Question 1

Find the square root of $4a^2 + b^2 + c^2 + 4ab - 2bc - 4ac$

Accepted Answer

### Concept & Formula This question requires factoring a quadratic polynomial with three variables. The underlying principle is the algebraic identity for the square of a trinomial: $$ (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx $$ ### Step-by-Step Solution * The given expression under the root is $4a^2 + b^2 + c^2 + 4ab - 2bc - 4ac$. * Rewrite the squared terms to identify our base variables: $(2a)^2 + (b)^2 + (c)^2$. * Next, observe the signs of the cross-product terms: $+4ab$, $-2bc$, $-4ac$. * Because the terms involving $c$ are negative, but the term with just $a$ and $b$ is positive, the $c$ term must carry the negative sign in our factored form. * Let $x = 2a$, $y = b$, and $z = -c$. * Verify the cross terms against the formula: $$ 2xy = 2(2a)(b) = 4ab $$ $$ 2yz = 2(b)(-c) = -2bc $$ $$ 2zx = 2(2a)(-c) = -4ac $$ * The expression factors perfectly into the square: $(2a + b - c)^2$. * Taking the square root removes the square exponent, leaving $(2a + b - c)$. ### Exam Strategy & Shortcut Use the "Sign Check" method to bypass the rigorous algebra. Look exclusively at the mixed terms: $+4ab$, $-2bc$, $-4ac$. The negative signs are attached only to pairs containing the variable $c$, while the $ab$ pair is strictly positive. This logically dictates that $a$ and $b$ share the same sign, and $c$ has the opposite sign. Only the option $2a + b - c$ satisfies this exact rule. ### Common Pitfall The most common error is ignoring the negative signs entirely and carelessly assuming the answer is $(2a + b + c)$ because all the square terms are positive. Always trace the negative terms in the expanded polynomial back to identify which base variable carries the minus sign. ### Final Answer Therefore, the correct answer is **$2a + b - c$**.

Question 2

$A$ is counting the numbers from $1$ to $31$ and $B$ from $31$ to $1$. $A$ is counting the odd numbers only. The speed of both is the same. What will be the number which will be pronounced by $A$ and $B$ together?

Accepted Answer

### Concept & Logic This problem can be solved by modeling the spoken numbers as two simultaneous Arithmetic Progressions (AP). Since their counting speed is the same, we need to find the $n$-th term where both sequences produce the exact same value. ### Step-by-Step Solution * $A$ counts odd numbers starting from 1: $1, 3, 5, 7, \dots$ This is an AP with first term $a = 1$ and common difference $d = 2$. The $n$-th term for $A$ is $1 + (n - 1)2 = 2n - 1$. * $B$ counts backwards from 31 normally: $31, 30, 29, 28, \dots$ This is an AP with first term $a = 31$ and common difference $d = -1$. The $n$-th term for $B$ is $31 + (n - 1)(-1) = 32 - n$. * Since they pronounce the same number at the same time, we equate the two $n$-th terms: $$ 2n - 1 = 32 - n $$ * Solve for $n$: $$ 3n = 33 $$ $$ n = 11 $$ * Substitute $n = 11$ back into either formula to find the pronounced number. Using $B$'s formula: $$ 32 - 11 = 21 $$ ### Exam Strategy & Shortcut Use relative speed logic. $A$ moves $+2$ per step, and $B$ moves $-1$ per step. Their relative gap closes by $3$ units per step. The initial gap between their starting numbers ($31$ and $1$) is $30$. It takes $30 / 3 = 10$ steps *after* the first pronunciation to meet. So they meet on the $11$-th pronunciation. For $B$, stepping down 10 times from 31 lands directly on 21. ### Common Pitfall A common mistake is drawing out the entire table manually, which consumes excessive time and invites simple counting errors under exam pressure. Rely on the mathematical progression to guarantee accuracy. ### Final Answer Therefore, the correct answer is **21**.

