Find the two-digit number.

I. The sum of the squares of the two digits is 26.

II. The ratio between the two-digit number and the sum of the digits of that number is 5:2.

III. The digit in the ten’s place is 4 less than the digit in the unit place.

Which of the following statements is/are sufficient to find the number?

Question

Find the two-digit number.

I. The sum of the squares of the two digits is 26.

II. The ratio between the two-digit number and the sum of the digits of that number is 5:2.

III. The digit in the ten’s place is 4 less than the digit in the unit place.

Which of the following statements is/are sufficient to find the number?

Accepted Answer

**Solution:**

Let the two-digit number be represented as $10x + y$, where $x$ is the digit in the ten's place and $y$ is the digit in the unit place.

1. From statement I:  
   $$x^2 + y^2 = 26$$

2. From statement II:  
   The ratio of the number to the sum of its digits is $5:2$, so
   $$\frac{10x + y}{x + y} = \frac{5}{2}$$

3. From statement III:  
   The digit in the ten's place is 4 less than the digit in the unit place, so
   $$x = y - 4$$

**Analysis:**

- Using statements I and III together:
  Substitute $x = y - 4$ into $x^2 + y^2 = 26$:
  $$(y - 4)^2 + y^2 = 26$$
  $$y^2 - 8y + 16 + y^2 = 26$$
  $$2y^2 - 8y + 16 = 26$$
  $$2y^2 - 8y - 10 = 0$$
  $$y^2 - 4y - 5 = 0$$

Solving the quadratic:
  $$y = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2}$$
  $$y = \frac{4 \pm 6}{2}$$

Possible values:
  - $y = 5$ (since digits must be between 0 and 9)
  - $y = -1$ (not valid)

So, $y = 5$, then $x = 5 - 4 = 1$.

Number is $10 	imes 1 + 5 = 15$.

- Using statements II and III together:
  Substitute $x = y - 4$ into the ratio equation:
  $$\frac{10(y - 4) + y}{(y - 4) + y} = \frac{5}{2}$$
  $$\frac{10y - 40 + y}{2y - 4} = \frac{5}{2}$$
  $$\frac{11y - 40}{2y - 4} = \frac{5}{2}$$

Cross-multiplied:
  $$2(11y - 40) = 5(2y - 4)$$
  $$22y - 80 = 10y - 20$$
  $$22y - 10y = -20 + 80$$
  $$12y = 60$$
  $$y = 5$$

Then $x = 5 - 4 = 1$, number is 15.

- Using statements I and II together:
  From II:
  $$\frac{10x + y}{x + y} = \frac{5}{2} \Rightarrow 2(10x + y) = 5(x + y)$$
  $$20x + 2y = 5x + 5y$$
  $$15x = 3y$$
  $$5x = y$$

Substitute $y = 5x$ into I:
  $$x^2 + (5x)^2 = 26$$
  $$x^2 + 25x^2 = 26$$
  $$26x^2 = 26$$
  $$x^2 = 1$$
  $$x = \pm 1$$

Since $x$ is a digit, $x=1$, then $y=5$.

Number is 15.

**Conclusion:**

Any two of the three statements together are sufficient to find the number uniquely as 15.

**Answer:** Any two of the three together are sufficient.

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