Place Value and Base

The position of a digit within a number is called its place, the value represented by a digit in that position is its place value, and the structural system used to represent these values in a numeral system is called the base.

\[r_0, \; r_1, \; r_2, \; r_3, \; \dots, \; r_{n-1}, \; r_n\]

Let each denotes a digit and $a$ denotes the base of the numeral system. The place values of an $(n + 1)$-digit number $(r_n, r_{n-1},\dots, r_2, r_1, r_0)_a$ in base $a$ (the base-$a$ positional system) are defined as follows:

$$ (r_n \; r_{n-1} \; \dots \; r_2 \; r_1 \; r_0)_a $$

$$
\begin{aligned}
&r_0 &\longrightarrow & a^0\text{ place} \\
&r_1 &\longrightarrow & a^1\text{ place} \\
&r_2 &\longrightarrow & a^2\text{ place} \\
&\ \ \vdots \\
&r_{n-1} &\longrightarrow & a^{n-1}\text{ place} \\
&r_n &\longrightarrow & a^n\text{ place} \\
\end{aligned}
$$

The expanded notation of the number $[(r_n, r_{n-1}, \dots, r_2, r_1, r_0)_a]$ is written as:

\[ r_n \cdot a^n + r_{n-1} \cdot a^{n-1} + \dots + r_2 \cdot a^2 + r_1 \cdot a^1 + r_0 \cdot a^0 \] This expression represents the expanded form of the number.

Example:

Let’s examine the number 1965 in the decimal system (base 10).

\[ \begin{array}{c c c c} 1 & 9 & 6 & 5 \\ \downarrow & \downarrow & \downarrow & \downarrow \\ 10^3 & 10^2 & 10^1 & 10^0 \\ \downarrow & \downarrow & \downarrow & \downarrow \\ 3 & 2 & 1 & 0 \\ \downarrow & \downarrow & \downarrow & \downarrow \\ 10^3 \, \text{thousands place} & 10^2 \, \text{hundreds place} & 10^1 \, \text{tens place} & 10^0 \, \text{ones place} \end{array} \]

In the number 1965, the place value of the digit 1 is one thousand, the place value of the digit 9 is nine hundred, the place value of the digit 6 is sixty, and the place value of the digit 5 is five.

$$ 1965= 1\cdot10^3 + 9\cdot10^2 + 6\cdot10^1+ 5\cdot10^0$$

Examples:

The expanded form of the number $(3412)_5$

\[(3412)_5 = 3 \cdot 5^3 + 4 \cdot 5^2 + 1 \cdot 5^1 + 2 \cdot 5^0 \]

The expanded form of the number $(20000)_3$

\[(20000)_3 = 2 \cdot 3^4\]

The expanded form of the decimal number $(384.217)$

\[ 384.217 = 3 \cdot 10^2 + 8 \cdot 10^1 + 4 \cdot 10^0 + 2 \cdot 10^{-1} + 1 \cdot 10^{-2} + 7 \cdot 10^{-3} \]

To the right of the radix point (decimal point), place values decrease by negative powers of the base. In this example, since the base is 10, the values decrease by powers of one-tenth.

The expanded form of the number $(432.21)_5$

\[ (432.21)_5 = 4 \cdot 5^2 + 3 \cdot 5^1 + 2 \cdot 5^0 + 2 \cdot 5^{-1} + 1 \cdot 5^{-2} \]

Examples:

Two-digit numbers represented as $ab$ and $aa$

\[ ab = 10 \cdot a + b \]

\[ aa = 10 \cdot a + a = 11 \cdot a \]

Three-digit numbers represented as $abc$ and $aaa$

\[ abc = 100 \cdot a + 10 \cdot b + c \] \[ aaa = 100 \cdot a + 10 \cdot a + a = 111 \cdot a \] These are analyzed using their expanded algebraic forms as shown above.