Question 3

Simplify $ \frac{789 	imes 789 	imes 789 + 211 	imes 211 	imes 211}{789 	imes 789 - 789 	imes 211 + 211 	imes 211} $

Accepted Answer

### Concept & Formula This large fraction is a classic substitution problem based on the algebraic identity for the sum of two cubes. By mapping the numbers to variables, the complex expression perfectly simplifies: $$ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $$ ### Step-by-Step Solution * Let $a = 789$ and $b = 211$. * The numerator consists of $(789)^3 + (211)^3$, which is $a^3 + b^3$. * The denominator consists of $(789)^2 - (789 	imes 211) + (211)^2$, which is $a^2 - ab + b^2$. * Substitute the variables into the fraction: $$ \frac{a^3 + b^3}{a^2 - ab + b^2} $$ * Expand the numerator using the sum of cubes identity: $$ \frac{(a + b)(a^2 - ab + b^2)}{a^2 - ab + b^2} $$ * The quadratic term $(a^2 - ab + b^2)$ cancels out completely from the top and bottom. * We are left with simply $(a + b)$. * Substitute the original numbers back: $$ 789 + 211 = 1000 $$ ### Exam Strategy & Shortcut Whenever you see a numerator with a sum of cubes ($x 	imes x 	imes x + y 	imes y 	imes y$) and a denominator with the corresponding quadratic, bypass the algebra entirely. The answer will always just be the sum of the two base numbers: $x + y$. In this case, $789 + 211 = 1000$. This should take you 3 seconds to solve. ### Common Pitfall A massive trap is attempting to multiply these 3-digit numbers manually. Examiners specifically design these questions to penalize students who rely on brute-force calculation rather than algebraic pattern recognition. ### Final Answer Therefore, the correct answer is **1000**.

Question 4

Simplify $ \frac{658 	imes 658 	imes 658 - 328 	imes 328 	imes 328}{658 	imes 658 + 658 	imes 328 + 328 	imes 328} $

Accepted Answer

### Concept & Formula This expression is structured around the algebraic identity for the difference of two cubes. By replacing the large numbers with variables, the expression collapses into a simple subtraction: $$ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $$ ### Step-by-Step Solution * Let $a = 658$ and $b = 328$. * The numerator is the difference of cubes: $(658)^3 - (328)^3 ightarrow a^3 - b^3$. * The denominator is a quadratic sum: $(658)^2 + (658 	imes 328) + (328)^2 ightarrow a^2 + ab + b^2$. * Rewrite the full expression algebraically: $$ \frac{a^3 - b^3}{a^2 + ab + b^2} $$ * Expand the numerator using the identity formula: $$ \frac{(a - b)(a^2 + ab + b^2)}{a^2 + ab + b^2} $$ * Cancel out the common quadratic factor $(a^2 + ab + b^2)$ from the numerator and denominator. * This leaves exactly: $(a - b)$. * Substitute the original numbers back in: $$ 658 - 328 = 330 $$ ### Exam Strategy & Shortcut Recognize the pattern instantly: if the numerator has a minus sign connecting the cubed terms ($x^3 - y^3$), the entire expression will always simplify to exactly $(x - y)$. You can skip all intermediate steps and just subtract the base numbers: $658 - 328 = 330$. ### Common Pitfall A frequent error is confusing the sum and difference formulas, accidentally adding the two numbers ($658 + 328 = 986$) instead of subtracting them. Always check the sign linking the cubic terms in the numerator to determine your final operation. ### Final Answer Therefore, the correct answer is **330**.

