Find the two-digit number given the following conditions:

A. Sum of the squares of the two digits of the two-digit number is 26.

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

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

Question

Find the two-digit number given the following conditions:

A. Sum of the squares of the two digits of the two-digit number is 26.

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

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

Accepted Answer

**Solution:**

Let the two-digit number be $10x + y$, where $x$ is the ten’s digit and $y$ is the unit’s digit.

1. From condition A:
   $$x^2 + y^2 = 26$$

2. From condition B:
   $$\frac{10x + y}{x + y} = \frac{5}{2}$$
   This implies:
   $$2(10x + y) = 5(x + y)$$
   $$20x + 2y = 5x + 5y$$
   $$20x - 5x = 5y - 2y$$
   $$15x = 3y$$
   $$5x = y$$

3. From condition C:
   $$x = y - 4$$

**Check sufficiency:**

- Using B alone:
  From $y = 5x$, substitute into the number:
  Digits must be single digits (0 to 9), so try values of $x$:
  - If $x=1$, then $y=5$, number = 15, sum of digits = 6, ratio = 15/6 = 2.5 = 5/2 correct.
  Check sum of squares: $1^2 + 5^2 = 1 + 25 = 26$ (matches A)
  Check C: $x = y - 4 ightarrow 1 = 5 - 4 = 1$ (matches C)

So B alone is sufficient to find the number 15.

- Using A and C together:
  From C: $x = y - 4$
  Substitute into A:
  $$(y - 4)^2 + y^2 = 26$$
  $$y^2 - 8y + 16 + y^2 = 26$$
  $$2y^2 - 8y + 16 = 26$$
  $$2y^2 - 8y - 10 = 0$$
  $$y^2 - 4y - 5 = 0$$
  $$(y - 5)(y + 1) = 0$$
  $$y = 5 	ext{ or } y = -1 	ext{ (not valid)}$$

So, $y = 5$, then $x = 5 - 4 = 1$
  Number = 15

Check B:
  $$\frac{15}{1 + 5} = \frac{15}{6} = \frac{5}{2}$$

So A and C together also give the number 15.

**Answer:** Either A and C together or B alone are sufficient to find the number.

Hence, option (c) is correct.

Answer

Answer

Answer