Grade 8
Subject Mathematics
Topic Number Theory
Study Guide: Flipping Digits
These digit flipping challenges are tough at first, but with practice you can become a pro.
Challenge 1
There is an unknown two digit number. The sum of its digits is 7. When these digits are reversed, the new number is 27 greater than the original number. What is the original number?
- We are told that the sum of the digits is 7.
- Let $t$ be the digit in the tens column.
- Let $u$ be the digit in the units (ones) column.
- Write an equation for the sum of these digits (ignoring place value).
\begin{aligned}{} \text{t} + \text{u} &= 7 &\textit{Equation 1} \end{aligned}
- We are told that the reversed number is 27 greater than the original.
- Write an equation for the difference of the two numbers (including place value).
Place Value
Understanding place value is essential for solving this problem. In the above equation we intentionally ignored place value, as $t$ and $u$ stand for specific digits and we wanted to create an equation for the sum of the digits, NOT for the value of the two-digit number as a whole.
In this next step we create another equation, this time including place value. For example, when the digit symbolized by $t$ is in the tens place it's actual value in the whole number is ($10 \times t)$ or $10t$, and when it's in the one's place it's value in the whole number is ($1 \times t$) or simply $t$.We will soon combine our two equations into a system of equations to solve the problem.
\begin{aligned}{}
(10 \times t) + (1 \times u) &= \textit{Value of original number} \\
(10 \times u) + (1 \times t) &= \textit{Value of reverse number} \\
27 &= \textit{Difference}
\end{aligned}
\begin{aligned}{}
\text{[Reversed]} &= \text{[Original]} + \text{[Difference]} \\
(10u + t) &= (10t + u) + (27) &\textit{Equation 2}
\end{aligned}
Combine like terms, and get all variables on the left side.
\begin{aligned}
\textit{[Reversed]} &= \textit{[Original]} + \textit{[Difference]} \\
(10\text{u} + \text{t}) &= (10\text{t} + \text{u}) + (27) &\textit{Combine like terms.} \\
9\text{u} -9\text{t} &= 27 &\textit{Divide all terms by 9.} \\
\text{u} - \text{t} &= 3 &\textit{Equation 2 simplified}
\end{aligned}
Create a system of equations and add to eliminate one variable.
$$
\left\{
\begin{aligned}{}
\text{t} + \text{u} &= 7 \\
\text{u} - \text{t} &= 3 \\
\end{aligned}
\right\}
$$
$$
\begin{aligned}
2\text{u} &=10 \\
\text{u} &=5
\end{aligned}
$$
Use $u$ to find $t$.
\begin{aligned} \\
\text{t} + {\color{teal}\text{u}} &=7 &\textit{Equation 1} \\
\text{t} + {\color{teal}(5)} &=7 &\textit{Replace }{\color{teal}\textit{u}} \textit{ with } {\color{teal}5} \\
\text{t} &= 2 &\textit{Simplified} \\
\end{aligned}
\begin{aligned}
&\text{(t, u)} = (2, 5) &\text{Original digits} \\
\hline
&25 &\text{Original number} \\
&52 &\text{Reversed number} \\
\hline
&2 + 5 = 7 & \textit{Sum of digits} \\
&25 + 27 = 52 & \textit{Proof}
\end{aligned}
Challenge 2
There is an unknown two digit number. The sum of its digits is 8. When these digits are reversed, the new number is 54 greater than the original number. What is the original number?
- We are told that the sum of the digits is 8.
- Let $t$ be the digit in the tens column.
- Let $u$ be the digit in the units (ones) column.
- Write an equation for the sum of these digits (ignoring place value).
\begin{aligned}{} \text{t} + \text{u} &= 8 &\textit{Equation 1} \end{aligned}
- We are told that the reversed number is 54 greater than the original.
- Write an equation for the difference of the two numbers (including place value).
Place Value
Understanding place value is essential for solving this problem. In the above equation we intentionally ignored place value, as $t$ and $u$ stand for specific digits and we wanted to create an equation for the sum of the digits, NOT for the value of the two-digit number as a whole.