Question 5

Simplify $ \frac{(893 + 786)^2 - (893 - 786)^2}{893 	imes 786} $

Accepted Answer

### Concept & Formula The expression is a direct application of the algebraic identity that relates the square of a sum and the square of a difference. By factoring it algebraically, the large numerical values become irrelevant: $$ (a + b)^2 - (a - b)^2 = 4ab $$ ### Step-by-Step Solution * Let $a = 893$ and $b = 786$. * Substitute these variables into the given mathematical expression: $$ \frac{(a + b)^2 - (a - b)^2}{ab} $$ * Expand the numerator using standard binomial identities: $$ (a^2 + b^2 + 2ab) - (a^2 + b^2 - 2ab) $$ * Distribute the negative sign and simplify the numerator: $$ a^2 + b^2 + 2ab - a^2 - b^2 + 2ab = 4ab $$ * Substitute the simplified numerator back into the fraction: $$ \frac{4ab}{ab} $$ * Cancel the common $ab$ terms from the numerator and denominator to get the final result: $$ 4 $$ ### Exam Strategy & Shortcut Recognize the $\frac{(a+b)^2 - (a-b)^2}{ab}$ pattern immediately. It will ALWAYS evaluate to $4$ because the variables completely cancel out. You can solve this question in one second without even reading the specific numbers provided in the numerator and denominator. ### Common Pitfall A common mistake is attempting to perform the actual arithmetic—adding and subtracting the numbers inside the parentheses, squaring the massive results, and then dividing. This wastes immense time and introduces a very high risk of basic calculation errors. ### Final Answer Therefore, the correct answer is **4**.

Question 6

Which of the following is NOT a prime number?

Accepted Answer

### Concept & Logic To verify if a number $N$ is prime, you do not need to test every number up to $N$. You only need to find the approximate square root of $N$, denoted as $\sqrt{N}$. Test for divisibility only by the prime numbers less than or equal to this square root. If it is not divisible by any of them, it is prime. ### Step-by-Step Solution * **Test Option A (241):** The square root is slightly less than 16 (since $16^2 = 256$). Prime numbers below 16 are 2, 3, 5, 7, 11, 13. 241 is not divisible by any of these. Thus, 241 is prime. * **Test Option B (337):** The square root is less than 19 (since $19^2 = 361$). Primes below 19 are 2, 3, 5, 7, 11, 13, 17. 337 is not divisible by any of these. Thus, 337 is prime. * **Test Option C (391):** The square root is less than 20 (since $20^2 = 400$). Primes below 20 are 2, 3, 5, 7, 11, 13, 17, 19. * Testing divisibility for 391: It fails for 2, 3, 5, 7, 11, and 13. * Let's test 17: $391 \div 17 = 23$. * Since 391 is evenly divisible by 17, it has factors other than 1 and itself. Therefore, 391 is NOT a prime number. ### Exam Strategy & Shortcut Use divisibility rules for 2, 3, 5, and 11 first to quickly eliminate obvious numbers. When testing larger primes like 13, 17, or 19, look at the unit digit. For example, to get a unit digit of 1 in 391 using the prime 17, you need to multiply 7 by a number ending in 3 (like 13 or 23). $17 	imes 20 = 340$, and $17 	imes 3 = 51$. $340 + 51 = 391$. ### Common Pitfall A massive time-waster is trying to divide a large number by every single odd number sequentially. Always stop checking once your prime divisor exceeds the approximate square root of the target number. ### Final Answer Therefore, the correct answer is **391**.

Question 7

If $\Delta$ stands for the operation 'adding first number to twice the second number', then find the value of $(1 \Delta 2) \Delta 3$.

Accepted Answer

### Concept & Logic The problem introduces a custom mathematical operator $\Delta$. The rule defined is $a \Delta b = a + 2b$. We must apply this rule following the standard order of operations, resolving parentheses first. ### Step-by-Step Solution **Given:** The operation is defined as adding the first number to twice the second number. Expression to evaluate: $(1 \Delta 2) \Delta 3$ **Calculation:** Step 1: Evaluate the inner bracket $(1 \Delta 2)$. Here, $a = 1$ and $b = 2$. $1 \Delta 2 = 1 + 2(2) = 1 + 4 = 5$ Step 2: Substitute this result back into the original expression. The expression becomes $5 \Delta 3$. Here, $a = 5$ and $b = 3$. $5 \Delta 3 = 5 + 2(3) = 5 + 6 = 11$ ### Exam Strategy & Shortcut Custom operator questions are direct substitution problems. The fastest approach is purely mental math. Quickly visualize $1 + 2(2) = 5$, then $5 + 2(3) = 11$. Do not write down the intermediate equations unless absolutely necessary, as it wastes time. ### Common Pitfall The most common mistake is ignoring the parentheses and processing the expression linearly from left to right without fully evaluating the bracket, or confusing the order by adding twice the first number to the second instead. Always resolve brackets first. ### Final Answer Therefore, the correct answer is **11**.

