Question 1

Find the number which is nearest to 3105 and is exactly divisible by 21.

Accepted Answer

### Concept & Strategy To find the nearest divisible number, you must calculate both the nearest lower multiple (by subtracting the remainder) and the nearest upper multiple (by adding the difference between the divisor and remainder), then compare which is closer. ### Step-by-Step Solution - **Given:** Target number = 3105, Divisor = 21. - **Calculation / Deduction:** Divide 3105 by 21 to find the remainder. - $3105 \div 21$ gives a remainder of 18. - Option 1 (Subtract): Subtract the remainder to find the lower multiple: $3105 - 18 = 3087$. The distance is 18. - Option 2 (Add): Add $(	ext{Divisor} - 	ext{Remainder})$ to find the upper multiple: $3105 + (21 - 18) = 3105 + 3 = 3108$. The distance is 3. - Comparing distances, 3 is strictly less than 18, so 3108 is nearer to 3105. ### Exam Strategy & Shortcut Compare the remainder to half of the divisor. If the remainder is greater than half the divisor ($18 > 10.5$), the *upper* multiple is closer, so add $(	ext{Divisor} - 	ext{Remainder})$. If it is less, the *lower* multiple is closer, so subtract the remainder. ### Common Pitfall A frequent mistake is automatically subtracting the remainder and picking that number (3087) without checking if the next multiple is physically closer on the number line. Always evaluate both directions. ### Final Answer Therefore, the correct answer is 3108.

Question 2

Find the smallest number of five digits which is exactly divisible by 476.

Accepted Answer

### Concept & Logic To find the smallest $N$-digit number divisible by a given divisor, start with the absolute smallest possible $N$-digit base number. Treat it as an addition problem because subtracting anything from the smallest $N$-digit number would reduce it to $N-1$ digits. ### Step-by-Step Solution - **Given:** Find the smallest 5-digit number divisible by 476. - **Calculation / Deduction:** The absolute smallest 5-digit number is 10000. - Divide 10000 by 476 to determine the remainder. - $10000 \div 476$ yields a quotient of 21 and a remainder of 4. - Since we cannot subtract (which would yield a 4-digit number like 9996), we must add to reach the next multiple. - Number to add = $	ext{Divisor} - 	ext{Remainder} = 476 - 4 = 472$. - Add this value to the base number: $10000 + 472 = 10472$. ### Exam Strategy & Shortcut Always write down the base number (like 10000) and immediately find the remainder. Apply the standard formula for finding the next multiple: $	ext{Base} + (	ext{Divisor} - 	ext{Remainder})$. Never use subtraction for smallest N-digit constraint questions. ### Common Pitfall Students often subtract the remainder (4) from 10000 to get 9996. While 9996 is perfectly divisible by 476, it is only a 4-digit number, thereby failing the primary constraint of the question. ### Final Answer Therefore, the correct answer is 10472.

Question 3

Find the greatest number of five digits which is exactly divisible by 47.

Accepted Answer

### Concept & Logic To find the greatest $N$-digit number divisible by a given divisor, divide the largest mathematically possible $N$-digit number by the divisor. The remainder represents the excess amount. Subtract this remainder from the original number to find the greatest divisible number. ### Step-by-Step Solution - **Given:** We need the largest 5-digit number divisible by 47. - **Calculation:** The absolute largest 5-digit number is 99999. - Divide 99999 by 47 to find the remainder. - $99999 = 47 	imes 2127 + 30$. - The remainder is 30. This means 99999 is 30 units greater than a perfect multiple of 47. - Subtract the remainder from the base number: $99999 - 30 = 99969$. ### Exam Strategy & Shortcut Start with 99999 and perform standard long division. The moment you find the final remainder (30), immediately subtract it from 99999. Never attempt to add the difference (divisor - remainder) for "greatest $N$-digit" questions, as adding even 1 to 99999 will flip it into a 6-digit number. ### Common Pitfall A common mistake is adding $(47 - 30)$ to 99999, which gives 100016. While 100016 is perfectly divisible by 47, it violates the constraint of being a 5-digit number. ### Final Answer Therefore, the correct answer is **99969**.

