Question 1

Find the remainder when $(397)^{3589} + 5$ is divided by $398$.

Accepted Answer

### Concept & Logic This problem is best solved using the Negative Remainder Theorem. When a number $N$ is divided by $N+1$, the remainder can be treated as $-1$. ### Step-by-Step Solution **Given:** Expression: $(397)^{3589} + 5$ Divisor: $398$ **Calculation / Deduction:** * We need to find the remainder of each term when divided by $398$. * First term: $397 \pmod{398}$. Since $397$ is exactly $1$ less than $398$, it leaves a remainder of $-1$. * Raise this negative remainder to the given power: $(-1)^{3589}$. * Since $3589$ is an odd number, $(-1)^{	ext{odd}} = -1$. * Second term: $5 \pmod{398}$ is simply $5$, as it is already smaller than the divisor. * Add the remainders together: $-1 + 5 = 4$. ### Exam Strategy & Shortcut Whenever the base is exactly $1$ less than the divisor, look at the power. If the power is odd, the base yields a remainder of $-1$. If the power is even, it yields $+1$. Here, the power is odd, so you get $-1$. Immediately add this to the $+5$ to get the final answer: $4$. This takes less than 5 seconds mentally. ### Common Pitfall A common mistake is trying to apply algebraic expansion formulas like $(x^n + a^n)$ which, while mathematically sound, are overly tedious and increase the chance of making a sign error compared to using direct negative remainders. ### Final Answer Therefore, the correct answer is 4.

Question 2

If $7^{126}$ is divided by $48$, find the remainder.

Accepted Answer

### Concept & Strategy When finding remainders of large powers, the goal is to rewrite the base so it is exactly $1$ more or $1$ less than a multiple of the divisor. This allows you to use $1$ or $-1$ as the remainder, making the large exponent trivial to calculate. ### Step-by-Step Solution **Given:** Expression: $7^{126}$ Divisor: $48$ **Calculation / Deduction:** * Look for a power of $7$ that is close to a multiple of $48$. We know $7^1 = 7$, and $7^2 = 49$. * Notice that $49$ is exactly $1$ more than the divisor $48$. * Rewrite the original expression in terms of $7^2$: $$ 7^{126} = (7^2)^{63} = (49)^{63} $$ * Now, apply the modular arithmetic. When $49$ is divided by $48$, the remainder is $1$. $$ 49 \equiv 1 \pmod{48} $$ * Substitute this remainder back into the exponent: $$ 1^{63} = 1 $$ ### Exam Strategy & Shortcut Instantly check the squares or cubes of the given base against the divisor. You see $7$ and $48$. You should immediately associate $7^2 = 49$. Since $49$ is $1$ greater than $48$, it yields a remainder of $+1$. $+1$ raised to any power is always $1$. ### Common Pitfall Students often try to find the cyclicity of $7$ by dividing powers of $7$ by $48$ manually (e.g., finding $7^1/48$, $7^2/48$, $7^3/48$, etc.). This is a massive time sink. Always look to match the base to $(D+1)$ or $(D-1)$ where $D$ is the divisor. ### Final Answer Therefore, the correct answer is 1.

Question 3

Find the remainder when $(257^{166} - 243^{166})$ is divided by $500$.

Accepted Answer

### Concept & Formula This question relies on standard divisibility rules for algebraic identities. Specifically, for an expression in the form $(x^n - a^n)$, it is always perfectly divisible by $(x + a)$ whenever $n$ is an **even** integer. ### Step-by-Step Solution **Given:** Expression: $257^{166} - 243^{166}$ Divisor: $500$ **Calculation / Deduction:** * Identify the format: The expression matches $x^n - a^n$ where $x = 257$, $a = 243$, and $n = 166$. * Check the exponent: The power $166$ is an even number. * Apply the rule: Because the power is even, the expression is perfectly divisible by $(x + a)$. * Calculate $(x + a)$: $$ 257 + 243 = 500 $$ * Since the expression is perfectly divisible by $500$, and our given divisor is also $500$, there is no remainder. ### Exam Strategy & Shortcut Look at the bases and the divisor. $257 + 243 = 500$. This matches the divisor perfectly. Then, look at the sign (minus) and the power (even). By the $(x^n - a^n)$ rule, an even power with a minus sign is divisible by the sum of the bases. Conclude instantly that the remainder is $0$. ### Common Pitfall A frequent error is confusing the parity rules. Students might mix up $x^n - a^n$ and $x^n + a^n$. Remember: 1. $(x^n - a^n)$ is divisible by $(x - a)$ for ALL $n$. 2. $(x^n - a^n)$ is divisible by $(x + a)$ ONLY for EVEN $n$. 3. $(x^n + a^n)$ is divisible by $(x + a)$ ONLY for ODD $n$. ### Final Answer Therefore, the correct answer is 0.