Question 8

Given that $1^2 + 2^2 + 3^2 + ..... + 10^2 = 385$, then find the value of $2^2 + 4^2 + 6^2 + ...... + 20^2$.

Accepted Answer

### Concept & Formula The problem requires finding the sum of a modified series based on a given standard series of squares. The key insight is factoring out a common multiplier from the target series. $$ (ab)^2 = a^2 	imes b^2 $$ ### Step-by-Step Solution **Given:** $1^2 + 2^2 + 3^2 + ... + 10^2 = 385$ **Calculation:** We need to find the value of: $2^2 + 4^2 + 6^2 + ... + 20^2$ Rewrite each term by factoring out $2$ from the base: $= (2 	imes 1)^2 + (2 	imes 2)^2 + (2 	imes 3)^2 + ... + (2 	imes 10)^2$ Apply the exponent rule $(ab)^2 = a^2b^2$: $= (2^2 	imes 1^2) + (2^2 	imes 2^2) + (2^2 	imes 3^2) + ... + (2^2 	imes 10^2)$ Factor out the common term $2^2$ (which is $4$): $= 4 	imes (1^2 + 2^2 + 3^2 + ... + 10^2)$ Substitute the given value of the original series ($385$): $= 4 	imes 385 = 1540$ ### Exam Strategy & Shortcut When you see a series where each base is a direct multiple of the base in a known series of squares, simply square that multiple and multiply it by the given sum. Here, the bases are doubled ($2x$), so the total sum increases by a factor of $2^2 = 4$. Mentally calculate $4 	imes 385 = 1540$. ### Common Pitfall A frequent mistake is factoring out $2$ instead of $2^2$, leading students to calculate $2 	imes 385 = 770$. Remember that when extracting a factor from a squared term, the extracted factor must also be squared. ### Final Answer Therefore, the correct answer is **1540**.

Question 9

Which of the following numbers is divisible by $3$?

Accepted Answer

### Concept & Logic The problem tests the standard divisibility rule for $3$. A number is perfectly divisible by $3$ if and only if the sum of all its individual digits is a multiple of $3$. ### Step-by-Step Solution **Calculation:** We evaluate the given options by summing their digits. Test Option A: $541326$ Sum of digits $= 5 + 4 + 1 + 3 + 2 + 6 = 21$ Since $21$ is divisible by $3$ ($21 / 3 = 7$), the number $541326$ is divisible by $3$. Test Option B: $5967013$ Sum of digits $= 5 + 9 + 6 + 7 + 0 + 1 + 3 = 31$ Since $31$ is not divisible by $3$, the number is not divisible by $3$. (We can stop testing options as we have found the correct one). ### Exam Strategy & Shortcut Instead of adding all digits, use the ''digit cancellation'' method. Quickly scan the number and cross out any digits or pairs of digits that intuitively add up to $3, 6,$ or $9$. For $541326$: cross out $3$, cross out $6$, cross out $5+4$ ($9$), and cross out $1+2$ ($3$). Since all digits are crossed out with no remainder, it is divisible by $3$. This takes two seconds per option. ### Common Pitfall Attempting to actually divide the entire large number by $3$ using long division. This wastes precious exam time and increases the likelihood of a calculation error. ### Final Answer Therefore, the correct answer is **541326**.

Question 10

What least value must be assigned to $*$ so that the number $197*5462$ is divisible by $9$?