Question 4

When a certain number is multiplied by 13, the product consists entirely of fives. Find the smallest such number.

Accepted Answer

### Concept & Strategy The problem implies an equation where $x 	imes 13 = 55555\dots$ To find the smallest integer $x$, we must perform long division of a sequence of 5s ($55555\dots$) by 13 until we reach a point where the remainder is exactly zero. ### Step-by-Step Solution - **Calculation / Deduction:** Set up a long division dividing a continuous string of 5s by 13. - Divide 55 by 13: Quotient is 4, Remainder is $55 - 52 = 3$. - Bring down the next 5 (making 35). Divide 35 by 13: Quotient is 2, Remainder is $35 - 26 = 9$. - Bring down the next 5 (making 95). Divide 95 by 13: Quotient is 7, Remainder is $95 - 91 = 4$. - Bring down the next 5 (making 45). Divide 45 by 13: Quotient is 3, Remainder is $45 - 39 = 6$. - Bring down the next 5 (making 65). Divide 65 by 13: Quotient is 5, Remainder is $65 - 65 = 0$. - Since the remainder is now 0, the division is complete. The quotient sequence forms our answer: 42735. ### Exam Strategy & Shortcut Perform standard division but watch the unit digits. Because the final division must leave a 0 remainder and the dividend brings down a 5, the last step of the division must involve a multiple of 13 that ends in 5. Since $13 	imes 5 = 65$, you know the final digit of your answer will be 5 once the remainder hits 6. ### Common Pitfall Miscalculating a single subtraction during the long division will completely alter the subsequent remainders, leading to a never-ending cycle or a completely wrong quotient. Double-check your basic multiplication table for 13. ### Final Answer Therefore, the correct answer is **42735**.

Question 5

When a certain number is multiplied by 18, the product consists entirely of 2's. What is the minimum number of 2's in the product?

Accepted Answer

### Concept & Logic The question asks for the smallest number of the form $2222\dots$ that is exactly divisible by 18. Since 18 is composed of co-prime factors 2 and 9 ($18 = 2 	imes 9$), the number must satisfy the divisibility rules for both 2 and 9. ### Step-by-Step Solution - **Calculation / Deduction:** A number consisting entirely of 2s is naturally even, so it is always divisible by 2. We only need to focus on the rule for 9. - For a number to be divisible by 9, the sum of its digits must be a multiple of 9. - Let $n$ be the number of 2s. The sum of the digits is $2 	imes n$, or $2n$. - We need the smallest integer $n$ such that $2n$ is a multiple of 9. - Testing multiples: 2, 4, 6, 8, 10, 12, 14, 16, 18... - The smallest multiple of 9 in this sequence is 18. - So, $2n = 18 \implies n = 9$. - The product requires exactly nine 2s (which is 222222222). ### Exam Strategy & Shortcut Do not perform long division of $22222\dots$ by 18! Use divisibility rules. Since 18 requires divisibility by 9, just ask yourself: "How many 2s do I need to add together to reach 9, 18, or 27?" Nine 2s equal 18, instantly giving you the answer. ### Common Pitfall Wasting precious exam time attempting to set up a long division of 22222... by 18. This brute-force method is prone to counting errors and takes significantly longer than applying the rule of 9. ### Final Answer Therefore, the correct answer is **9**.

Question 6

Find the smallest number which when multiplied by 9 gives the product as 1 followed by a certain number of 7s only.