Question 4

Find a common factor of $(127^{127} + 97^{127})$ and $(127^{97} + 97^{97})$.

Accepted Answer

### Concept & Formula The problem tests the divisibility property of the algebraic sum of powers. The fundamental rule states that any expression in the form $(x^n + a^n)$ is perfectly divisible by $(x + a)$ for all **odd** values of $n$. ### Step-by-Step Solution **Given:** Expression 1: $127^{127} + 97^{127}$ Expression 2: $127^{97} + 97^{97}$ **Calculation / Deduction:** * **Analyze Expression 1:** It is in the form $x^n + a^n$ with $x = 127$, $a = 97$, and $n = 127$. Since the exponent ($127$) is an odd number, it is perfectly divisible by $(x + a)$. $$ 127 + 97 = 224 $$ * So, $224$ is a factor of the first expression. * **Analyze Expression 2:** It is also in the form $x^n + a^n$ with $x = 127$, $a = 97$, and $n = 97$. Since the exponent ($97$) is also an odd number, it is perfectly divisible by $(x + a)$. $$ 127 + 97 = 224 $$ * So, $224$ is a factor of the second expression as well. * Since $224$ divides both expressions, it is their common factor. ### Exam Strategy & Shortcut When you see $(x^{	ext{odd}} + y^{	ext{odd}})$, immediately calculate $x + y$. Here, $127 + 97 = 224$. Since both expressions feature odd powers over the exact same bases, $224$ is guaranteed to be the common factor. Ignore the different exponents entirely. ### Common Pitfall A trap students fall into is trying to find a relationship between the exponents ($127$ and $97$), perhaps trying to find the HCF of the powers. For the $(x^n + a^n)$ rule, the specific value of the exponent does not matter, only its odd/even parity matters. ### Final Answer Therefore, the correct answer is 224.

Question 5

A $99$-digit number is formed by writing the first $59$ natural numbers one after the other as $1234567891011121314\dots5859$. Find the remainder obtained when the above number is divided by $16$.

Accepted Answer

### Concept & Logic To check for divisibility by powers of $2$, you only need to examine the last digits of the number. * Divisibility by $2^1$ ($2$): Check the last $1$ digit. * Divisibility by $2^2$ ($4$): Check the last $2$ digits. * Divisibility by $2^3$ ($8$): Check the last $3$ digits. * Divisibility by $2^4$ ($16$): Check the last $4$ digits. ### Step-by-Step Solution **Given:** The number is $12345678910\dots575859$. The divisor is $16$. **Calculation / Deduction:** * Since $16 = 2^4$, the remainder of the entire $99$-digit number divided by $16$ will be exactly the same as the remainder of just its last $4$ digits divided by $16$. * The sequence ends with $57, 58, 59$. The last four digits form the number $5859$. * Now, simply divide $5859$ by $16$. $$ 5859 = 16 	imes 366 + 3 $$ * Let's break down the division mentally: * $16 	imes 300 = 4800$ (Remainder: $1059$) * $16 	imes 60 = 960$ (Remainder: $99$) * $16 	imes 6 = 96$ (Remainder: $3$) * The final remainder is $3$. ### Exam Strategy & Shortcut Do not let the "$99$-digit number" phrase intimidate you; it is just filler. Identify the divisor ($16$), recall the rule of $16$ (last $4$ digits), extract those digits ($5859$), and perform standard division. ### Common Pitfall A common mistake is attempting to use modular arithmetic on blocks of numbers or trying to find a pattern in the remainders of $1, 2, 3\dots$, which is a massive waste of time and will not yield the correct answer. Stick to the basic divisibility rules. ### Final Answer Therefore, the correct answer is 3.

Question 6

What is the place value of $5$ in $3254710$?

