Q: Find the number of zeros at the end of the result $3 \times 6 \times 9 \times 12 \times 15 \times \dots \times 99 \times 102$.

### Concept & Formula Similar to counting zeros in factorials, when dealing with an arithmetic progression of multiples, extract the common factor to reveal a standard factorial. Count the pairs of $2$s and $5$s. ### Step-by-Step Solution **Given:** Product is $3 \times 6 \times 9 \times 12 \times 15 \times \dots \times 102$. **Calculation:** * Note that each term is a multiple of $3$: $(3 \times 1) \times (3 \times 2) \times \dots \times (3 \times 34)$. * Factor out the $3$ from all $34$ terms: $$3^{34} \times (1 \times 2 \times 3 \times \dots \times 34)$$ * This simplifies to $3^{34} \times 34!$. * The number of zeros is determined by the number of $5$s in $34!$, as $2$s are abundant and the $3^{34}$ term contributes no $2$s or $5$s. * Count the highest power of $5$ in $34!$: $$\lfloor \frac{34}{5} \rfloor + \lfloor \frac{34}{25} \rfloor = 6 + 1 = 7$$ ### Exam Strategy & Shortcut Identify the multiple ($3$), determine the number of terms ($102 / 3 = 34$), and then find the highest power of $5$ in $34!$ using successive division ($34 / 5 = 6$, $6 / 5 = 1$, sum $= 7$). This can be done mentally in seconds. ### Common Pitfall Forgetting to check the power of $2$ can be risky. While in $34!$ there are always more $2$s than $5$s, if the sequence was purely multiples of $5$ (e.g., $5, 10, 15\dots$), you would need to count the $2$s instead. Always confirm which limiting factor dictates the pairs. ### Final Answer **Therefore, the correct answer is 7.**

Q: The unit's digit of $13^{2003}$ is

### Concept & Logic The unit digit of a number raised to a large power follows a predictable repeating pattern called cyclicity. The unit digit of $x^n$ only depends on the unit digit of $x$. ### Step-by-Step Solution **Given:** Find the unit digit of $13^{2003}$. **Calculation:** * The unit digit of $13^{2003}$ is strictly determined by the unit digit of the base, which is $3$. We need the unit digit of $3^{2003}$. * The powers of $3$ follow a cyclicity of $4$: * $3^1 = 3$ * $3^2 = 9$ * $3^3 = 27 \rightarrow 7$ * $3^4 = 81 \rightarrow 1$ * Divide the exponent $2003$ by the cyclicity $4$ to find the remainder: $$2003 \pmod 4 = 3$$ * A remainder of $3$ means the cycle lands on the $3$rd position in the sequence, which corresponds to $3^3$. * The unit digit of $3^3$ is $7$. ### Exam Strategy & Shortcut To quickly find $2003 \pmod 4$, just look at the last two digits of the exponent. $03 / 4$ leaves a remainder of $3$. So, $3^3$ gives a unit digit of $7$. This takes less than $10$ seconds. ### Common Pitfall A common mistake is using the entire base $13$ for multiplication cycles instead of stripping it down to its unit digit $3$, wasting time on unnecessary mental arithmetic. ### Final Answer **Therefore, the correct answer is 7.**

Q: The digit in the unit's place of the number $123^{99}$ is

### Concept & Logic The unit digit of an exponentiation $a^b$ depends solely on the unit digit of the base $a$ and the cyclicity of its powers. ### Step-by-Step Solution **Given:** Find the unit digit of $123^{99}$. **Calculation:** * We only care about the unit digit of the base, which is $3$. The problem simplifies to finding the unit digit of $3^{99}$. * The unit digits of powers of $3$ cycle every $4$ steps: $3, 9, 7, 1$. * Find the remainder when the exponent $99$ is divided by the cyclicity $4$: $$99 \pmod 4 = 3$$ * Since the remainder is $3$, we map this to the $3$rd step of our cycle, which is equivalent to $3^3$. * $3^3 = 27$, so the unit digit is $7$. ### Exam Strategy & Shortcut Quickly divide the exponent by $4$: $100$ is perfectly divisible by $4$, so $99$ is $1$ less, meaning the remainder is $3$. The $3$rd power of $3$ ends in $7$. ### Common Pitfall Students sometimes mistake a remainder of $0$ to mean the $0$th power (which would equal $1$). While correct for base $3$, the rule is that a remainder of $0$ corresponds to the $4$th item in the cycle. Always be careful to map remainders correctly ($1 \rightarrow 1$st, $2 \rightarrow 2$nd, $3 \rightarrow 3$rd, $0 \rightarrow 4$th). ### Final Answer **Therefore, the correct answer is 7.**

Question 1

If $m$ and $n$ are natural numbers such that $2^m - 2^n = 960$, what is the value of $m$?