Accepted Answer

### Concept & Logic We are looking for a multiplier that produces a number of the format $1777\dots7$, which is a multiple of 9. The divisibility rule for 9 dictates that the sum of the digits of a number must be a multiple of 9 for the entire number to be divisible by 9. ### Step-by-Step Solution - **Calculation / Deduction:** Let the number of 7s be $n$. The sum of the digits is $1 + 7n$. - Test values for $n$ to find the smallest sum that is a multiple of 9: - $n = 1$: Sum = $1 + 7 = 8$ - $n = 2$: Sum = $1 + 14 = 15$ - $n = 3$: Sum = $1 + 21 = 22$ - $n = 4$: Sum = $1 + 28 = 29$ - $n = 5$: Sum = $1 + 35 = 36$ - 36 is divisible by 9. Therefore, the smallest valid product requires five 7s, making the number 177777. - To find the multiplier (the original number), divide the product by 9: $$ 177777 \div 9 = 19753 $$ ### Exam Strategy & Shortcut Alternatively, use option elimination. Multiply the given options by 9 and check the last few digits. $19753 	imes 9$: $3 	imes 9 = 27$ (ends in 7). $5 	imes 9 + 2 = 47$ (ends in 7). This confirms the sequence of 7s without having to find the large target number first. ### Common Pitfall A frequent error is successfully identifying the target number (177777) but selecting it as the answer (if it were an option). The question asks for the *smallest number which when multiplied*, meaning you must find the multiplier by dividing by 9. ### Final Answer Therefore, the correct answer is **19753**.

Question 7

What is the unit's digit in the product? $81 	imes 82 	imes 83 	imes \dots 	imes 89$

Accepted Answer

### Concept & Logic The unit digit of a large multiplication expression depends entirely on the product of the individual unit digits. If any combination of those unit digits produces a zero, the unit digit for the entire massive product will definitively be zero. ### Step-by-Step Solution - **Calculation / Deduction:** The given sequence is $81 	imes 82 	imes 83 	imes 84 	imes 85 	imes \dots 	imes 89$. - Extract only the unit digits to evaluate: $1 	imes 2 	imes 3 	imes 4 	imes 5 	imes \dots 	imes 9$. - Notice that the sequence contains a 2 (from 82) and a 5 (from 85). - When multiplied, $2 	imes 5 = 10$, which ends in a 0. - Any integer multiplied by a number ending in 0 will result in a product that also ends in 0. ### Exam Strategy & Shortcut Rule of thumb: Whenever you are asked for the unit digit of a continuous sequence of 5 or more consecutive numbers, the answer is always **0**. This is because any sequence of 5 numbers guarantees the presence of at least one even number (ends in 2, 4, 6, 8) and one multiple of 5, which combine to create a trailing zero. ### Common Pitfall Students sometimes start manually multiplying the sequence ($1 	imes 2 = 2$, $2 	imes 3 = 6$, $6 	imes 4 = 24$, etc.) which wastes valuable time and introduces the possibility of arithmetic mistakes before reaching the 5. Always scan the series for a 2 and a 5 first. ### Final Answer Therefore, the correct answer is **0**.

Question 8

Find the unit's digit in the product $(2467)^{153} 	imes (341)^{72}$.

Accepted Answer

### Concept & Formula To find the unit digit of a number with a large exponent, use the cyclicity rule. Extract the unit digit of the base. - The cyclicity of 7 is 4 (it repeats the pattern 7, 9, 3, 1). - The cyclicity of 1 is 1 (it is always 1 regardless of the power). ### Step-by-Step Solution - **Given:** We need to find the unit digit of $7^{153} 	imes 1^{72}$. - **Calculation (Base 1):** For $1^{72}$, the unit digit is inherently 1 because 1 multiplied by itself any number of times remains 1. - **Calculation (Base 7):** For $7^{153}$, we divide the exponent 153 by the cyclicity of 4 to find the position in the cycle. - $153 \div 4$ yields a quotient of 38 and a remainder of 1. - A remainder of 1 corresponds to the 1st power in the cycle: $7^1 = 7$. - **Final Product:** Multiply the resulting unit digits from both parts: $7 	imes 1 = 7$. ### Exam Strategy & Shortcut To quickly find the remainder when dividing an exponent by 4, you only need to divide the last two digits of the exponent. For 153, just divide 53 by 4. $53 = 13 	imes 4 + 1$. The remainder is 1. This instantly tells you to evaluate $7^1$. ### Common Pitfall A common mistake is taking the remainder of the base (dividing 2467 by 4) instead of the exponent. The cyclicity rule dictates that the exponent is divided by the length of the cycle, while the base is stripped down to just its unit digit. ### Final Answer Therefore, the correct answer is **7**.

Question 9

Find the unit's digit in $(264)^{102} + (264)^{103}$.