Accepted Answer

### Concept & Logic The **place value** of a digit in a number depends on its position. It is calculated by multiplying the digit's **face value** by its positional multiplier. $$	ext{Place Value} = 	ext{Face Value} 	imes 	ext{Position Value}$$ ### Step-by-Step Solution **Given:** The number is $3254710$ and we need the place value of the digit $5$. **Calculation:** 1. Identify the position of $5$ starting from the right (units, tens, hundreds, thousands, ten-thousands). 2. The digit $5$ is in the **ten-thousands** position. 3. Multiply the face value ($5$) by the position value ($10000$). $$5 	imes 10000 = 50000$$ ### Exam Strategy & Shortcut Locate the digit $5$ and simply replace all digits to its right with zeros. Since there are four digits after $5$ in $3254710$, add four zeros to get $50000$ instantly. ### Common Pitfall Students often confuse **place value** with **face value** and mistakenly choose $5$ (Option A). Always remember that place value includes the trailing zeros based on position. ### Final Answer Therefore, the correct answer is **50000**.

Question 7

The face value of $8$ in the number $458926$ is

Accepted Answer

### Concept & Logic The **face value** of any digit in a number is simply the actual value of the digit itself, entirely independent of where it is positioned in the number. ### Step-by-Step Solution **Given:** We need the face value of the digit $8$ in the number $458926$. **Deduction:** 1. The face value is the pure numerical value of the digit. 2. Therefore, regardless of $8$ being in the thousands place, its face value remains strictly $8$. ### Exam Strategy & Shortcut Whenever a question asks for "face value," immediately select the digit in question without looking at the whole number or doing any calculations. ### Common Pitfall The most frequent mistake is calculating the **place value** instead of the face value, which would lead to choosing $8000$ (Option C). Always read carefully between "face" and "place". ### Final Answer Therefore, the correct answer is **8**.

Question 8

The sum of the place values of $3$ in the number $503535$ is

Accepted Answer

### Concept & Logic To find the sum of multiple place values for a repeating digit, determine the positional value of each occurrence separately and add them together. ### Step-by-Step Solution **Given:** The number is $503535$ and we need the sum of the place values of all $3$s. **Calculation:** 1. Locate the first $3$ (from left to right): It is in the thousands place. Its place value is $3000$. 2. Locate the second $3$: It is in the tens place. Its place value is $30$. 3. Add the extracted place values together: $$3000 + 30 = 3030$$ ### Exam Strategy & Shortcut Write down the original number and simply replace every digit that is *not* a $3$ with a $0$. $503535 \rightarrow 003030$. Dropping the leading zeros gives you the answer $3030$ instantly. ### Common Pitfall A common trap is adding the **face values** instead of place values ($3 + 3 = 6$) or missing a positional zero (getting $3300$). Always align numbers properly when adding visually. ### Final Answer Therefore, the correct answer is **3030**.

Question 9

The difference between the place values of $7$ and $3$ in the number $527435$ is

Accepted Answer

### Concept & Logic Find the individual positional values (place values) of the specific digits and subtract the smaller value from the larger value. ### Step-by-Step Solution **Given:** The number is $527435$. We need the difference between the place values of $7$ and $3$. **Calculation:** 1. The digit $7$ is in the thousands place. Its place value is $7000$. 2. The digit $3$ is in the tens place. Its place value is $30$. 3. Subtract the two values: $$7000 - 30 = 6970$$ ### Exam Strategy & Shortcut Use the unit digit trick for fast subtraction. You are calculating $7000 - 30$. The last two digits will be $00 - 30 = 70$. Scan the options for a number ending in $70$. Option (d) is the only match, saving you full calculation time. ### Common Pitfall Students might rush and subtract the **face values** ($7 - 3 = 4$), falling for Option A. Always pause to confirm whether the question asks for "place" or "face" value. ### Final Answer Therefore, the correct answer is **6970**.

Question 10

The difference between the local value and the face value of $7$ in the numeral $32675149$ is

Accepted Answer

### Concept & Logic "Local value" is a synonym for **place value**. The question requires finding the place value of a digit and subtracting its constant **face value**. ### Step-by-Step Solution **Given:** The numeral is $32675149$. We need the difference between the local (place) value and face value of the digit $7$. **Calculation:** 1. The digit $7$ is in the ten-thousands place (four digits follow it). Its local value is $70000$. 2. The face value of the digit $7$ is exactly $7$. 3. Calculate the difference: $$70000 - 7 = 69993$$ ### Exam Strategy & Shortcut When subtracting a single digit from a perfect power of 10 base (like $70000 - 7$), the last digits of the answer will always be derived from $100 - 7 = 93$. Quickly look for the option that ends in $93$. ### Common Pitfall Failing to recognize that "local value" means "place value." If you assume local value means face value, you get $7 - 7 = 0$, which isn't an option, causing unnecessary panic during an exam. ### Final Answer Therefore, the correct answer is **69993**.

