X = -2 then

The digit tens place = -2

The digit ones place = (-2)² = 4

The two digit number = -24

The number is obtained by reversing the digits = -42

On adding 54 to -24 then

54-24 = 30

But the result will not be -42

So, the number can't be -24

Answer 2

Answer:

Given :-

• In two digit number the digit at units place is equal to square of digit at ten's place.
• If 54 is added to the number the digits get interchange.

To Find :-

• What is the required numbers.

Solution :-

Let,

$\mapsto \bf Ten's\: Place =\: x$

$\mapsto \bf Unit's\: Place =\: y$

Hence, the required number will be :

$\leadsto \sf\bold{\green{Required\: Numbers =\: 10x + y}}$

According to the question :

$\bigstar$ The digit at units place is equal to square of digit at ten's place.

So,

$\small \implies \bf \bigg\{Unit's\: Place\bigg\} =\: \bigg\{Ten's\: Place\bigg\}^2$

$\implies \sf\bold{\purple{y =\: x^2\: ------\: (Equation\: No\: 1)}}$

Again,

$\bigstar$ If 54 is added to the number the digits get interchange.

So,

$\implies \bf 10x + y + 54 =\: 10y + x$

$\implies \sf 10x - x + y - 10y + 54 =\: 0$

$\implies \sf 9x - 9y + 54 =\: 0$

$\implies \sf 9(x - y + 6) =\: 0$

$\implies \sf\bold{\purple{x - y + 6 =\: 0\: ------\: (Equation\: No\: 1)}}$

By putting the equation no 1 in the equation no 2 we get,

$\implies \bf x - y + 6 =\: 0$

$\implies \sf x - x^2 + 6 =\: 0$

$\implies \sf - x^2 + x + 6 =\: 0$

$\implies \sf - (x^2 - x - 6) =\: 0$

$\implies \sf x^2 - x - 6 =\: 0$

$\implies \sf x^2 - (3 - 2)x - 6 =\: 0$

$\implies \sf x^2 - 3x + 2x - 6 =\: 0$

$\implies \sf x(x - 3) + 2(x - 3) =\: 0$

$\implies \sf (x - 3)(x + 2) =\: 0$

$\implies \bf x - 3 =\: 0$

$\implies \sf\bold{\blue{x =\: 3}}$

Or,

$\implies \bf x + 2 =\: 0$

$\implies \sf\bold{\blue{x =\: - 2}}\: \: \bigg\lgroup \sf\bold{\pink{Numbers\: can't\: be\: negetive}}\bigg\rgroup$

So, the value of x is 3 .

Again, by putting the value x in the equation no 1 we get,

$\implies \bf y =\: x^2$

$\implies \sf y =\: (3)^2$

$\implies \sf y =\: (3 \times 3)$

$\implies \sf\bold{\blue{y =\: 9}}$

Hence, the required number is :

$\dashrightarrow \bf Required\: Number =\: 10x + y$

$\dashrightarrow \sf Required\: Number =\: 10(3) + 9$

$\dashrightarrow \sf Required\: Number =\: 30 + 9$

$\dashrightarrow \sf\boxed{\bold{\red{Required\: Number =\: 39}}}$

$\sf\bold{\pink{\underline{\therefore\: The\: required\: numbers\: is\: 39\: .}}}$

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