Accepted Answer

### Concept & Logic This problem relies on the standard divisibility rule for $9$. A number is divisible by $9$ if the total sum of its digits is a multiple of $9$. ### Step-by-Step Solution **Given:** The number is $197*5462$. Let the unknown digit $*$ be $x$. **Calculation:** Step 1: Calculate the sum of the known digits. Sum $= 1 + 9 + 7 + x + 5 + 4 + 6 + 2$ Sum $= 34 + x$ Step 2: Determine the least value of $x$. For $(34 + x)$ to be divisible by $9$, it must equal the next multiple of $9$ that is greater than or equal to $34$. The multiples of $9$ are $9, 18, 27, 36, 45...$ The immediate next multiple is $36$. So, $34 + x = 36$ $x = 36 - 34 = 2$ ### Exam Strategy & Shortcut Use the ''casting out nines'' technique to save time and reduce mental fatigue. Ignore any $9$s or combinations of digits that sum to $9$. For $197*5462$: - Ignore $9$ - Ignore $7 + 2 = 9$ - Ignore $5 + 4 = 9$ You are left with $1 + 6 + x = 7 + x$. For $7 + x$ to be divisible by $9$, $x$ must be $2$. ### Common Pitfall Students often make simple arithmetic errors when adding a long string of numbers sequentially (like getting $33$ instead of $34$). Using the cancellation method completely prevents this pitfall by keeping the numbers small. ### Final Answer Therefore, the correct answer is **2**.

Question 11

Which of the following numbers is divisible by $4$?

Accepted Answer

### Concept & Logic The divisibility rule for $4$ states that a number is divisible by $4$ if the number formed by its last two digits is divisible by $4$. ### Step-by-Step Solution **Given:** The options provided are large numerical values to test for divisibility. **Calculation:** Test option A ($67920594$): The last two digits form $94$. Since $94$ is not perfectly divisible by $4$ ($94 / 4 = 23.5$), the entire number is not divisible by $4$. Test option B ($618703572$): The last two digits form $72$. Since $72$ is perfectly divisible by $4$ ($72 / 4 = 18$), the entire number is divisible by $4$. ### Exam Strategy & Shortcut Do not waste time looking at the entire string of numbers. Instantly focus your eyes only on the tens and units place. If checking $72$ mentally is slow, halve it twice: half of $72$ is $36$, and half of $36$ is $18$. Since $18$ is a whole number, it is divisible by $4$. ### Common Pitfall A common mistake is applying the divisibility rule for $3$ (summing the digits) or attempting to divide the entire multi-digit number by $4$. Always use the specific rule for the specific divisor. ### Final Answer Therefore, the correct answer is **618703572**.

Question 12

Which digits should come in place of $*$ and $\$$ if the number $62684*\$$ is divisible by both $8$ and $5$?

Accepted Answer

### Concept & Logic This problem requires applying the divisibility rules for both $5$ and $8$. A number is divisible by $5$ if its last digit is $0$ or $5$. A number is divisible by $8$ if the number formed by its last three digits is divisible by $8$. ### Step-by-Step Solution **Given:** The number is $62684*\$$. It must be divisible by $5$ and $8$. **Calculation:** Step 1: Determine $\$$ using the rule for $5$. The last digit $\$$ must be $0$ or $5$. However, an even divisor like $8$ can only divide even numbers. A number ending in $5$ is odd. Thus, $\$$ must be $0$. The number is now $62684*0$. Step 2: Determine $*$ using the rule for $8$. The last three digits form the number $4*0$. We need $4*0$ to be divisible by $8$. Substitute $*$ with digits from the options: If $*$ is $4$, the number is $440$. $440 / 8 = 55$. This is perfectly divisible. ### Exam Strategy & Shortcut Immediately deduce that the final digit of any number divisible by an even number must be even. Therefore, $\$$ cannot be $5$, it must be $0$. This allows you to instantly eliminate options B and D. Then, plug in the remaining possible values for $*$ into the last three digits ($4*0$) to find the match. ### Common Pitfall Students often try to test for the divisibility of $8$ before securing the unit digit. Fixing the most restrictive condition first (the unit digit) drastically reduces the number of possibilities you have to check. ### Final Answer Therefore, the correct answer is **4 and 0**.

Question 13

Is the number $4832718$ divisible by $11$?