Question 11

The sum of the greatest and smallest number of five digits is

Accepted Answer

### Concept & Logic The greatest $n$-digit number is formed entirely of $9$s, and the smallest $n$-digit number consists of a $1$ followed by zeros. To find their sum, we simply add the two constructed numbers. ### Step-by-Step Solution **Calculation:** 1. Determine the greatest 5-digit number: This is formed by five $9$s, which is $99999$. 2. Determine the smallest 5-digit number: This is formed by a $1$ followed by four $0$s, which is $10000$. 3. Calculate their sum: $$99999 + 10000 = 109999$$ ### Exam Strategy & Shortcut When adding $10000$ to any 5-digit number, you only need to increment the ten-thousands digit by $1$. Since $9 + 1 = 10$, write $10$ in front of the remaining $9999$ to get $109999$ instantly. ### Common Pitfall A frequent error is assuming the "smallest" 5-digit number is formed by repeating ones ($11111$), which would incorrectly lead to $99999 + 11111 = 111110$. ### Final Answer Therefore, the correct answer is **109,999**.

Question 12

If the largest three-digit number is subtracted from the smallest five-digit number, then the remainder is

Accepted Answer

### Concept & Logic This problem requires correctly identifying boundary numbers based on digit counts and performing basic subtraction. ### Step-by-Step Solution **Calculation:** 1. Formulate the smallest 5-digit number: A $1$ followed by four zeros ($10000$). 2. Formulate the largest 3-digit number: Three $9$s ($999$). 3. Subtract the 3-digit number from the 5-digit number: $$10000 - 999 = 9001$$ ### Exam Strategy & Shortcut Instead of standard subtraction with borrowing, use the round-number approximation. Treat $999$ as $(1000 - 1)$. Equation becomes: $10000 - (1000 - 1) = 9000 + 1 = 9001$. This eliminates calculation errors completely. ### Common Pitfall Miscounting the zeros in the smallest 5-digit number (using $1000$ instead of $10000$) will lead to $1000 - 999 = 1$, which is a trap option deliberately provided as Option A. ### Final Answer Therefore, the correct answer is **9001**.

Question 13

The smallest number of $5$ digits beginning with $3$ and ending with $5$ will be

Accepted Answer

### Concept & Logic To construct the absolute minimum value for a number with fixed boundaries (starting and ending digits), all free internal positions must be filled with the lowest possible digit, which is $0$. ### Step-by-Step Solution **Deduction:** 1. The number has 5 places: [Ten-Thousands] [Thousands] [Hundreds] [Tens] [Units]. 2. Condition 1: It begins with $3$. So the Ten-Thousands place is $3$. (Format: $3 \_ \_ \_ \_$) 3. Condition 2: It ends with $5$. So the Units place is $5$. (Format: $3 \_ \_ \_ 5$) 4. Condition 3: It must be the smallest possible number. Therefore, fill the Thousands, Hundreds, and Tens places with the minimum digit $0$. 5. The final number is $30005$. ### Exam Strategy & Shortcut You do not need to construct the number manually. Simply scan the multiple-choice options. Since all options start with $3$ and end with $5$, compare their magnitudes directly: $30005 < 30015 < 30025 < 31005$. Option C is strictly the smallest. ### Common Pitfall A common mistake is repeating the starting digit in the middle (e.g., $33335$) or using $1$s ($31115$) instead of $0$s, failing to realize that $0$ is permissible and necessary for a minimum value. ### Final Answer Therefore, the correct answer is **30005**.

Question 14

What is the minimum number of four digits formed by using the digits $2$, $4$, $0$, $7$?