Note: When performing operations on the digits of a number, the place values of those digits must always be taken into account. For instance, increasing the tens digit of a number by 3 increases the total value of the number by $3 \times 10 = 30$. Decreasing the hundreds digit by 5 decreases the total value of the number by $5 \times 100 = 500$. If the tens digit is increased by 7 and the ones digit is decreased by 3, the net change in the number’s value is an increase of $7 \times 10 – 3 \times 1 = 67$.

Example:

When the digits of a two-digit number are reversed, the value of the number increases by 63. Find the absolute value of the difference between the digits of this number.

Let the original two-digit number be $ab$. When its digits are reversed, the new number is $ba$. Since the value of the number increases by 63, we can set up the following equation:

\[
\begin{align}
ba &= ab + 63 \\
\Rightarrow 10 \cdot b + a &= 10 \cdot a + b + 63 \\
\Rightarrow 9 \cdot |b – a| &= 63 \\
|b – a| &= 7
\end{align}
\]

Example:

When a two-digit number $ab$ is added to the two-digit number obtained by reversing its digits, the sum is 99. Find the maximum possible value of the number $ab$.

\[ \begin{align} ab + ba &= 99 \Rightarrow 10 \cdot a + b + 10 \cdot b + a = 99 \\ \Rightarrow 11 \cdot (a + b) &= 99 \\ \Rightarrow a + b &= 9 \end{align} \]

To maximize $ab$, the tens digit $a$ must be as large as possible. Since $a + b = 9$, if we choose $a = 9$ and $b = 0$, the reversed number $ba$ would be $09$, which is not a valid two-digit number. Therefore, choosing $a = 8$ and $b = 1$ gives the maximum valid value for $ab$, which is $81$.

Question 14:

The sum of four distinct three-digit natural numbers with distinct digits is 768. What is the maximum possible value for the largest of these numbers?

\[ \text{A) } 458 \quad \text{B) } 459 \quad \text{C) } 460 \quad \text{D) } 461 \quad \text{E) } 462 \]

Solution:

To maximize the largest number, we must select the other three numbers to be as small as possible. Since each number must have distinct digits and all four numbers must be distinct from one another, the three smallest possible numbers are $102$, $103$, and $104$:

\[ \text{Maximum Number} = 768 – (102 + 103 + 104) \]
\[ = 459 \]

$\textbf{Correct Answer: B} $

Question 15:

When the tens and thousands digits of a four-digit natural number $abcd$ are swapped, the value of the number increases by 5940. Find the value of the difference $c – a$.

\[ \text{A) } 3 \quad \text{B) } 4 \quad \text{C) } 5 \quad \text{D) } 6 \quad \text{E) } 7 \]

Solution:

\[\text{Since } cbad = abcd + 5940,\] we expand both sides using positional notation and simplify:

\[ \begin{aligned} 990 \cdot (c – a) &= 5940 \\
\Rightarrow c – a &= 6
\end{aligned}
\]

$\textbf{Correct Answer: D} $

Question 16:

A six-digit number is formed by writing a three-digit number twice side-by-side. How many times greater is the new number than the original three-digit number?

\[ \text{A) } 99 \quad \text{B) } 100 \quad \text{C) } 101 \quad \text{D) } 1000 \quad \text{E) } 1001 \]

Solution:

Let the original three-digit number be $abc$, and the newly formed six-digit number be $abcabc$. We express the relationship as:

\[\begin{aligned}
&abcabc – abc = x \cdot abc \\
&\Rightarrow abc000 = x \cdot abc \\
&\Rightarrow x = 1000 \quad \text{times greater.}
\end{aligned}
\]

$\textbf{Correct Answer: D} $

Question 17:

Let $a$, $b$, and $c$ be distinct digits.

\[
\begin{array}{r}
\text{abc} \\
\text{bca} \\
+ \text{cab} \\
\hline 2109
\end{array}
\]

Given the column addition above, find the sum $a + b + c$.

\[ \text{A) } 9 \quad \text{B) } 10 \quad \text{C) } 19 \quad \text{D) } 20 \quad \text{E) } 21 \]

