Question

In a two-digit positive number, the digit in the unit's place is equal to the square of the digit in the ten's place and the difference between the number and the number obtained by interchanging the digits is 54, then the sum of its digit is

Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

GiveN:

• The digit in the unit's place is equal to the square of the digit in the ten's place.
• The difference between the number and the number obtained by interchanging the digits is 54

To FinD:

• Sum of the digits?

Step-by-Step Explanation:

Let the digit in ten's place be x and the digit in unit's place be y. Then the no. will be 10x + y.

According to condition -1)

❒ Digit in the unit's place = Digit in the ten's place

⇒ y = x² ------------(1)

According to condition -2)

As the unit digit is more than the tens digit, the number formed by interchanging the digits is greater than the original number. Then,

❒ Reversed no. - Original no. = 54

⇒ 10y + x - (10x + y) = 54

⇒ 10y + x - 10x - y = 54

⇒ 9y - 9x = 54

⇒ y - x = 6

⇒ y = x + 6 -----------(2)

The LHS of Equation (1) and Equation (2) is same. So,

⇒ x² = x + 6

⇒ x² - x - 6 = 0

⇒ x² - 3x + 2x - 6 = 0

⇒ x(x - 3) + 2(x - 3) = 0

⇒ (x + 2)(x - 3) = 0

Then, x = -2 or 3.

Digit can't be negative.

Hence:

• x = 3 and y = 9
• The sum of the digits is 3 + 9 = 12 (Ans)

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large{\green{\underline \bold{\tt{Given :-}}}}$

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• In a two-digit positive number, the digit in the unit's place is equal to the square of the digit in the ten's place

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• Difference between the number and the number obtained by interchanging the digits is 54

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$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

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• Sum of its digit

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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• Let the unit digit be "y"

• Let the ten's digit be "x"

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$\purple{\underline \bold{According \: to \: the \ question :}}$

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The digit in the unit's place is equal to the square of the digit in the ten's place

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➠ x² = y -------- (1)

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$\underline{\bold{\texttt{Original number :}}}$

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➠ 10x + y

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$\underline{\bold{\texttt{Reversed number :}}}$

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➠ 10y + x

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Given that difference between the number and the number obtained by interchanging the digits is 54

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So,

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➜ 10y + x - (10x + y) = 54

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➜ 10y + x - 10x - y = 54

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➜ -9x + 9y = 54

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⟮ Dividing the above equation by "9" ⟯

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We got,

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➜ -x + y = 6 -------- (2)

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⟮ Putting y = x² from (1) to (2) ⟯

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➜ -x + y = 6

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➜ -x + x² = 6

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➜ x² - x - 6 = 0

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⟮ Splitting the middle term ⟯

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➜ x² + 2x - 3x - 6 = 0

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➜ x(x + 2) -3(x + 2) = 0

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➜ (x + 2)(x - 3) = 0

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• x = -2

• x = 3

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But it is given that the number is a positive number ,

Thus x = 3

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• Hence the ten's digit is 3

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⟮ Putting x = 3 in (1) ⟯

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➜ x² = y

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➜ 3² = y

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➨ y = 9

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• Hence the one's digit is 9

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$\underline{\bold{\texttt{Sum of its digit :}}}$

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• Ten's digit = 3

• One's digit = 9

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Sum of digits = 3 + 9

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➨ 12

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• Hence the sum of digits is 12

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