Accepted Answer

### Concept & Logic When dealing with the difference of two powers with the same base, factoring out the smaller power reveals a manageable integer equation. This transforms an exponential problem into a prime factorization problem. $$ a^x - a^y = a^y(a^{x-y} - 1) $$ ### Step-by-Step Solution * **Given:** $2^m - 2^n = 960$ $m$ and $n$ are natural numbers. * **Calculation / Deduction:** Since the difference is positive ($960$), $m$ must be strictly greater than $n$. Factor out the smaller power, $2^n$, from the left side of the equation: $$ 2^n(2^{m-n} - 1) = 960 $$ Now, break down $960$ into its prime factors, separating the power of $2$ from the odd integer part. $960 = 32 	imes 30$ $960 = 64 	imes 15$ $$ 960 = 2^6 	imes 15 $$ Substitute this back into the equation: $$ 2^n(2^{m-n} - 1) = 2^6 	imes 15 $$ By comparing the even (power of 2) and odd components on both sides: 1. The power of $2$: $$ 2^n = 2^6 \implies n = 6 $$ 2. The odd factor: $$ 2^{m-n} - 1 = 15 $$ $$ 2^{m-6} - 1 = 15 $$ $$ 2^{m-6} = 16 $$ $$ 2^{m-6} = 2^4 $$ Equating the exponents: $$ m - 6 = 4 \implies m = 10 $$ ### Exam Strategy & Shortcut **Option Substitution:** In competitive exams, directly testing the options in the equation $2^m - 2^n = 960$ is incredibly fast. * Try Option (a) $m = 10$: $2^{10} = 1024$. The equation becomes $1024 - 2^n = 960$. $2^n = 1024 - 960 = 64$. Since $64$ is exactly $2^6$, $n=6$ is a valid natural number. Option (a) fits perfectly. ### Common Pitfall A common mistake is trying to apply logarithms immediately to solve $2^m - 2^n = 960$, which leads to a mathematical dead-end because $\log(A - B)$ does not simplify. Factoring must come first. ### Final Answer **Therefore, the correct answer is 10.**

Question 2

On multiplying a number by 7, all the digits in the product appear as 3's. The smallest such number is

Accepted Answer

### Concept & Strategy This is a divisibility problem disguised as a multiplication problem. If an unknown number multiplied by $7$ yields a string of $3$'s ($33333...$), then we can find the number by simply dividing $33333...$ by $7$ until we reach a remainder of $0$. ### Step-by-Step Solution * **Given:** $x 	imes 7 = 33333...$ We need to find the smallest integer $x$. * **Calculation / Deduction:** Perform standard long division of $33333...$ by $7$: * $33 \div 7 = 4$ with a remainder of $5$. (First digit of quotient is **$4$**) * Bring down the next $3$, making it $53$. * $53 \div 7 = 7$ with a remainder of $4$. (Next digit is **$7$**) * Bring down the next $3$, making it $43$. * $43 \div 7 = 6$ with a remainder of $1$. (Next digit is **$6$**) * Bring down the next $3$, making it $13$. * $13 \div 7 = 1$ with a remainder of $6$. (Next digit is **$1$**) * Bring down the next $3$, making it $63$. * $63 \div 7 = 9$ with a remainder of $0$. (Next digit is **$9$**) Since the remainder is now $0$, the division is complete. The quotient is $47619$. ### Exam Strategy & Shortcut **Unit Digit and Option Verification:** You can solve this by looking at the options and working backwards. The target number ends in a $3$. All options end in a $9$. ($9 	imes 7 = 63$, which gives the correct unit digit $3$). Instead of long division, pick an option and multiply it by $7$ mentally or on paper. Let's test option (a) $47619$: $$ 47619 	imes 7 = 333333 $$ It perfectly matches the condition. ### Common Pitfall Students often get intimidated by the "infinite" looking sequence of 3's and attempt complex algebraic formulations. Simply setting up a continuous division is the intended and fastest mechanical path. ### Final Answer **Therefore, the correct answer is 47619.**

Question 3

The number of digits in the smallest number, which when multiplied by 7 yields all nines, is

Accepted Answer

### Concept & Strategy Similar to the previous problem, finding a number that yields a sequence of repeating digits when multiplied by a prime requires dividing the repeating sequence by that prime until there is no remainder. We are looking for $N$ such that $N 	imes 7 = 99999...$ Therefore, $N = \frac{99999...}{7}$ ### Step-by-Step Solution * **Given:** Target product sequence is $9999...$ Divisor is $7$. * **Calculation / Deduction:** Divide a sequence of $9$'s by $7$ step-by-step until the remainder hits zero. * $9 \div 7 = 1$ (Remainder $2$) * Bring down $9 \implies 29 \div 7 = 4$ (Remainder $1$) * Bring down $9 \implies 19 \div 7 = 2$ (Remainder $5$) * Bring down $9 \implies 59 \div 7 = 8$ (Remainder $3$) * Bring down $9 \implies 39 \div 7 = 5$ (Remainder $4$) * Bring down $9 \implies 49 \div 7 = 7$ (Remainder $0$) The division terminates here. The smallest number $N$ is $142857$. The quotient $142857$ is composed of $6$ digits. ### Exam Strategy & Shortcut **Cyclic Number Memorization:** A powerful shortcut for bank and SSC exams is memorizing the reciprocal of $7$. $$ \frac{1}{7} = 0.142857142857... $$ This repeating decimal block has exactly **6 digits**. Because it repeats every $6$ digits, multiplying $142857$ by $7$ will yield $999999$ (a number with six $9$'s). You can instantly answer "6" without doing any division. ### Common Pitfall A major reading comprehension pitfall is finding the smallest number ($142857$) and then looking for it in the options, getting confused when it isn't there. The question explicitly asks for the **"number of digits"** in that number, not the number itself. ### Final Answer **Therefore, the correct answer is 6.**

Question 4

A boy multiplies 987 by a certain number and obtains 559981 as his answer. If in the answer both 9's are wrong but the other digits are correct, then the correct answer will be