Accepted Answer

### Concept & Logic The divisibility rule for $11$ states that a number is divisible by $11$ if the difference between the sum of its digits at odd positions and the sum of its digits at even positions is either $0$ or a multiple of $11$. ### Step-by-Step Solution **Given:** The number to evaluate is $4832718$. **Calculation:** Step 1: Calculate the sum of digits at odd places (starting from the rightmost digit). Odd places (1st, 3rd, 5th, 7th): $8 + 7 + 3 + 4 = 22$ Step 2: Calculate the sum of digits at even places. Even places (2nd, 4th, 6th): $1 + 2 + 8 = 11$ Step 3: Find the difference between these two sums. $$ 22 - 11 = 11 $$ Since the difference is $11$, which is a multiple of $11$, the entire number is divisible by $11$. ### Exam Strategy & Shortcut Instead of summing them separately, you can pair adjacent digits and find their differences, then sum those differences. For $4832718$: $(4-8) + (3-2) + (7-1) + 8 = -4 + 1 + 6 + 8 = 11$. Since $11$ is divisible by $11$, the number is divisible. This method keeps the numbers much smaller during mental calculation. ### Common Pitfall A frequent error is miscounting the positions (even/odd) or simply adding all the digits together, confusing the rule of $11$ with the rule of $3$ or $9$. ### Final Answer Therefore, the correct answer is **Yes, it is divisible by 11**.

Question 14

Is the number $52563744$ divisible by $24$?

Accepted Answer

### Concept & Logic To check the divisibility of a large number by a composite number (like $24$), break the divisor into co-prime factors. Two numbers are co-prime if their highest common factor (HCF) is $1$. $$ 24 = 3 	imes 8 $$ ### Step-by-Step Solution **Given:** The number is $52563744$. We need to verify its divisibility by $24$. **Calculation:** Step 1: Verify divisibility by $3$. Sum of digits = $5 + 2 + 5 + 6 + 3 + 7 + 4 + 4 = 36$. Since $36$ is divisible by $3$, the number is divisible by $3$. Step 2: Verify divisibility by $8$. Check the last three digits: $744$. $744 / 8 = 93$. Since $744$ is divisible by $8$, the entire number is divisible by $8$. Step 3: Conclusion. Since $52563744$ is divisible by both $3$ and $8$ (which are co-prime), it is divisible by their product, $24$. ### Exam Strategy & Shortcut Always select co-prime pairs for this rule. Once established, use the digital sum rule for $3$ (crossing out $3$s, $6$s, and $9$s to speed it up) and the last-three-digits rule for $8$. ### Common Pitfall The most dangerous trap is splitting $24$ into factors that are not co-prime, such as $4$ and $6$. A number can be divisible by both $4$ and $6$ but not divisible by $24$ (for example, $12$). Factors must have an HCF of $1$. ### Final Answer Therefore, the correct answer is **Yes, because it is divisible by both 3 and 8**.

Question 15

What are the values of $M$ and $N$ respectively if $M39048458N$ is divisible by both 8 and 11, where $M$ and $N$ are single-digit integers?

Accepted Answer

### Concept & Formula To determine if a large number is divisible by composite combinations, use the co-prime factors divisibility rules. For a number to be divisible by 88 (since 8 and 11 are co-prime), it must satisfy both rules: - Divisibility by 8: The number formed by the last 3 digits must be divisible by 8. - Divisibility by 11: The difference between the sum of digits at even places and odd places must be 0 or a multiple of 11. ### Step-by-Step Solution - **Given:** The number $M39048458N$ is divisible by both 8 and 11. - **Calculation (Divisibility by 8):** The last three digits form the number $58N$. For $58N$ to be divisible by 8, $N$ must be 4 because 584 is exactly divisible by 8 ($584 / 8 = 73$). - **Calculation (Divisibility by 11):** Sum of digits at even places: $8 + 4 + 4 + 9 + M = 25 + M$ Sum of digits at odd places: $4 + 5 + 8 + 0 + 3 = 20$ Difference: $(25 + M) - 20 = M + 5$ For this difference to be divisible by 11 (and since $M$ is a single-digit integer), $M + 5$ must equal 11. $$ M + 5 = 11 \implies M = 6 $$ ### Exam Strategy & Shortcut Always check the rule that affects the fewest digits first! The divisibility rule for 8 only looks at the last 3 digits ($58N$). This immediately gives $N = 4$. If you glance at the options, any choice where the second value is not 4 can be eliminated instantly, saving you from calculating the 11 divisibility rule altogether if only one option remains. ### Common Pitfall Students frequently confuse "even/odd digits" with "even/odd places". The divisibility rule for 11 requires summing the digits based on their position index (1st, 2nd, 3rd, etc.), not whether the numeric value of the digit itself is even or odd. ### Final Answer Therefore, the correct answer is 6, 4.

