Exercise 38

In a two-digit number, the ones digit is twice the tens digit, and if we add the number with the result of inverting its digits, we obtain 66. If \(U\) represents the ones digit and \(D\) represents the tens digit in the original number, then:

  1. \( D = 2U; \; U + D = 6 \)

  2. \( U = 2D; \; U + D = 6 \)

  3. \( D = 2U; \; U + D = 66 \)

  4. \( U = 2D; \; U + D = 66 \)

  5. \( D = 2U; \; 11U + 11D = 66 \)

Try to solve it before checking the answer.

  1. \( U = 2D; \; U + D = 6 \)

The number is written as follows:

\(DU\),   which means   \( 10D + U \)

Since \(U\) doubles \(D\), it follows that \( U = 2D \); therefore, the number is:

\( 10D + 2D = 12D \hspace{4em} \boldsymbol{(1)} \)

The number that results from inverting the figures of \(DU\) is \(UD\), which means \( 10U + D \).

But since \( U = 2D \), then the inverted number is \( 10(2D) + D \); that is:

\( 20D + D = 21D \hspace{4em} \boldsymbol{(2)} \)

On the other hand, since the sum of the number with inverted figures is 66, then:

\[ \begin{aligned} &12D + 21D = 66 \\[.5em] &\hspace{
  1. \( y^2 < x^2 \)
5em}\Rightarrow 33D = 66 \\[.5em] &\hspace{5em}\Rightarrow D = 2 \\[.5em] &\hspace{5em}\Rightarrow \boldsymbol{ D = 2 \quad \text{ y } \quad U = 4 } \end{aligned} \]
\[ \begin{aligned} 12D + 21D = 66 &\Rightarrow 33D = 66 \\[.5em] &\Rightarrow D = 2 \\[.5em] &\Rightarrow \boldsymbol{ D = 2 \quad \text{ y } \quad U = 4 } \end{aligned} \]

Hence,   \( \boldsymbol{ U = 2D } \)   y   \( \boldsymbol{ U + D = 6 } \).

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