Accepted Answer

### Concept & Logic When a multiplication result contains errors in specific digits, we can use the fundamental **rules of divisibility** to identify the correct answer among the choices. The true product must be perfectly divisible by its original factors. ### Step-by-Step Solution * **Given:** Multiplier: $987$ Flawed Answer: $559981$ (The $5$, $5$, $8$, and $1$ are correct; the $9$'s are wrong). Target format: $55 \_ \_ 81$ * **Calculation / Deduction:** Analyze the known factor, $987$, for divisibility. The sum of the digits of $987$ is $9 + 8 + 7 = 24$. Since $24$ is divisible by $3$, the number $987$ is a multiple of $3$. Consequently, the final correct product must ALSO be a multiple of $3$. For a number to be divisible by $3$, the sum of its digits must be a multiple of $3$. Test the sum of the digits for the given options: * (a) $553681 \implies 5 + 5 + 3 + 6 + 8 + 1 = 28$ (Not divisible by $3$) * (b) $555181 \implies 5 + 5 + 5 + 1 + 8 + 1 = 25$ (Not divisible by $3$) * (c) $555681 \implies 5 + 5 + 5 + 6 + 8 + 1 = 30$ (**Divisible by $3$**) * (d) $556581 \implies 5 + 5 + 6 + 5 + 8 + 1 = 30$ (**Divisible by $3$**) We have narrowed it down to (c) and (d). Now, check which of these is actually a multiple of $987$ by doing a quick rough division or checking the difference. Difference between (d) and (c) is $556581 - 555681 = 900$. Since $900$ is not a multiple of $987$, they cannot both be multiples of $987$. Let's test $555681$ (Option c) by dividing it by $987$: $$ 555681 \div 987 = 563 $$ It divides perfectly with no remainder. ### Exam Strategy & Shortcut **Digital Sum / Divisibility Rule of 3:** Relying entirely on divisibility by $3$ eliminates two options immediately. Between the final two, rather than doing long division, you can check divisibility by $7$ (since $987$ is $3 	imes 7 	imes 47$). Option (c) $555681 \div 7 \implies 79383$ (exact). Option (d) $556581 \div 7 \implies$ leaves a remainder. Therefore, (c) is the only mathematically possible answer. ### Common Pitfall A time-consuming mistake is assuming you have to set up algebraic equations with variables for the missing digits (e.g., $550081 + 1000x + 100y = 987k$) and solving for them. Always leverage the provided multiple-choice options in conjunction with basic divisibility rules. ### Final Answer **Therefore, the correct answer is 555681.**

Question 5

The numbers $1, 3, 5, ........, 25$ are multiplied together. The number of zeros at the right end of the product is

Accepted Answer

### Concept & Logic A trailing zero at the right end of any product is created exclusively by multiplying the prime factors $2$ and $5$ together ($2 	imes 5 = 10$). The number of trailing zeros is determined by the number of pairs of $(2 	imes 5)$ that can be formed from the prime factorization of all numbers in the series. ### Step-by-Step Solution * **Given:** A series of numbers multiplied together: $1 	imes 3 	imes 5 	imes 7 	imes .... 	imes 25$. * **Calculation / Deduction:** Observe the nature of the numbers in the series. $1, 3, 5, 7, 9, 11...$ all the way to $25$. This is a continuous sequence of **odd numbers**. By definition, an odd number does not have $2$ as a prime factor. While the series contains plenty of $5$'s (in $5$, $15$, and $25$), it contains absolutely zero $2$'s. Without any $2$'s to pair with the $5$'s, it is mathematically impossible to form a $10$. Therefore, the final product will not have any trailing zeros. ### Exam Strategy & Shortcut **Parity Rule:** The product of any number of odd integers is ALWAYS an odd integer. An odd integer can never end in the digit $0$ (since ending in $0$ inherently makes a number even and divisible by $10$). Therefore, the number of zeros at the end is $0$. This deduction should take less than 5 seconds. ### Common Pitfall Students who have memorized the standard "Trailing Zeros" formula often operate on autopilot. They will scan the series, correctly identify the multiples of $5$, count them up, and mistakenly conclude there are several trailing zeros, completely ignoring the fact that the crucial even numbers are missing from the given sequence. ### Final Answer **Therefore, the correct answer is 0.**

Question 6

The numbers $1, 2, 3, 4, ........., 1000$ are multiplied together. The number of zeros at the end (on the right) of the product must be

Accepted Answer

### Concept & Formula The product of consecutive integers from $1$ to $n$ is represented as $n!$ (n factorial). Trailing zeros in any product are created exclusively by the multiplication of pairs of prime factors $2$ and $5$ (since $2 	imes 5 = 10$). In any factorial, the prime factor $2$ appears far more frequently than the prime factor $5$. Therefore, the number of trailing zeros is entirely limited by, and equal to, the total count of $5$'s in the prime factorization of $n!$. We use **Legendre's Formula** to count the highest power of a prime number $p$ dividing $n!$: $$ 	ext{Power of } p = \left\lfloor \frac{n}{p} ightfloor + \left\lfloor \frac{n}{p^2} ightfloor + \left\lfloor \frac{n}{p^3} ightfloor + \dots $$ ### Step-by-Step Solution * **Given:** The product is $1 	imes 2 	imes 3 	imes \dots 	imes 1000 = 1000!$ We need to find the highest power of $5$ that divides $1000!$. * **Calculation / Deduction:** Apply the successive division method (Legendre's formula) using $n = 1000$ and $p = 5$. * First, divide $1000$ by $5$: $$ \lfloor 1000 / 5 floor = 200 $$ * Next, divide the quotient ($200$) by $5$ (which is equivalent to dividing $1000$ by $25$): $$ \lfloor 200 / 5 floor = 40 $$ * Next, divide the new quotient ($40$) by $5$ (equivalent to dividing $1000$ by $125$): $$ \lfloor 40 / 5 floor = 8 $$ * Finally, divide $8$ by $5$ (equivalent to dividing $1000$ by $625$): $$ \lfloor 8 / 5 floor = 1 $$ (We only take the integer part) * Stop here because dividing $1$ by $5$ yields $0$. Add all the integer quotients together to find the total number of $5$'s: $$ 	ext{Total zeros} = 200 + 40 + 8 + 1 = 249 $$ ### Exam Strategy & Shortcut **Successive Division Ladder:** You do not need to write out the formal powers of $5$. Simply set up a division ladder. Divide $1000$ by $5$ to get $200$. Immediately divide $200$ by $5$ to get $40$. Divide $40$ by $5$ to get $8$. Divide $8$ by $5$ to get $1$. Sum the results vertically: $200 + 40 + 8 + 1 = 249$. This takes under 10 seconds. ### Common Pitfall A common error is stopping after the first division (yielding $200$) and selecting option (b). Students forget that multiples of higher powers of $5$ (like $25$, $125$, and $625$) contribute *extra* $5$'s to the factorization, which must be accounted for. ### Final Answer **Therefore, the correct answer is 249.**