In this next step we create another equation, this time including place value. For example, when the digit symbolized by $t$ is in the tens place it's actual value in the whole number is ($10 \times t)$ or $10t$, and when it's in the one's place it's value in the whole number is ($1 \times t$) or simply $t$.We will soon combine our two equations into a system of equations to solve the problem.
\begin{aligned}{}
(10 \times t) + (1 \times u) &= \textit{Value of original number} \\
(10 \times u) + (1 \times t) &= \textit{Value of reverse number} \\
54 &= \textit{Difference}
\end{aligned}
\begin{aligned}{}
\text{[Reversed]} &= \text{[Original]} + \text{[Difference]} \\
(10u + t) &= (10t + u) + (54) &\textit{Equation 2}
\end{aligned}
Combine like terms, and get all variables on the left side.
\begin{aligned}
\textit{[Reversed]} &= \textit{[Original]} + \textit{[Difference]} \\
(10\text{u} + \text{t}) &= (10\text{t} + \text{u}) + (54) &\textit{Combine like terms.} \\
9\text{u} -9\text{t} &= 54 &\textit{Divide all terms by 9.} \\
\text{u} - \text{t} &= 6 &\textit{Equation 2 simplified}
\end{aligned}
Create a system of equations and add to eliminate one variable.
$$
\left\{
\begin{aligned}{}
\text{t} + \text{u} &= 8 \\
\text{u} - \text{t} &= 6 \\
\end{aligned}
\right\}
$$
$$
\begin{aligned}
2\text{u} &=14 \\
\text{u} &=7
\end{aligned}
$$
Use $u$ to find $t$.
\begin{aligned} \\
\text{t} + {\color{teal}\text{u}} &=8 &\textit{Equation 1} \\
\text{t} + {\color{teal}(7)} &=8 &\textit{Replace }{\color{teal}\textit{u}} \textit{ with } {\color{teal}7} \\
\text{t} &= 1 &\textit{Simplified} \\
\end{aligned}
\begin{aligned}
&\text{(t, u)} = (1, 7) &\text{Original digits} \\
\hline
&17 &\text{Original number} \\
&71 &\text{Reversed number} \\
\hline
&1 + 7 = 8 & \textit{Sum of digits} \\
&17 + 54 = 71 & \textit{Proof}
\end{aligned}
Challenge 3
There is an unknown two digit number. The sum of its digits is 8. When these digits are reversed, the new number is 36 less than the original number. What is the original number?
- We are told that the sum of the digits is 8.
- Let $t$ be the digit in the tens column.
- Let $u$ be the digit in the units (ones) column.
- Write an equation for the sum of these digits (ignoring place value).
\begin{aligned}{} \text{t} + \text{u} &= 8 &\textit{Equation 1} \end{aligned}
- We are told that the reversed number is 36 less than the original.
- Write an equation for the difference of the two numbers (including place value).
Place Value
Understanding place value is essential for solving this problem. In the above equation we intentionally ignored place value, as $t$ and $u$ stand for specific digits and we wanted to create an equation for the sum of the digits, NOT for the value of the two-digit number as a whole.
In this next step we create another equation, this time including place value. For example, when the digit symbolized by $t$ is in the tens place it's actual value in the whole number is ($10 \times t)$ or $10t$, and when it's in the one's place it's value in the whole number is ($1 \times t$) or simply $t$.We will soon combine our two equations into a system of equations to solve the problem.
\begin{aligned}{}
(10 \times t) + (1 \times u) &= \textit{Value of original number} \\
(10 \times u) + (1 \times t) &= \textit{Value of reverse number} \\
-36 &= \textit{Difference}
\end{aligned}
\begin{aligned}{}
\text{[Reversed]} &= \text{[Original]} - \text{[Difference]} \\
(10u + t) &= (10t + u) + (-36) &\textit{Equation 2}
\end{aligned}
Combine like terms, and get all variables on the left side.
\begin{aligned}
\textit{[Reversed]} &= \textit{[Original]} - \textit{[Difference]} \\
(10\text{u} + \text{t}) &= (10\text{t} + \text{u}) + (-36) &\textit{Combine like terms.} \\
9\text{u} -9\text{t} &= -36 &\textit{Divide all terms by 9.} \\
\text{u} - \text{t} &= -4 &\textit{Equation 2 simplified}
\end{aligned}
Create a system of equations and add to eliminate one variable.
