A two-digit number is 9 more than four times of the number obtained by interchanging its digits. If the product of digits in the two-digit number is 8, then what is the number?

Question

Prepp Team · Accepted Answer

This problem asks us to find a specific two-digit number based on two conditions related to its digits and the number formed by interchanging them.

Understanding the Problem Components

We are looking for a two-digit number.
The first condition relates the number to the number formed by interchanging its digits. Specifically, the original number is 9 more than four times the reversed number.
The second condition states that the product of the digits is 8.

Representing the Two-Digit Number Algebraically

Let the tens digit of the two-digit number be '$a$' and the units digit be '$b$'. The value of the two-digit number can be expressed as $10a + b$. For example, if the number is 73, then $a=7$ and $b=3$, and the value is $10(7) + 3 = 73$.

The number formed by interchanging the digits (swapping the tens and units digits) would have '$b$' as the tens digit and '$a$' as the units digit. The value of this reversed number is $10b + a$.

Translating Conditions into Equations

Let's translate the two given conditions into mathematical equations:

Condition 1: "A two-digit number is 9 more than four times of the number obtained by interchanging its digits." This translates to: $$10a + b = 4(10b + a) + 9$$
Condition 2: "The product of digits in the two-digit number is 8." This translates to: $$a \times b = 8$$

Solving the Equations Systematically

We now have a system of two equations with two variables:

$10a + b = 4(10b + a) + 9$
$a \times b = 8$

Simplifying the First Equation:

Let's simplify the first equation:

$10a + b = 40b + 4a + 9$

Now, group the terms with '$a$' and '$b$' on one side:

$10a - 4a + b - 40b = 9$

$6a - 39b = 9$

We can simplify this equation by dividing all terms by 3:

$$2a - 13b = 3 \quad (\text{Equation i})$$

Using the Second Equation:

The second equation is:

$$a \times b = 8 \quad (\text{Equation ii})$$

From Equation (ii), we can express '$a$' in terms of '$b$' (since the digits must be non-zero for their product to be 8):

$$a = \frac{8}{b}$$

Substitution and Solving for 'b':

Substitute the expression for '$a$' from Equation (ii) into Equation (i):

$2\left(\frac{8}{b}\right) - 13b = 3$

$$\frac{16}{b} - 13b = 3$$

To eliminate the fraction, multiply the entire equation by '$b$' (remember $b \ne 0$):

$16 - 13b^2 = 3b$

Rearrange this into a standard quadratic equation form ($Ax^2 + Bx + C = 0$):

$13b^2 + 3b - 16 = 0$

We can solve this quadratic equation for '$b$'. Let's try factoring. We need two numbers that multiply to $13 \times (-16) = -208$ and add to $3$. The numbers $16$ and $-13$ satisfy these conditions ($16 \times -13 = -208$ and $16 + (-13) = 3$).

$13b^2 + 16b - 13b - 16 = 0$

Factor by grouping:

$b(13b + 16) - 1(13b + 16) = 0$

$(b - 1)(13b + 16) = 0$

This gives two possible solutions for '$b$':

$b - 1 = 0 \implies b = 1$
$13b + 16 = 0 \implies b = -\frac{16}{13}$

Since '$b$' represents a digit, it must be a non-negative integer between 0 and 9. Therefore, the only valid value for '$b$' is $1$.

Finding the Value of 'a':

Now, use the value $b=1$ and Equation (ii) ($a \times b = 8$) to find '$a$':

$a \times 1 = 8$

$a = 8$

The digits are $a=8$ and $b=1$. Both are valid digits.

Constructing the Two-Digit Number

The two-digit number is formed using the digits $a=8$ (tens digit) and $b=1$ (units digit).

The number is $10a + b = 10(8) + 1 = 80 + 1 = 81$.

Verifying the Solution

Let's check if the number 81 satisfies both original conditions:

Condition 2 Check (Product of digits): The digits are 8 and 1. Their product is $8 \times 1 = 8$. This condition is satisfied.
Condition 1 Check (Relation with reversed number): The number is 81. The number obtained by interchanging its digits is 18. Is 81 equal to 9 more than four times 18? Calculate $4 \times 18 + 9$: $4 \times 18 = 72$ $72 + 9 = 81$ So, $81 = 4(18) + 9$. This condition is also satisfied.

Alternative Method: Checking the Options

We can also verify the given options:

Option 1: 81 Digits are 8 and 1. Product = $8 \times 1 = 8$. (Condition 2 met). Reversed number = 18. Check: $4 \times 18 + 9 = 72 + 9 = 81$. (Condition 1 met). This is the correct number.
Option 2: 42 Digits are 4 and 2. Product = $4 \times 2 = 8$. (Condition 2 met). Reversed number = 24. Check: $4 \times 24 + 9 = 96 + 9 = 105$. $42 \ne 105$. (Condition 1 not met).
Option 3: 24 Digits are 2 and 4. Product = $2 \times 4 = 8$. (Condition 2 met). Reversed number = 42. Check: $4 \times 42 + 9 = 168 + 9 = 177$. $24 \ne 177$. (Condition 1 not met).
Option 4: 18 Digits are 1 and 8. Product = $1 \times 8 = 8$. (Condition 2 met). Reversed number = 81. Check: $4 \times 81 + 9 = 324 + 9 = 333$. $18 \ne 333$. (Condition 1 not met).

Both methods confirm that the number satisfying the conditions is 81.

Final Answer

The two-digit number that meets both conditions is 81.

Prepp Team · Answer

42

Prepp Team · Answer

24

Prepp Team · Answer

18