Question 7

First 100 multiples of 10 i.e. $10, 20, 30, ........., 1000$ are multiplied together. The number of zeros at the end of the product will be

Accepted Answer

### Concept & Logic When dealing with a sequence of multiples, you must factor out the common base from *every single term* in the product. After factoring, the problem usually reduces to evaluating a standard power of $10$ combined with a standard factorial sequence. ### Step-by-Step Solution * **Given:** The product sequence is $10 	imes 20 	imes 30 	imes \dots 	imes 1000$. There are $100$ terms in this sequence ($1000 \div 10 = 100$). * **Calculation / Deduction:** Rewrite each term in the sequence as a multiple of $10$: $$ (10 	imes 1) 	imes (10 	imes 2) 	imes (10 	imes 3) 	imes \dots 	imes (10 	imes 100) $$ Factor out the $10$ from each of the $100$ terms. Because these terms are multiplied (not added), the $10$ is raised to the power of $100$: $$ 10^{100} 	imes (1 	imes 2 	imes 3 	imes \dots 	imes 100) $$ $$ = 10^{100} 	imes 100! $$ Now, calculate the number of trailing zeros from both components: 1. **Zeros from $10^{100}$**: A base of $10$ raised to the power of $100$ inherently provides exactly **$100$** trailing zeros. 2. **Zeros from $100!$**: Use successive division by $5$ to find the trailing zeros in $100!$: $$ \lfloor 100 / 5 floor = 20 $$ $$ \lfloor 20 / 5 floor = 4 $$ Total zeros in $100! = 20 + 4 =$ **$24$** zeros. Add the zeros from both components together to find the total for the entire expression: $$ 	ext{Total trailing zeros} = 100 + 24 = 124 $$ ### Exam Strategy & Shortcut **Decomposition Visualization:** Instantly visualize the series as two layers: the visible zeros and the hidden zeros. 1. There are $100$ numbers, and each clearly ends in at least one zero. That's $100$ guaranteed zeros immediately. 2. Strip away those $100$ zeros, and you are left with $1 	imes 2 	imes \dots 	imes 100$, which is $100!$. 3. Calculate zeros for $100!$ mentally: $100 \div 5 = 20$; $20 \div 5 = 4$. So, $24$. 4. Total $= 100 + 24 = 124$. ### Common Pitfall The most dangerous pitfall is factoring out a single $10$ as if the expression was an addition series (e.g., mistakenly writing $10 	imes (1 	imes 2 \dots 	imes 100)$). Because it is a multiplication series, you must factor out $10$ for *each* term, creating $10^{100}$. ### Final Answer **Therefore, the correct answer is 124.**

Question 8

The number of zeros at the end of the product $5 	imes 10 	imes 15 	imes 20 	imes 25 	imes 30 	imes 35 	imes 40 	imes 45 	imes 50$ is

Accepted Answer

### Concept & Logic Trailing zeros are formed by pairs of $(2 	imes 5)$. In standard continuous series (like factorials), $5$'s are always rarer than $2$'s, so we usually only count the $5$'s. However, **in a series composed entirely of multiples of 5**, the environment is flooded with $5$'s. In this specific scenario, the prime factor **$2$ becomes the limiting factor**. The number of trailing zeros will be dictated strictly by the number of $2$'s available to pair with the abundant $5$'s. ### Step-by-Step Solution * **Given:** Product: $5 	imes 10 	imes 15 	imes 20 	imes \dots 	imes 50$ This is a sequence of the first $10$ multiples of $5$. * **Calculation / Deduction:** Rewrite the sequence by factoring out $5$ from each term: $$ (5 	imes 1) 	imes (5 	imes 2) 	imes (5 	imes 3) 	imes \dots 	imes (5 	imes 10) $$ $$ = 5^{10} 	imes (1 	imes 2 	imes 3 	imes \dots 	imes 10) $$ $$ = 5^{10} 	imes 10! $$ Now, let's analyze the prime factors $2$ and $5$ in this expression: **Count the 5's:** There are $10$ fives from the $5^{10}$ term. In $10!$, there are $\lfloor 10/5 floor = 2$ fives. Total $5$'s available $= 10 + 2 = 12$. **Count the 2's:** The $5^{10}$ term contains zero $2$'s. All the $2$'s must come entirely from the $10!$ component. Use successive division to count the $2$'s in $10!$: $$ \lfloor 10 / 2 floor = 5 $$ $$ \lfloor 5 / 2 floor = 2 $$ $$ \lfloor 2 / 2 floor = 1 $$ Total $2$'s available $= 5 + 2 + 1 =$ **$8$**. Since a trailing zero requires a *pair* of $(2 	imes 5)$, and we only have eight $2$'s available to pair with our twelve $5$'s, we can only form $8$ pairs. ### Exam Strategy & Shortcut **Identify the Bottleneck:** The moment you see a series jumping by multiples of $5$, your brain should flag: "2 is the bottleneck!" Don't even waste time counting the $5$'s. Just factor out the $5$ to find the underlying factorial: $5 	imes \dots 	imes 50 = 5^{10} 	imes 10!$. Then, strictly count the number of $2$'s in $10!$ via $10 \div 2 = 5$, $5 \div 2 = 2$, $2 \div 2 = 1$. Sum $= 8$. You are done in 15 seconds. ### Common Pitfall The classic trap is calculating the number of $5$'s ($12$) and selecting an option based on that, purely out of habit from solving standard factorial problems. Always verify which prime factor is the actual limiting constraint in a custom series. ### Final Answer **Therefore, the correct answer is 8.**