Question 16

Find the number of digits in the smallest number which is made up of digits 1 and 0 only and is divisible by 225.

Accepted Answer

### Concept & Logic When a number is divided by a composite number like 225, break it down into its co-prime factors. $$ 225 = 9 	imes 25 $$ A number is divisible by 225 if it is independently divisible by both 9 and 25. - Divisibility by 25: The last two digits must be 00, 25, 50, or 75. - Divisibility by 9: The sum of all its digits must be a multiple of 9. ### Step-by-Step Solution - **Given:** The smallest number made strictly of the digits 1 and 0, which is divisible by 225. - **Calculation (Divisibility by 25):** Since the number can only contain 1s and 0s, the last two digits must be 00 to satisfy the divisibility rule for 25. - **Calculation (Divisibility by 9):** The sum of the digits must be a multiple of 9. Since the available digits are only 1 and 0, the smallest way to get a digit sum of 9 is to use exactly nine 1s. - **Deduction:** Combining these two conditions, the smallest number consists of nine 1s followed by two 0s to satisfy both rules simultaneously. This number is 11111111100. - Total digits = 9 (from the 1s) + 2 (from the 0s) = 11 digits. ### Exam Strategy & Shortcut Treat the conditions modularly. The "00" at the end handles the 25 divisibility and contributes zero to the digit sum. Thus, you only need to count how many 1s are needed to hit a sum of 9. Nine 1s plus two 0s equals 11 digits. No complex division is required. ### Common Pitfall A major mistake is finding the correct number (11111111100) but then forgetting to answer the specific question asked, which is the *number of digits*, not the number itself. Always re-read the final sentence of the prompt before selecting an option. ### Final Answer Therefore, the correct answer is 11.

Question 17

If the number $3422213pq$ is divisible by 99, find the missing digits $p$ and $q$.

Accepted Answer

### Concept & Formula To check divisibility by 99, we use its co-prime factors, 9 and 11. $$ 99 = 9 	imes 11 $$ The number must satisfy both divisibility rules: 1. Sum of digits must be a multiple of 9. 2. The difference between the sum of alternating digits must be 0 or a multiple of 11. ### Step-by-Step Solution - **Given:** The number $3422213pq$ is divisible by 99. - **Calculation (Divisibility by 9):** Sum of digits: $3 + 4 + 2 + 2 + 2 + 1 + 3 + p + q = 17 + p + q$. For this to be a multiple of 9, $17 + p + q$ can be 18 or 27. This simplifies to $p + q = 1$ (Eq. i) or $p + q = 10$ (Eq. ii). - **Calculation (Divisibility by 11):** Sum of odd places: $q + 3 + 2 + 2 + 3 = q + 10$ Sum of even places: $p + 1 + 2 + 4 = p + 7$ Difference: $(q + 10) - (p + 7) = q - p + 3$ For this to be 0 or 11: $q - p = -3 \implies p - q = 3$ (Eq. iii) OR $q - p + 3 = 11 \implies q - p = 8$ (Eq. iv) - **Deduction:** If $p + q = 1$, it does not yield single-digit integer solutions with (iii) or (iv). Solving $p + q = 10$ and $q - p = 8$: Adding both equations gives $2q = 18 \implies q = 9$. Substituting $q = 9$ into $p + q = 10$ gives $p = 1$. ### Exam Strategy & Shortcut Use option elimination! Instead of solving simultaneous equations, test the given options using the rule of 9. The sum of known digits is 17. Testing $p=1, q=9$: Sum = $17 + 1 + 9 = 27$ (Divisible by 9). Check the alternating difference for 11: $(9+10) - (1+7) = 19 - 8 = 11$ (Divisible by 11). Both rules pass instantly. ### Common Pitfall Setting up the alternating difference equation backwards or miscalculating basic addition when combining variables and constants. Always strictly group the alternating places before finding the difference to avoid sign errors. ### Final Answer Therefore, the correct answer is 1, 9.