Solution:

Expanding each of the three-digit numbers into its positional components yields:

\[
\begin{array}{r}
100\cdot a + 10\cdot b + c \\
100\cdot b + 10\cdot c + a \\
+ 100\cdot c + 10\cdot a + b \\
\hline 111 \cdot (a + b + c)
\end{array}
\]
\[
\begin{aligned}
&111 \cdot (a + b + c) = 2109 \\
\\
&\Rightarrow a + b + c = \dfrac{2109}{111} = 19
\end{aligned}\]

$\textbf{Correct Answer: C} $

Question 18:

In the multiplication problem below, each dot and the letters $a$, $b$, and $c$ represent a single digit:

\[ \begin{array}{c@{}c}
\;\;\;\;ab\\
\times\;\;6c\\
\hline
\;\;…\\
+450\;\;\;\;\\
\hline
\;\;4725\\\
\end{array} \]

Based on this calculation, find the sum $a + b + c$.

\[ \text{A) } 14 \quad \text{B) } 15 \quad \text{C) } 16 \quad \text{D) } 27 \quad \text{E) } 28 \]

Solution:

The partial product obtained by multiplying the first factor $ab$ by $6$ is $450$. Therefore:

\[ 6 \cdot ab = 450 \Rightarrow ab = 75 \]

We rewrite the multiplication template with the known value:

\[ \begin{array}{c@{}c}
\;\;\;\;75 & \quad \text{1st Factor} \\
\times\;\;\;6c &\quad \text{2nd Factor} \\
\hline
\;\;xyz\\
+\;\;450\;\;\;\;\\
\hline
\;\;4725\\\
\end{array} \]

\[
\begin{array}{c@{}c}
\;\;\;\;xyz& \\
+\;\;4500 &\\
\hline
\;\;4725\\
\\
\\
xyz = 4725 – 4500 \Rightarrow xyz = 225 \\
\text{Since } 75 \cdot c = xyz = 225 \Rightarrow c = \dfrac{225}{75} \Rightarrow c = 3 \\
\text{From } ab = 75, \text{ we determine that } a = 7 \text{ and } b = 5. \\
\text{Thus, } a + b + c = 7 + 5 + 3 = 15.
\end{array}
\]

$\textbf{Correct Answer: B} $

Question 19:

Let $xyz$ be a three-digit natural number. Find the sum of all odd numbers that can be formed satisfying the conditions $x = y + 2$ and $y = z + 2$.

\[
\text{A) 2159} \quad \text{B) 2259} \quad \text{C) 2616} \quad \text{D) 2915} \quad \text{E) 3210}
\]

Solution:

The values of the digits $x$ and $y$ depend directly on the choice of $z$. Since the three-digit number must be odd, $z$ must be an odd digit. Choosing valid odd values for $z$ ($1, 3,$ or $5$ are acceptable, as any odd digit greater than $5$ would force $x$ and $y$ to exceed single-digit limits), we evaluate the corresponding numbers and compute their sum:

\[
531 + 753 + 975 = 2259
\]

$\textbf{Correct Answer: B} $

Question 20:

Given that $a \neq b$, and $abb$ and $baa$ are three-digit natural numbers, evaluate the rational expression $\dfrac{abb – baa}{b – a}$.

\[\text{A) 89 \quad B) -89 \quad C) 91 \quad D) -91 \quad E) 101}\]

Solution:

\[
\frac{abb – baa}{b – a} = \frac{(100a + 10b + b) – (100b + 10a + a)}{b – a} \\
= \frac{89a – 89b}{b – a} \\
= \frac{89 \cdot (a – b)}{b – a} = 89 \times \frac{a – b}{b – a} = -89
\]

$\textbf{Correct Answer: B} $

Question 21:

Let $2mn4$ and $1mn7$ be four-digit natural numbers. Find the result of the following subtraction:

\[ \begin{array}{r} 2mn4 \\ – 1mn7 \\ \hline \cdots \end{array} \]

\[\text{A) 1004 \quad B) 1003 \quad C) 997 \quad D) 907 \quad E) 903}\]