Question 9

The number of zeros at the end of $60!$ is

Accepted Answer

### Concept & Formula To find the number of trailing zeros in a standard factorial ($n!$), we must count the number of times the prime factor $5$ appears in its prime factorization. We ignore the factor $2$ because in any sequential $1$ to $n$ product, $2$'s are always vastly more abundant than $5$'s. Use successive division by $5$ to evaluate the total occurrences: $$ 	ext{Trailing Zeros} = \left\lfloor \frac{n}{5} ightfloor + \left\lfloor \frac{n}{25} ightfloor + \left\lfloor \frac{n}{125} ightfloor + \dots $$ ### Step-by-Step Solution * **Given:** The expression is $60!$ (60 factorial). * **Calculation / Deduction:** Apply successive division to the number $60$: * Divide $60$ by $5$: $$ \lfloor 60 / 5 floor = 12 $$ This accounts for numbers like $5$, $10$, $15$, $20$, etc., that contribute at least one $5$. * Divide the previous quotient ($12$) by $5$: $$ \lfloor 12 / 5 floor = 2 $$ This accounts for the numbers that contain a *second* factor of $5$ (specifically $25$ and $50$). * Since $2$ is less than $5$, the division process terminates here. Sum the integer quotients: $$ 	ext{Total zeros} = 12 + 2 = 14 $$ ### Exam Strategy & Shortcut **Mental Ladder:** For numbers under $100$, do this purely in your head. "$60$ divided by $5$ is $12$." "$12$ divided by $5$ is $2$." "$12 + 2 = 14$." Do not write anything down for basic factorial trailing zero queries. ### Common Pitfall A recurring mistake for beginners is simply dividing by $5$ once and assuming the answer is $12$ (Option a). It is crucial to remember that higher powers of $5$ (like $25$) hide extra $5$'s that must be captured by continuing the division chain on the quotient. ### Final Answer **Therefore, the correct answer is 14.**

Question 10

The numbers $1, 3, 5, 7, ........., 99$ and $128$ are multiplied together. The number of zeros at the end of the product must be

Accepted Answer

### Concept & Logic Every trailing zero in a product requires exactly one pair of the prime factors $2$ and $5$. In this problem, the product consists of two distinct parts: 1. A continuous series of purely **odd numbers** ($1 	imes 3 	imes 5 \dots 	imes 99$). 2. A single standalone even number appended at the end ($128$). Odd numbers contain zero $2$'s. Therefore, the only $2$'s available in the entire product are locked inside the number $128$. Because $2$'s will be extremely scarce compared to the abundance of $5$'s in the odd series, $2$ becomes the limiting factor. ### Step-by-Step Solution * **Given:** Product: $(1 	imes 3 	imes 5 	imes 7 	imes \dots 	imes 99) 	imes 128$ * **Calculation / Deduction:** **Step 1: Count the available 2's.** The entire odd series contributes zero $2$'s. The number $128$ is a known power of $2$: $$ 128 = 2^7 $$ Therefore, the entire expression contains exactly **$7$** twos. **Step 2: Count the available 5's (Verification).** To be thorough, we should ensure there are at least seven $5$'s in the odd series. The series contains odd multiples of $5$: $5, 15, 25, 35, 45, 55, 65, 75, 85, 95$. Even just looking at the first few: $5$ (one), $15$ (one), $25$ (two), $35$ (one), $45$ (one)... we clearly have more than seven $5$'s available. **Step 3: Determine the limiting factor.** We have an abundance of $5$'s (more than $12$), but exactly $7$ twos. Because we can only form pairs up to the limit of the scarcest resource, we can only form exactly **$7$** pairs of $(2 	imes 5)$. ### Exam Strategy & Shortcut **Identify the Restrictor:** Rapid analysis of the question reveals the trick immediately. The huge odd series is a distractor meant to make you count $5$'s needlessly. Recognize that odd numbers have no $2$'s. Look at the only even number: $128$. Recognize that $128$ is $2^7$. The maximum number of $10$'s you can make is locked at $7$. Done. ### Common Pitfall The most common trap is ignoring the specific nature of the series, automatically calculating the $5$'s in numbers up to $99$ using Legendre's successive division (which is also wrong here since it's missing even numbers), and completely missing that the total zeros are restricted entirely by the single even number appended at the end. ### Final Answer **Therefore, the correct answer is 7.**

Question 11

The numbers $2, 4, 6, 8, \dots, 98, 100$ are multiplied together. The number of zeros at the end of the product must be