Accepted Answer

### Concept & Logic To form the smallest possible number from a given set of unique digits, arrange the digits in ascending order. However, a number cannot have $0$ as its leading digit, so the smallest non-zero digit must take the first position. ### Step-by-Step Solution **Given:** The digits are $2, 4, 0, 7$. We need a 4-digit number. **Calculation:** 1. Sort the given digits in purely ascending order: $0, 2, 4, 7$. 2. Placing $0$ at the beginning ($0247$) creates a 3-digit number. Therefore, swap the $0$ with the next smallest digit ($2$). 3. The leading digit becomes $2$. The $0$ immediately follows it. 4. Place the remaining digits in ascending order: $4$, then $7$. 5. The resulting sequence is $2047$. ### Exam Strategy & Shortcut First, eliminate any option that doesn't use the exact set of given digits (e.g., Option B uses two $2$s). Out of the remaining valid permutations ($2047$, $2407$, $2470$), simply identify the numerically smallest value. ### Common Pitfall The most common mistake is ignoring the rule that 4-digit numbers cannot start with $0$, leading students to search for $0247$. When it's missing, they might panic and guess randomly. ### Final Answer Therefore, the correct answer is **2047**.

Question 15

All natural numbers and $0$ are called the ........ numbers.

Accepted Answer

### Concept & Logic Number systems are built by progressively expanding sets. The set of **Natural numbers** starts at $1$ ($1, 2, 3...$). When exactly one element, $0$, is added to this set, the resulting collection defines the **Whole numbers**. ### Step-by-Step Solution **Deduction:** 1. Recall the definition of Natural Numbers ($\mathbb{N}$): $\{1, 2, 3, 4, ...\}$ 2. The question asks for the set that combines $\mathbb{N}$ and $\{0\}$. 3. The combined set is $\{0, 1, 2, 3, 4, ...\}$. 4. By mathematical definition, this exact set is known as the Whole Numbers ($\mathbb{W}$). ### Exam Strategy & Shortcut Memorize the hierarchy of basic number systems: Natural (starts at $1$) $\rightarrow$ Whole (adds $0$) $\rightarrow$ Integer (adds negatives). The phrase "and $0$" immediately flags the jump from Natural to Whole. ### Common Pitfall Students frequently confuse "integers" with "whole numbers." Remember that integers also include negative numbers ($-1, -2$), whereas the union of just natural numbers and zero stops exactly at the boundary of whole numbers. ### Final Answer Therefore, the correct answer is **whole**.

Question 16

Consider the following statements about natural numbers: (1) There exists a smallest natural number. (2) There exists a largest natural number. (3) Between two natural numbers, there is always a natural number. Which of the above statements is/are c…

Accepted Answer

### Concept & Logic Analyze the fundamental properties of the set of Natural Numbers, denoted mathematically as $\mathbb{N} = \{1, 2, 3, \dots\}$. ### Step-by-Step Solution **Deduction:** 1. Statement (1): The set of natural numbers begins at $1$. Therefore, $1$ is the smallest natural number. This statement is **True**. 2. Statement (2): Natural numbers extend infinitely upward. For any number $n$, there is always $n+1$. There is no largest natural number. This statement is **False**. 3. Statement (3): Consider two consecutive natural numbers like $1$ and $2$. There are no integer or natural values between them. This statement is **False**. 4. Consequently, only the first statement is correct. ### Exam Strategy & Shortcut To quickly falsify broad mathematical statements, search for a single basic counterexample. For statement 3, simply pick $1$ and $2$. Since there's no whole count between them, the statement immediately fails, letting you eliminate options quickly. ### Common Pitfall Students often confuse the properties of natural numbers with real or rational numbers. While it is true that there are infinite real numbers between any two real numbers, this density property does not apply to discrete, countable sets like natural numbers. ### Final Answer Therefore, the correct answer is **Only 1**.

Question 17

Every rational number is also

Accepted Answer

### Concept & Logic The Real Number system encompasses all continuous values on the number line, broadly divided into two mutually exclusive, primary sets: Rational numbers and Irrational numbers. ### Step-by-Step Solution **Deduction:** 1. Rational numbers ($\mathbb{Q}$) are any numbers that can be expressed as a fraction $\frac{p}{q}$ (where $p, q$ are integers and $q 
eq 0$). 2. Real numbers ($\mathbb{R}$) include all rational numbers AND all irrational numbers. 3. By definition, every single rational number is automatically classified as a subset of the real number system. 4. Conversely, integers, natural numbers, and whole numbers are subsets of rational numbers, so a rational number (like $\frac{1}{2}$) is not necessarily an integer or whole number. ### Exam Strategy & Shortcut Visualize the number system hierarchy: Real numbers act as the "universal container" for basic math. If a number is established as rational, it must be contained within the broader "real number" category. ### Common Pitfall Assuming the reverse hierarchy or confusing subsets. For example, believing that every rational number is an integer. Fractions like $\frac{3}{4}$ are rational, but they are clearly not integers. ### Final Answer Therefore, the correct answer is **a real number**.