$$
\left\{
\begin{aligned}{}
\text{t} + \text{u} &= 8 \\
\text{u} - \text{t} &= -4 \\
\end{aligned}
\right\}
$$
$$
\begin{aligned}
2\text{u} &=4 \\
\text{u} &=2
\end{aligned}
$$
Use $u$ to find $t$.
\begin{aligned} \\
\text{t} + {\color{teal}\text{u}} &=8 &\textit{Equation 1} \\
\text{t} + {\color{teal}(2)} &=8 &\textit{Replace }{\color{teal}\textit{u}} \textit{ with } {\color{teal}2} \\
\text{t} &= 6 &\textit{Simplified} \\
\end{aligned}
\begin{aligned}
&\text{(t, u)} = (6, 2) &\text{Original digits} \\
\hline
&62 &\text{Original number} \\
&26 &\text{Reversed number} \\
\hline
&6 + 2 = 8 & \textit{Sum of digits} \\
&62 + (-36) = 26 & \textit{Proof}
\end{aligned}
Challenge 4
There is an unknown two digit number. The sum of its digits is 17. When these digits are reversed, the new number is 9 less than the original number. What is the original number?
- We are told that the sum of the digits is 17.
- Let $t$ be the digit in the tens column.
- Let $u$ be the digit in the units (ones) column.
- Write an equation for the sum of these digits (ignoring place value).
\begin{aligned}{} \text{t} + \text{u} &= 17 &\textit{Equation 1} \end{aligned}
- We are told that the reversed number is 9 less than the original.
- Write an equation for the difference of the two numbers (including place value).
Place Value
Understanding place value is essential for solving this problem. In the above equation we intentionally ignored place value, as $t$ and $u$ stand for specific digits and we wanted to create an equation for the sum of the digits, NOT for the value of the two-digit number as a whole.
In this next step we create another equation, this time including place value. For example, when the digit symbolized by $t$ is in the tens place it's actual value in the whole number is ($10 \times t)$ or $10t$, and when it's in the one's place it's value in the whole number is ($1 \times t$) or simply $t$.We will soon combine our two equations into a system of equations to solve the problem.
\begin{aligned}{}
(10 \times t) + (1 \times u) &= \textit{Value of original number} \\
(10 \times u) + (1 \times t) &= \textit{Value of reverse number} \\
-9 &= \textit{Difference}
\end{aligned}
\begin{aligned}{}
\text{[Reversed]} &= \text{[Original]} - \text{[Difference]} \\
(10u + t) &= (10t + u) + (-9) &\textit{Equation 2}
\end{aligned}
Combine like terms, and get all variables on the left side.
\begin{aligned}
\textit{[Reversed]} &= \textit{[Original]} - \textit{[Difference]} \\
(10\text{u} + \text{t}) &= (10\text{t} + \text{u}) + (-9) &\textit{Combine like terms.} \\
9\text{u} -9\text{t} &= -9 &\textit{Divide all terms by 9.} \\
\text{u} - \text{t} &= -1 &\textit{Equation 2 simplified}
\end{aligned}
Create a system of equations and add to eliminate one variable.
$$
\left\{
\begin{aligned}{}
\text{t} + \text{u} &= 17 \\
\text{u} - \text{t} &= -1 \\
\end{aligned}
\right\}
$$
$$
\begin{aligned}
2\text{u} &=16 \\
\text{u} &=8
\end{aligned}
$$
Use $u$ to find $t$.
\begin{aligned} \\
\text{t} + {\color{teal}\text{u}} &=17 &\textit{Equation 1} \\
\text{t} + {\color{teal}(8)} &=17 &\textit{Replace }{\color{teal}\textit{u}} \textit{ with } {\color{teal}8} \\
\text{t} &= 9 &\textit{Simplified} \\
\end{aligned}
\begin{aligned}
&\text{(t, u)} = (9, 8) &\text{Original digits} \\
\hline
&98 &\text{Original number} \\
&89 &\text{Reversed number} \\
\hline
&9 + 8 = 17 & \textit{Sum of digits} \\
&98 + (-9) = 89 & \textit{Proof}
\end{aligned}
Source: https://class.ronliskey.com/study/mathematics-8/flipping-digits/