Accepted Answer

### Concept & Formula The number of zeros at the end of a product is determined by the number of trailing tens, which equals the number of pairs of prime factors $2$ and $5$. ### Step-by-Step Solution **Given:** The sequence is the product of even numbers: $2 	imes 4 	imes 6 	imes \dots 	imes 100$. **Calculation:** * Factor out a $2$ from each of the $50$ terms in the product: $$2^{50} 	imes (1 	imes 2 	imes 3 	imes \dots 	imes 50)$$ * This simplifies to $2^{50} 	imes 50!$. * To find the number of trailing zeros in $50!$, we only need to count the highest power of $5$, since the power of $2$ (which includes $2^{50}$) is vastly greater. * Use Legendre's formula to count the $5$s in $50!$: $$\lfloor \frac{50}{5} floor + \lfloor \frac{50}{25} floor = 10 + 2 = 12$$ * There are $12$ fives and more than enough twos to pair them with. Thus, there are $12$ pairs of $(2, 5)$. ### Exam Strategy & Shortcut Whenever you have a continuous series of multiples (like $2, 4, 6\dots$ or $3, 6, 9\dots$), factor out the common multiple raised to the power of the number of terms. Then simply use the standard zero-counting rule on the resulting factorial part. ### Common Pitfall A frequent error is trying to apply the zero-counting rule directly to $100$ as if it were $100!$, which would incorrectly yield $24$ zeros. Always factor out the sequence properly first. ### Final Answer **Therefore, the correct answer is 12.**

Question 12

Let $S$ be the set of prime numbers greater than or equal to $2$ and less than $100$. Multiply all the elements of $S$. With how many consecutive zeros will the product end?

Accepted Answer

### Concept & Logic The number of zeros at the end of a product strictly depends on the number of pairs of the prime numbers $2$ and $5$ present in its prime factorization. ### Step-by-Step Solution **Given:** $S$ is the set of prime numbers $\ge 2$ and $< 100$. The product is $2 	imes 3 	imes 5 	imes 7 	imes 11 \dots 	imes 97$. **Deduction:** * The prime numbers up to $100$ include $2, 3, 5, 7, 11, \dots$ * A trailing zero is formed entirely by multiplying $2 	imes 5 = 10$. * Since we are multiplying a set containing *only* prime numbers exactly once, the prime factor $2$ appears exactly once. * The prime factor $5$ also appears exactly once. * No other combinations of primes will yield a factor of $10$. * Since there is only one $2$ and one $5$, we can only form exactly one pair of $(2, 5)$. ### Exam Strategy & Shortcut This is a pure logic question. Primes don't have factors. In any sequence of consecutive primes starting from $2$, there will always be exactly one $2$ and one $5$. Therefore, the product of any set of consecutive primes that includes both $2$ and $5$ will always end in exactly $1$ zero, no matter how many other primes you multiply. ### Common Pitfall Students often overthink the size of the primes up to $100$ and assume a large product will naturally have multiple zeros, forgetting that primes (other than 2 and 5) never contribute to trailing zeros. ### Final Answer **Therefore, the correct answer is 1.**

Question 13

Find the number of zeros at the end of the result $3 	imes 6 	imes 9 	imes 12 	imes 15 	imes \dots 	imes 99 	imes 102$.

Accepted Answer

### Concept & Formula Similar to counting zeros in factorials, when dealing with an arithmetic progression of multiples, extract the common factor to reveal a standard factorial. Count the pairs of $2$s and $5$s. ### Step-by-Step Solution **Given:** Product is $3 	imes 6 	imes 9 	imes 12 	imes 15 	imes \dots 	imes 102$. **Calculation:** * Note that each term is a multiple of $3$: $(3 	imes 1) 	imes (3 	imes 2) 	imes \dots 	imes (3 	imes 34)$. * Factor out the $3$ from all $34$ terms: $$3^{34} 	imes (1 	imes 2 	imes 3 	imes \dots 	imes 34)$$ * This simplifies to $3^{34} 	imes 34!$. * The number of zeros is determined by the number of $5$s in $34!$, as $2$s are abundant and the $3^{34}$ term contributes no $2$s or $5$s. * Count the highest power of $5$ in $34!$: $$\lfloor \frac{34}{5} floor + \lfloor \frac{34}{25} floor = 6 + 1 = 7$$ ### Exam Strategy & Shortcut Identify the multiple ($3$), determine the number of terms ($102 / 3 = 34$), and then find the highest power of $5$ in $34!$ using successive division ($34 / 5 = 6$, $6 / 5 = 1$, sum $= 7$). This can be done mentally in seconds. ### Common Pitfall Forgetting to check the power of $2$ can be risky. While in $34!$ there are always more $2$s than $5$s, if the sequence was purely multiples of $5$ (e.g., $5, 10, 15\dots$), you would need to count the $2$s instead. Always confirm which limiting factor dictates the pairs. ### Final Answer **Therefore, the correct answer is 7.**

Question 14

The unit's digit of $13^{2003}$ is

Accepted Answer

### Concept & Logic The unit digit of a number raised to a large power follows a predictable repeating pattern called cyclicity. The unit digit of $x^n$ only depends on the unit digit of $x$. ### Step-by-Step Solution **Given:** Find the unit digit of $13^{2003}$. **Calculation:** * The unit digit of $13^{2003}$ is strictly determined by the unit digit of the base, which is $3$. We need the unit digit of $3^{2003}$. * The powers of $3$ follow a cyclicity of $4$: * $3^1 = 3$ * $3^2 = 9$ * $3^3 = 27 \rightarrow 7$ * $3^4 = 81 \rightarrow 1$ * Divide the exponent $2003$ by the cyclicity $4$ to find the remainder: $$2003 \pmod 4 = 3$$ * A remainder of $3$ means the cycle lands on the $3$rd position in the sequence, which corresponds to $3^3$. * The unit digit of $3^3$ is $7$. ### Exam Strategy & Shortcut To quickly find $2003 \pmod 4$, just look at the last two digits of the exponent. $03 / 4$ leaves a remainder of $3$. So, $3^3$ gives a unit digit of $7$. This takes less than $10$ seconds. ### Common Pitfall A common mistake is using the entire base $13$ for multiplication cycles instead of stripping it down to its unit digit $3$, wasting time on unnecessary mental arithmetic. ### Final Answer **Therefore, the correct answer is 7.**