Solution:

Expanding both numbers into their respective positional components allows the algebraic terms to cancel out during subtraction:

\[ \begin{array}{r} (2000 + 100m + 10n + 4) \\ – (1000 + 100m + 10n + 7) \\ \hline 1000 – 3 = 997 \end{array} \]

$\textbf{Correct Answer: C} $

Question 22:

Let $a$ and $b$ be single digits. Given the vertical subtraction:

\[ \begin{array}{r} ab3 \\ – ab \\ \hline 408 \end{array} \]

Find the product $a \cdot b$.

\[\text{A) 15 \quad B) 18 \quad C) 20 \quad D) 24 \quad E) 25}\]

Solution:

Method 1

Expanding the numbers according to their place value definitions yields:

\[
\begin{array}{l}
100\cdot a + 10\cdot b + 3 – (10\cdot a + b) = 408\\
\Rightarrow 90\cdot a + 9\cdot b = 405 \\
\Rightarrow 9\cdot (10\cdot a + b) = 405\\
\Rightarrow ab = 45 \\
\text{Hence, } a = 4, \quad b = 5, \quad \text{and } a \cdot b = 20.
\end{array}
\]

Method 2

\[ \begin{array}{l} ab3 \\ – ab \\ \hline 408 \end{array} \]

Analyzing the ones column, we observe that subtracting $b$ from $3$ yields $8$, which requires regrouping (borrowing) from the tens column. Borrowing $1$ ten converts the $3$ into $13$. From the column equation $13 – b = 8$, we deduce that $b = 5$. Moving to the tens column, since $1$ was borrowed, the remaining value is $(b – 1)$. The column equation $(b – 1) – a = 0$ simplifies to $(5 – 1) – a = 0$, yielding $a = 4$.

Consequently, the product $a \cdot b = 4 \cdot 5 = 20$.

$\textbf{Correct Answer: C} $

Question 23:

A student was asked to calculate the product of a certain three-digit number and 75. The student found a result of 8850. Upon checking the calculation, the teacher noticed that the student mistook the tens digit 7 of the three-digit number for a 1.

Find the correct product of the original multiplication problem.

\[\text{A) 9300 \quad B) 9750 \quad C) 10500 \quad D) 11400 \quad E) 13350}\]

Solution:

Method 1

Because the student misread the tens digit as a 1 instead of a 7, the value used was smaller by 6 tens. Thus, the calculated product is lower than the actual product by:

\[
\begin{array}{l}
(7 – 1) \cdot 10 \cdot 75 = 4500 \\
\text{Adding this deficit back yields the correct product:} \\
8850 + 4500 = 13350.
\end{array}
\]

Method 2

Let $x$ represent the incorrect three-digit number read by the student. The calculation performed was:

$$x \cdot 75 = 8850 \quad \Rightarrow \quad x = 118.$$

Since the tens digit was misread as 1 instead of 7, the correct three-digit number is 178. The correct product is:

$$178 \cdot 75 = 13350.$$

$\textbf{Correct Answer: E} $

Note:

Let a represent the base of the numeral system, and let x, y, and z represent valid digits in that system.

The expression $(xyz)_a$ denotes a three-digit positional number in base $a$. In this positional system, the base $a$ must be an integer greater than 1, and strictly greater than any individual digit ($a > x$, $a > y$, and $a > z$).

The smallest possible integer base for a positional numeral system is base 2 (binary system).
A system with base a utilizes exactly a distinct digits, ranging from 0 to a – 1.
Numbers expressed in the standard decimal system (base 10) omit explicit base notation.

Example:

The set of valid digits in base 2 is $\{0, 1\}$
The set of valid digits in base 5 is $\{0, 1, 2, 3, 4\}$
The set of valid digits in base 10 is $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$
The set of valid digits in base 12 is $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B\}$, where the alphanumeric digit A represents a value of 10 and B represents a value of 11.