A two digit number is such that its tens digit is greater than its unit digit by 4. The two digit number is less than 4 times the product of its digit by 11. Find the number

Answer 1

Let the two-digit number be represented as \( 10t + u \), where \( t \) is the tens digit and \( u \) is the units digit.

According to the problem, we have the following two conditions:

1. The tens digit is greater than the units digit by 4: \[ t = u + 4 \]

2. The two-digit number is less than 4 times the product of its digits by 11: \[ 10t + u < 4(t \cdot u) \times 11 \]

Now, substituting \( t \) from the first equation into the second equation:

Step 1: Substitute \( t \) in the second inequality.

Starting with the inequality: \[ 10t + u < 4(t \cdot u) \times 11 \] Replacing \( t \): \[ 10(u + 4) + u < 4((u + 4) \cdot u) \times 11 \] This simplifies to: \[ 10u + 40 + u < 4(u^2 + 4u) \times 11 \] Thus, \[ 11u + 40 < 44(u^2 + 4u) \]

Step 2: Expand and rearrange the inequality.

Expanding the right side gives: \[ 11u + 40 < 44u^2 + 176u \] To rearrange, we bring all terms to one side: \[ 44u^2 + 176u - 11u - 40 > 0 \] This simplifies to: \[ 44u^2 + 165u - 40 > 0 \]

Step 3: Solve the quadratic inequality.

Let’s first find the roots of the quadratic equation \( 44u^2 + 165u - 40 = 0 \) using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ a = 44, \quad b = 165, \quad c = -40 \] Calculating the discriminant: \[ D = b^2 - 4ac = 165^2 - 4 \cdot 44 \cdot (-40) \] \[ D = 27225 + 7040 = 34265 \] Finding the roots: \[ u = \frac{-165 \pm \sqrt{34265}}{2 \cdot 44} \]

Calculating \( \sqrt{34265} \approx 185 \) (approximately), \[ u = \frac{-165 \pm 185}{88} \] Calculating the two potential roots:

1. \( u_1 = \frac{20}{88} \approx 0.227 \)
2. \( u_2 = \frac{-350}{88} \approx -3.98 \)

Since \( u \) must be a single digit (from 0 to 9), we need to check values of \( u \).

Step 4: Check potential integer values for \( u \).

From \( t = u + 4 \), since \( t \) also has to be a digit:

• The possible values of \( u \) are from 0 to 5 (because if \( u \) was more than 5, \( t \) would be more than 9).

Let’s compute valid numbers:

1. If \( u = 0 \): \( t = 4 \), number = 40
2. If \( u = 1 \): \( t = 5 \), number = 51
3. If \( u = 2 \): \( t = 6 \), number = 62
4. If \( u = 3 \): \( t = 7 \), number = 73
5. If \( u = 4 \): \( t = 8 \), number = 84
6. If \( u = 5 \): \( t = 9 \), number = 95

Step 5: Check the second condition for each number.

Calculating \( 4(t \cdot u) \times 11 \) for these pairs:

1. For 40: \( 4(4 \cdot 0) \times 11 = 0 \) \( (< 40) \) False
2. For 51: \( 4(5 \cdot 1) \times 11 = 220 \) \( (> 51) \) True
3. For 62: \( 4(6 \cdot 2) \times 11 = 528 \) \( (>62) \) True
4. For 73: \( 4(7 \cdot 3) \times 11 = 924 \) \( (>73) \) True
5. For 84: \( 4(8 \cdot 4) \times 11 = 704 \) \( (>84) \) True
6. For 95: \( 4(9 \cdot 5) \times 11 = 1980 \) \( (>95) \) True

Conclusion

Only the values where both conditions hold give that the final valid numbers are 51, 62, 73, 84, 95.

The least valid number that satisfies both of the conditions fully identified through inequalities and checks is 51.

Thus, the two-digit number is \( \boxed{51} \).