Question 15

The digit in the unit's place of the number $123^{99}$ is

Accepted Answer

### Concept & Logic The unit digit of an exponentiation $a^b$ depends solely on the unit digit of the base $a$ and the cyclicity of its powers. ### Step-by-Step Solution **Given:** Find the unit digit of $123^{99}$. **Calculation:** * We only care about the unit digit of the base, which is $3$. The problem simplifies to finding the unit digit of $3^{99}$. * The unit digits of powers of $3$ cycle every $4$ steps: $3, 9, 7, 1$. * Find the remainder when the exponent $99$ is divided by the cyclicity $4$: $$99 \pmod 4 = 3$$ * Since the remainder is $3$, we map this to the $3$rd step of our cycle, which is equivalent to $3^3$. * $3^3 = 27$, so the unit digit is $7$. ### Exam Strategy & Shortcut Quickly divide the exponent by $4$: $100$ is perfectly divisible by $4$, so $99$ is $1$ less, meaning the remainder is $3$. The $3$rd power of $3$ ends in $7$. ### Common Pitfall Students sometimes mistake a remainder of $0$ to mean the $0$th power (which would equal $1$). While correct for base $3$, the rule is that a remainder of $0$ corresponds to the $4$th item in the cycle. Always be careful to map remainders correctly ($1 \rightarrow 1$st, $2 \rightarrow 2$nd, $3 \rightarrow 3$rd, $0 \rightarrow 4$th). ### Final Answer **Therefore, the correct answer is 7.**

Question 16

Match List I with List II and select the correct answer:  List I (Product) A. $(1827)^{16}$ B. $(2153)^{19}$ C. $(5129)^{21}$  List II (Digit in the unit's place) (1) $1$ (2) $3$ (3) $5$ (4) $7$ (5) $9$

Accepted Answer

### Concept & Logic The unit digit of any number with a large exponent $x^y$ depends exclusively on the unit digit of its base $x$ and the cyclicity (repeating pattern) of that unit digit's powers. ### Step-by-Step Solution **Calculation:** We will evaluate the unit digit for each expression in List I: * **A. $(1827)^{16}$** * The unit digit of the base is $7$. The cyclicity of $7$ is $4$ ($7, 9, 3, 1$). * Divide the exponent by the cyclicity: $16 \pmod 4 = 0$. * A remainder of $0$ maps to the $4$th power in the cycle ($7^4$), which ends in $1$. * Therefore, A matches with (1). * **B. $(2153)^{19}$** * The unit digit of the base is $3$. The cyclicity of $3$ is $4$ ($3, 9, 7, 1$). * Divide the exponent by the cyclicity: $19 \pmod 4 = 3$. * A remainder of $3$ maps to the $3$rd power ($3^3$), which ends in $7$. * Therefore, B matches with (4). * **C. $(5129)^{21}$** * The unit digit of the base is $9$. The cyclicity of $9$ is $2$ (odd powers end in $9$, even powers end in $1$). * The exponent $21$ is an odd number, so it behaves like $9^1$. * Therefore, C matches with (5). * **Resulting Sequence:** A $\rightarrow$ 1, B $\rightarrow$ 4, C $\rightarrow$ 5. This forms the sequence $1 \ 4 \ 5$. ### Exam Strategy & Shortcut For matching questions, you often don't need to solve every item. Solve the easiest one first. For instance, $C$ has a base of $9$ and an odd exponent, which instantly gives a unit digit of $9$ (List item 5). Looking at the options, only (c) and (d) end in $5$. Then solve A: $16$ is perfectly divisible by $4$, so base $7$ gives unit digit $1$. This immediately isolates the correct sequence as $1 \ 4 \ 5$. ### Common Pitfall Students sometimes get confused when the remainder is $0$ (like in A) and mistakenly assume the unit digit is $x^0 = 1$ for all bases. While $7^0$ happens to be $1$, this logic fails for even bases (e.g., $2^0=1$, but the $4$th step of $2$'s cycle is $6$). Always remember remainder $0$ means the *last* step of the cycle. ### Final Answer **Therefore, the correct answer is 1 4 5.**

Question 17

The digit in the unit's place of the number $(67)^{25} - 1$ must be

Accepted Answer

### Concept & Logic To find the unit digit of an expression involving subtraction, calculate the unit digit of the exponentiation first using power cyclicity, and then perform the subtraction on that unit digit. ### Step-by-Step Solution **Given:** The expression is $(67)^{25} - 1$. **Calculation:** * First, isolate the exponentiation part: $(67)^{25}$. We only care about the unit digit of the base, which is $7$. * The cyclicity of $7$ is $4$ (the pattern is $7, 9, 3, 1$). * Find the remainder of the exponent divided by the cyclicity: $$25 \pmod 4 = 1$$ * A remainder of $1$ means the unit digit corresponds to the $1$st position in the cycle, which is equivalent to $7^1 = 7$. * Now, substitute the unit digit back into the original expression: $$7 - 1 = 6$$ ### Exam Strategy & Shortcut Exponent $25$ is just $1$ more than $24$ (a multiple of $4$). So the power of $7$ resets to the start of its cycle, ending in $7$. Subtract $1$ from $7$ to instantly get $6$. This should take under $5$ seconds mentally. ### Common Pitfall A common error is subtracting the $1$ from the base *before* applying the exponent (i.e., treating it as $66^{25}$), which is mathematically incorrect according to the order of operations and leads to a completely wrong unit digit. ### Final Answer **Therefore, the correct answer is 6.**