Question 18

The number $\pi$ is

Accepted Answer

### Concept & Logic A number is strictly categorized as **irrational** if it cannot be expressed as a simple fraction of two integers, resulting in a decimal expansion that goes on forever without repeating. ### Step-by-Step Solution **Deduction:** 1. The mathematical constant $\pi$ (Pi) represents the ratio of a circle's circumference to its diameter. 2. Despite originating as a physical ratio, its exact mathematical value cannot be written as a simple fraction $\frac{p}{q}$. 3. The decimal expansion of $\pi$ ($3.14159265...$) is infinite and never falls into a periodic repeating pattern. 4. Thus, it fits the strict mathematical definition of an irrational number. ### Exam Strategy & Shortcut Memorize the most famous irrational constants: $\pi$, $e$, and the square roots of non-perfect squares. If you see $\pi$ in any classification question, instantly select "irrational." ### Common Pitfall Choosing "a fraction" or "a rational number" because students are routinely taught to use $\frac{22}{7}$ for manual calculations. This is a dangerous trap; $\frac{22}{7}$ is merely a convenient approximation, not the exact value of $\pi$. ### Final Answer Therefore, the correct answer is **an irrational number**.

Question 19

$\sqrt{2}$ is a/an

Accepted Answer

### Concept & Logic The square root of any prime number, or any positive integer that is not a perfect square, cannot be expressed as a finite or repeating decimal. These are the hallmark traits of **irrational numbers**. ### Step-by-Step Solution **Deduction:** 1. Check the integer under the root symbol: $2$. 2. The number $2$ is not a perfect square (like $1, 4, 9, 16$). 3. Because it is not a perfect square, $\sqrt{2}$ cannot be written as a simple fraction $\frac{p}{q}$ consisting of integers. 4. Its actual decimal value is approximately $1.41421356...$ which extends infinitely without repeating. 5. Therefore, it is mathematically classified as an irrational number. ### Exam Strategy & Shortcut Learn this absolute rule: Any radical $\sqrt{x}$ where $x$ is a positive integer but not a perfect square is automatically irrational. Since $2$ isn't a perfect square, pick "irrational" immediately. ### Common Pitfall Confusing irrational numbers with fractional approximations. Some test-takers try to estimate $\sqrt{2}$ as $1.4$ or $\frac{7}{5}$ to simplify math, and then incorrectly label the root itself as a rational number based on that approximation. ### Final Answer Therefore, the correct answer is **irrational number**.

Question 20

The number $\sqrt{3}$ is

Accepted Answer

### Concept & Logic The square root of any non-perfect square is an irrational number. The defining characteristic of all irrational numbers is that their decimal form expands infinitely without forming a repeating sequence (an **infinite non-recurring decimal**). ### Step-by-Step Solution **Deduction:** 1. The given number is $\sqrt{3}$. 2. Check if the integer $3$ is a perfect square. It is not. Therefore, by definition, $\sqrt{3}$ is an irrational number. 3. Next, map the properties of rational vs. irrational numbers to decimal forms. Rational numbers have decimals that either terminate (finite) or repeat (infinite recurring). 4. Irrational numbers strictly have decimals that never terminate and never repeat in a block pattern. 5. Therefore, the decimal expansion of $\sqrt{3}$ must inherently be infinite and non-recurring. ### Exam Strategy & Shortcut Recognize synonymous terminology: "Irrational" is perfectly synonymous with "infinite non-recurring decimal." Once you know $\sqrt{3}$ is irrational, the specific decimal descriptor is locked in without calculation. ### Common Pitfall Selecting "equal to $1.732$" (Option C). While $1.732$ is routinely used for practical calculations in geometry, it is a rounded truncation. The actual value continues forever without repeating, meaning it is not strictly *equal* to $1.732$. ### Final Answer Therefore, the correct answer is **an infinite non-recurring decimal**.

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