Accepted Answer

### Concept & Logic The unit digit of a number raised to a power depends entirely on the unit digit of its base and the cyclicity of its powers. ### Step-by-Step Solution **Given:** The expression is $(264)^{102} + (264)^{103}$. **Calculation / Deduction:** * We only need to consider the unit digit of the base, which is $4$. Thus, we evaluate the unit digit of $4^{102} + 4^{103}$. * The powers of $4$ follow a simple even/odd pattern: $4^1 = 4$, $4^2 = 16$ (ends in $6$), $4^3 = 64$ (ends in $4$). * Rule: $4^{	ext{odd}}$ ends in $4$, and $4^{	ext{even}}$ ends in $6$. * Since $102$ is an even power, $4^{102}$ yields a unit digit of $6$. * Since $103$ is an odd power, $4^{103}$ yields a unit digit of $4$. * Adding these together: $6 + 4 = 10$. * The unit digit of $10$ is $0$. ### Exam Strategy & Shortcut Memorize the cyclicity of $4$ and $9$. Instead of dividing the exponent by $4$ (the standard cyclicity method), just look at whether the exponent is even or odd. You can instantly deduce the unit digits as $6$ and $4$, giving a sum ending in $0$. ### Common Pitfall Students often try to calculate standard cyclicity by dividing the large exponents by $4$ to find the remainder. While this works, it wastes valuable time compared to the direct even/odd shortcut for numbers ending in $4$ or $9$. ### Final Answer Therefore, the correct answer is 0.

Question 10

Find the total number of prime factors in the expression $(4)^{11} 	imes (7)^5 	imes (11)^2$.

Accepted Answer

### Concept & Logic Prime factorization requires converting all composite bases in an expression into their prime factors before summing the exponents. ### Step-by-Step Solution **Given:** Expression: $(4)^{11} 	imes (7)^5 	imes (11)^2$. **Calculation / Deduction:** * Identify the prime and composite bases. Here, $7$ and $11$ are already prime numbers, but $4$ is composite. * Convert $4$ to its prime base: $4 = 2^2$. * Substitute this back into the expression: $$ (2^2)^{11} 	imes (7)^5 	imes (11)^2 $$ * Apply the exponent rule $(a^m)^n = a^{m 	imes n}$: $$ 2^{22} 	imes 7^5 	imes 11^2 $$ * The total number of prime factors is simply the sum of these prime exponents: $22 + 5 + 2 = 29$. ### Exam Strategy & Shortcut To find total prime factors quickly, just multiply the exponent of any composite number by the count of its prime components. Since $4 = 2 	imes 2$ (two components), its exponent contribution is $11 	imes 2 = 22$. Then just add the rest: $22 + 5 + 2 = 29$. ### Common Pitfall The most common mistake is ignoring the composite base and directly adding the given exponents ($11 + 5 + 2 = 18$). Always verify that every base is a prime number before summing. ### Final Answer Therefore, the correct answer is 29.

Question 11

What is the number of zeros at the end of the product of the numbers from $1$ to $100$?

Accepted Answer

### Concept & Logic The number of trailing zeros in a factorial or continuous product is determined by the number of $(2 	imes 5)$ pairs. Because multiples of $2$ are far more frequent than multiples of $5$, the bottleneck is the highest power of $5$. ### Step-by-Step Solution **Given:** The product is $1 	imes 2 	imes 3 	imes \dots 	imes 100$, which is $100!$. **Calculation / Deduction:** * We need to find the highest power of $5$ in $100!$. * We use Legendre's formula (successive division) to count the multiples of $5$, $25$, $125$, etc. $$ \left\lfloor \frac{100}{5} ightfloor + \left\lfloor \frac{100}{5^2} ightfloor $$ * Calculate the terms: $$ \frac{100}{5} = 20 $$ $$ \frac{100}{25} = 4 $$ * Add the results: $20 + 4 = 24$. ### Exam Strategy & Shortcut Use the successive division method mentally. Divide $100$ by $5$ to get $20$. Then divide that quotient ($20$) by $5$ to get $4$. Since $4$ is less than $5$, stop. Add the quotients: $20 + 4 = 24$. ### Common Pitfall A frequent error is stopping after the first division (yielding $20$) and forgetting that multiples of $25$ (like $25, 50, 75, 100$) contain an extra $5$ that must be counted. Always continue dividing the quotient until it is less than $5$. ### Final Answer Therefore, the correct answer is 24.