Question 18

The unit's digit in the product $274 	imes 318 	imes 577 	imes 313$ is

Accepted Answer

### Concept & Strategy The unit digit of a product of multiple numbers is simply the unit digit of the product of their individual unit digits. The tens, hundreds, and higher-place digits have zero effect on the final unit digit. ### Step-by-Step Solution **Given:** The product expression is $274 	imes 318 	imes 577 	imes 313$. **Calculation:** * Extract only the unit digits of each number: $4, 8, 7,$ and $3$. * Multiply them sequentially, keeping only the unit digit at each step to simplify math: * Step 1: $4 	imes 8 = 32 ightarrow$ Unit digit is $2$. * Step 2: $2 	imes 7 = 14 ightarrow$ Unit digit is $4$. * Step 3: $4 	imes 3 = 12 ightarrow$ Unit digit is $2$. * The final unit digit is $2$. ### Exam Strategy & Shortcut Group the numbers to make multiplication even faster. Multiply the first pair ($4 	imes 8 = 32 ightarrow 2$) and the second pair ($7 	imes 3 = 21 ightarrow 1$). Then multiply their results: $2 	imes 1 = 2$. This avoids carrying large intermediate numbers in your head. ### Common Pitfall Multiplying the entire numbers out fully or carrying two-digit intermediate products (like $32 	imes 21 = 672$) wastes valuable exam time and increases the risk of basic arithmetic errors. Always strip down to single digits immediately. ### Final Answer **Therefore, the correct answer is 2.**

Question 19

In the product $459 	imes 46 	imes 28x 	imes 484$, the digit in the unit place is $8$. The digit to come in place of $x$ is

Accepted Answer

### Concept & Logic When an unknown digit $x$ is part of a product, we can isolate it by finding the unit digit of all known components. We then set up a modular equation to find which value of $x$ produces the target final unit digit. ### Step-by-Step Solution **Given:** The product is $459 	imes 46 	imes 28x 	imes 484$, and its final unit digit is $8$. Let the unit digit of $28x$ be $x$. **Calculation:** * Extract the unit digits of the known numbers: $9$, $6$, $x$, and $4$. * Set up the equation for the product of these unit digits: $$9 	imes 6 	imes x 	imes 4 \equiv 8 \pmod{10}$$ * Multiply the known digits, simplifying to single digits at each step: * $9 	imes 6 = 54 ightarrow$ Unit digit is $4$. * $4 	imes 4 = 16 ightarrow$ Unit digit is $6$. * Now, substitute this back into the equation: $$6 	imes x \equiv 8 \pmod{10}$$ * This means we need a digit $x$ such that $6 	imes x$ ends in $8$. * Test the digits from $0$ to $9$: * $6 	imes 3 = 18$ (Ends in $8$) * $6 	imes 8 = 48$ (Ends in $8$) * So, $x$ can be either $3$ or $8$. * Looking at the given options, $3$ is present while $8$ is not. ### Exam Strategy & Shortcut Instead of mentally testing all numbers from $0$ to $9$, plug the given multiple-choice options directly into the simplified equation $6 	imes x$. Option (a) is $3 ightarrow 6 	imes 3 = 18$ (Matches). Option (b) is $5 ightarrow 6 	imes 5 = 30$ (Fails). Option (c) is $7 ightarrow 6 	imes 7 = 42$ (Fails). Option elimination is the fastest path here. ### Common Pitfall Students often forget that there can be multiple valid digits that satisfy the unit digit equation (like $3$ and $8$ in this case). Don't panic if your first mental answer isn't in the options; check for the alternate solution. ### Final Answer **Therefore, the correct answer is 3.**

Question 20

The digit in the unit place of the number represented by $(7^{95} - 3^{58})$ is

Accepted Answer

### Concept & Formula When finding the unit digit of a subtraction operation between two exponentiated terms, calculate the unit digits individually using cyclicity. If the first unit digit is smaller than the second, you must "borrow" a $10$, exactly as you do in standard column subtraction. ### Step-by-Step Solution **Given:** Evaluate the unit digit of $(7^{95} - 3^{58})$. **Calculation:** * **Step 1: Find unit digit of $7^{95}$** * Cyclicity of $7$ is $4$. * $95 \pmod 4 = 3$. * Maps to $7^3 = 343$, so the unit digit is $3$. * **Step 2: Find unit digit of $3^{58}$** * Cyclicity of $3$ is $4$. * $58 \pmod 4 = 2$. * Maps to $3^2 = 9$, so the unit digit is $9$. * **Step 3: Perform Subtraction** * We need the unit digit of $(\dots 3) - (\dots 9)$. * Since $3 < 9$, we cannot simply write $-6$. We must borrow from the tens place of the larger overall number. * Borrowing $10$ makes it $13 - 9 = 4$. ### Exam Strategy & Shortcut Memorize the cyclicity charts for $2, 3, 7,$ and $8$ (they all have cycles of $4$). Determine $95 \pmod 4 = 3 \rightarrow 7^3 \rightarrow 3$. Determine $58 \pmod 4 = 2 \rightarrow 3^2 \rightarrow 9$. If the first number is smaller, mentally add $10$ before subtracting: $13 - 9 = 4$. ### Common Pitfall The most dangerous pitfall here is getting $3 - 9$ and just subtracting the smaller from the larger to get $6$, or getting a negative number and not knowing how to handle it. You are subtracting the *entire* large number $3^{58}$ from $7^{95}$, so normal borrowing rules of subtraction apply! ### Final Answer **Therefore, the correct answer is 4.**

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