Question 18

$x$ is a positive integer such that $x^2 + 12$ is exactly divisible by $x$. Find all the possible values of $x$.

Accepted Answer

### Concept & Logic When a variable divides a polynomial, split the polynomial into individual fractions. For the final expression to be an integer, every resulting fraction must simplify to an integer. ### Step-by-Step Solution - **Calculation / Deduction:** Split the given algebraic expression into two separate fractions: $$ \frac{x^2 + 12}{x} = \frac{x^2}{x} + \frac{12}{x} = x + \frac{12}{x} $$ - Since $x$ is a positive integer, the first term $x$ is always an integer. - For the entire expression to be an integer, the second term $\frac{12}{x}$ must also result in a whole number. - This means $x$ must be a factor of 12. - The positive factors of 12 are 1, 2, 3, 4, 6, and 12. ### Exam Strategy & Shortcut Whenever you see a pattern like $\frac{x^2 + k}{x}$, instantly ignore the $x^2$ term. The possible integer values for $x$ will always simply be the positive factors of the constant $k$. ### Common Pitfall Students often forget that 1 and the number itself (12) are valid factors. Always list factor pairs systematically ($1 	imes 12$, $2 	imes 6$, $3 	imes 4$) to avoid missing the boundary values. ### Final Answer Therefore, the correct answer is 1, 2, 3, 4, 6, 12.

Question 19

Find the smallest number to be added to 1000 so that 45 divides the sum exactly.

Accepted Answer

### Concept & Formula When finding the smallest number to add to a dividend so it is exactly divisible by a divisor, the required number is the difference between the divisor and the remainder. $$ 	ext{Number to add} = 	ext{Divisor} - 	ext{Remainder} $$ ### Step-by-Step Solution - **Given:** Dividend = 1000, Divisor = 45. - **Calculation / Deduction:** Divide 1000 by 45 to find the remainder. - $1000 \div 45$ gives a quotient of 22 and a remainder of 10. - The remainder 10 represents the excess above the previous multiple of 45. - To reach the next multiple of 45, we must bridge the gap between the remainder and the divisor. - Required number = $45 - 10 = 35$. ### Exam Strategy & Shortcut Execute the division quickly focusing only on the remainder. Once you see a remainder of 10, mentally subtract it from 45. There's no need to calculate the exact quotient or the final divisible number (1035) unless explicitly asked. ### Common Pitfall The most common error is selecting the remainder (10) as the answer. Subtracting 10 makes the number divisible, but the question asks what must be *added*. Always apply the divisor minus remainder logic for addition questions. ### Final Answer Therefore, the correct answer is 35.

Question 20

What least number must be subtracted from 2000 to get a number exactly divisible by 17?

Accepted Answer

### Concept & Logic When asked for the least number to subtract from a given value to make it exactly divisible, the answer is simply the remainder obtained when dividing the given value by the divisor. ### Step-by-Step Solution - **Given:** Dividend = 2000, Divisor = 17. - **Calculation / Deduction:** Perform standard division of 2000 by 17. - $2000 = 17 	imes 117 + 11$. - The division yields a remainder of 11. - This remainder is the exact extra amount that prevents 2000 from being a perfect multiple. Removing it gives a perfectly divisible number. ### Exam Strategy & Shortcut For subtraction-based divisibility questions, your calculation stops the moment you find the remainder. The remainder itself is your final answer. Do no further arithmetic. ### Common Pitfall Students sometimes confuse the rules for adding and subtracting. Subtracting the remainder from the divisor ($17 - 11 = 6$) is the rule for finding what to *add*. For subtraction, just use the direct remainder. ### Final Answer Therefore, the correct answer is 11.

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