Question 12

What is the number of zeros at the end of the product $5^5 	imes 10^{10} 	imes 15^{15} 	imes \dots 	imes 125^{125}$?

Accepted Answer

### Concept & Strategy Trailing zeros require pairs of $(2 	imes 5)$. In standard factorials, $2$ is abundant, so we count $5$s. However, in this specific product, every term is a multiple of $5$ raised to large powers, meaning $5$s are highly abundant. The bottleneck here is actually the highest power of $2$. ### Step-by-Step Solution **Given:** Product $N = 5^5 	imes 10^{10} 	imes \dots 	imes 125^{125}$. **Calculation / Deduction:** * We must find the highest power of $2$ in this product. Only terms that are even will contribute a factor of $2$. These are $10^{10}, 20^{20}, 30^{30}, \dots, 120^{120}$. * The power of $2$ contributed is the sum of their exponents, but we must also account for multiples of $4$ ($2^2$), $8$ ($2^3$), and $16$ ($2^4$). * **Multiples of 2:** $10 + 20 + 30 + \dots + 120 = 10(1 + 2 + \dots + 12) = 10 	imes \frac{12 	imes 13}{2} = 780$. * **Multiples of 4:** Provide an additional factor. These are $20^{20}, 40^{40}, \dots, 120^{120}$. Sum: $20 + 40 + \dots + 120 = 20(1 + 2 + \dots + 6) = 20 	imes \frac{6 	imes 7}{2} = 420$. * **Multiples of 8:** Provide a third factor. These are $40^{40}, 80^{80}, 120^{120}$. Sum: $40 + 80 + 120 = 240$. * **Multiples of 16:** Provide a fourth factor. The only term is $80^{80}$. Sum: $80$. * Total power of $2 = 780 + 420 + 240 + 80 = 1520$. ### Exam Strategy & Shortcut Whenever a sequence is heavily weighted with multiples of $5$, instantly switch your focus to finding the total power of $2$. Group the exponents using Arithmetic Progression (AP) sums to calculate the totals rapidly instead of adding them manually. ### Common Pitfall Assuming that the power of $5$ dictates the number of trailing zeros in *every* problem. If you try to find the power of $5$ here, you will get a massive number, leading you to an incorrect assumption and a wrong answer. Always identify the limiting prime factor first. ### Final Answer Therefore, the correct answer is 1520.

Question 13

On dividing $15968$ by a certain number, the quotient is $89$ and the remainder is $37$. Find the divisor.

Accepted Answer

### Concept & Formula The problem relies on the fundamental division algorithm, which relates the four components of division: $$ 	ext{Dividend} = (	ext{Divisor} 	imes 	ext{Quotient}) + 	ext{Remainder} $$ ### Step-by-Step Solution **Given:** * Dividend = $15968$ * Quotient = $89$ * Remainder = $37$ **Calculation / Deduction:** * We need to isolate the Divisor. Rearranging the division formula: $$ 	ext{Divisor} = \frac{	ext{Dividend} - 	ext{Remainder}}{	ext{Quotient}} $$ * Substitute the given values into the formula: $$ 	ext{Divisor} = \frac{15968 - 37}{89} $$ * Subtract the remainder from the dividend: $$ 	ext{Divisor} = \frac{15931}{89} $$ * Perform the division: $15931 \div 89 = 179$. ### Exam Strategy & Shortcut To avoid long division, use the unit digit method to identify the correct option. The numerator $15931$ ends in $1$. The denominator $89$ ends in $9$. You need a number that, when multiplied by $9$, gives a unit digit of $1$. Since $9 	imes 9 = 81$, the divisor must end in $9$. If only one option ends in $9$, pick it instantly. If multiple do, use quick approximation ($16000 \div 90 \approx 177$), pointing directly to $179$. ### Common Pitfall A common mistake is forgetting to subtract the remainder before dividing (i.e., doing $15968 \div 89$), which will result in a messy decimal and cause confusion during the exam. ### Final Answer Therefore, the correct answer is 179.

Question 14

A number when divided by $114$, leaves remainder $21$. If the same number is divided by $19$, find the remainder.

Accepted Answer

### Concept & Logic If a number divided by $D_1$ leaves a remainder $R_1$, and we subsequently divide the same number by $D_2$ (where $D_2$ is a direct factor of $D_1$), the new remainder is found simply by dividing $R_1$ by $D_2$. ### Step-by-Step Solution **Given:** * Initial divisor ($D_1$) = $114$ * Initial remainder ($R_1$) = $21$ * New divisor ($D_2$) = $19$ **Calculation / Deduction:** * First, verify that the new divisor is a factor of the initial divisor. $114 \div 19 = 6$. Since it divides perfectly, the shortcut property holds. * Let the original number be $N$. We can express it as: $$ N = 114k + 21 $$ * Rewrite $114$ as $19 	imes 6$: $$ N = 19(6k) + 19 + 2 $$ * Factor out $19$: $$ N = 19(6k + 1) + 2 $$ * This shows that when $N$ is divided by $19$, the quotient is $(6k + 1)$ and the remainder is $2$. ### Exam Strategy & Shortcut Always immediately check if the first divisor is a multiple of the second. $114$ is $19 	imes 6$. Because it is a perfect multiple, you can completely ignore the first divisor. Simply take the given remainder ($21$) and divide it by the new divisor ($19$). The remainder of $21 \div 19$ is exactly $2$. ### Common Pitfall A common mistake is assuming the problem cannot be solved because the original number is unknown, prompting students to arbitrarily guess values for the quotient ($k$). While setting $k=1$ works, it wastes time compared to the direct remainder division shortcut. **Therefore, the correct answer is 2.**

Question 15

A number being successively divided by $3$, $5$ and $8$ leaves remainders $1$, $4$ and $7$ respectively. Find the respective remainders if the order of divisors be reversed.

Accepted Answer

### Concept & Strategy Successive division means that the quotient of the first division becomes the dividend for the second division, and so forth. To find the original number, you must work backward from the final quotient (which we assume is $1$ to find the smallest valid number) using the formula: $	ext{Dividend} = (	ext{Divisor} 	imes 	ext{Quotient}) + 	ext{Remainder}$. ### Step-by-Step Solution **Given:** * Successive divisors: $3$, $5$, $8$ * Successive remainders: $1$, $4$, $7$ **Calculation / Deduction:** * Let the original number be $x$, and the subsequent quotients be $y$ and $z$. * **Step 1: Reconstruct the number from the bottom up.** * Assume the final quotient after dividing by $8$ is $1$. * Find $z$ (dividend for the 3rd division): $$ z = (8 	imes 1) + 7 = 15 $$ * Find $y$ (dividend for the 2nd division): $$ y = (5 	imes z) + 4 = (5 	imes 15) + 4 = 79 $$ * Find $x$ (the original number): $$ x = (3 	imes y) + 1 = (3 	imes 79) + 1 = 238 $$ * **Step 2: Reverse the divisors ($8$, $5$, $3$) and perform successive division on $238$.** * $238 \div 8$: Quotient is $29$, Remainder is $6$. * $29 \div 5$: Quotient is $5$, Remainder is $4$. * $5 \div 3$: Quotient is $1$, Remainder is $2$. * The respective remainders are $6$, $4$, and $2$. ### Exam Strategy & Shortcut Use a quick visual chain calculation. Start from $1$, multiply by the last divisor, add the last remainder. Repeat moving upwards. $((1 	imes 8 + 7) 	imes 5 + 4) 	imes 3 + 1 = 238$. Once you have $238$, rapidly divide it downwards by the reversed sequence. ### Common Pitfall The most common error is confusing "successive division" with standard division. Students frequently divide the original number directly by $8$, then directly by $5$, then directly by $3$, instead of dividing the *resulting quotients*. **Therefore, the correct answer is 6, 4, 2.**

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