Question 1

$1904 	imes 1904 = x$

Accepted Answer

### Concept & Formula When multiplying a number by itself (finding the square), and the number is close to a round base like 1900, we can use the algebraic expansion formula for the square of a binomial sum. $$(a + b)^2 = a^2 + 2ab + b^2$$ ### Step-by-Step Solution * Express 1904 as the sum of a base number and a smaller number: $$1904 = 1900 + 4$$ * Apply the binomial square formula: $$(1900 + 4)^2 = (1900)^2 + 2(1900)(4) + (4)^2$$ * Calculate each term individually: $$(1900)^2 = 3610000$$ $$2 	imes 1900 	imes 4 = 15200$$ $$(4)^2 = 16$$ * Sum the terms together: $$3610000 + 15200 + 16 = 3625216$$ ### Exam Strategy & Shortcut Instead of full calculation, use the last two digits approach. The last two digits of the square of $1900 + x$ will be determined by the last two digits of $x^2$. Here, $x = 4$, and $4^2 = 16$. The answer must end in 16. This eliminates options (b) and (d). Next, estimate the size. $1900^2 = 3610000$. The addition of the middle term $2ab$ ($15200$) pushes the answer into the $3625000$ range, easily confirming option (c) over the inflated option (a). ### Common Pitfall Resorting to traditional vertical multiplication for large four-digit numbers takes too long and frequently leads to carrying or addition errors across the multiple rows of zeroes. ### Final Answer **Therefore, the correct answer is 3625216.**

Question 2

$1397 	imes 1397 = x$

Accepted Answer

### Concept & Formula When finding the square of a number close to a large base (like 1400), we utilize the algebraic identity for the square of a binomial difference to drastically reduce calculation time. $$(a - b)^2 = a^2 - 2ab + b^2$$ ### Step-by-Step Solution * Rewrite 1397 as a subtraction from a round base: $$1397 = 1400 - 3$$ * Substitute this into our squaring equation: $$(1400 - 3)^2 = (1400)^2 - 2(1400)(3) + (3)^2$$ * Evaluate each component carefully: $$(1400)^2 = 1960000$$ $$2 	imes 1400 	imes 3 = 8400$$ $$(3)^2 = 9$$ * Perform the final arithmetic: $$1960000 - 8400 + 9 = 1951609$$ ### Exam Strategy & Shortcut Use the unit digit and estimation to eliminate distractors in seconds. The unit digit of $1397$ is $7$, so the unit digit of the square is the unit digit of $7 	imes 7 = 49$, which is $9$. This leaves all options in play. Look at the magnitude: $1400^2 = 1960000$. Since $1397$ is slightly less than $1400$, the answer must be slightly less than $1960000$. Option (a) $1951609$ fits perfectly. Option (c) is vastly too large ($18$ million), and (d) is over $2$ million. Option (b) is too close to $1960000$ to account for subtracting $8400$. ### Common Pitfall A major error is adding the $2ab$ term instead of subtracting it when using the $(a-b)^2$ formula, which would yield a much larger incorrect answer. Always respect the minus sign! ### Final Answer **Therefore, the correct answer is 1951609.**

Question 3

$107 	imes 107 + 93 	imes 93 = x$

Accepted Answer

### Concept & Formula This expression is in the form of $a^2 + b^2$. Notice that the numbers $107$ and $93$ are symmetrically distributed around a base of $100$. This allows us to use a powerful combined algebraic identity. $$(x + y)^2 + (x - y)^2 = 2(x^2 + y^2)$$ ### Step-by-Step Solution * Identify the symmetric base and variance: The base $x = 100$, and the variance $y = 7$. $$107 = 100 + 7$$ $$93 = 100 - 7$$ * Map this to our identity: $$(100 + 7)^2 + (100 - 7)^2 = 2 	imes ((100)^2 + (7)^2)$$ * Calculate the much simpler inner terms: $$(100)^2 = 10000$$ $$(7)^2 = 49$$ * Perform the final multiplication: $$2 	imes (10000 + 49) = 2 	imes 10049 = 20098$$ ### Exam Strategy & Shortcut Use the unit digits of the squares. $107^2$ ends in $9$ (since $7^2=49$). $93^2$ ends in $9$ (since $3^2=9$). The sum $9 + 9 = 18$, meaning the final unit digit must be $8$. This doesn't narrow it down enough, so check the last two digits. Using base rules, the last two digits of $(100+7)^2$ are $49$, and $(100-7)^2$ are $49$. Sum them: $49 + 49 = 98$. The answer must end in $98$. Only options (c) $20098$ and (d) $21908$ remain. Since $100^2 + 100^2 = 20000$, the answer should be extremely close to $20000$, pointing directly to $20098$. ### Common Pitfall Failing to recognize the symmetry around 100 and stubbornly calculating $107 	imes 107$ and $93 	imes 93$ separately, burning precious exam minutes on arithmetic. ### Final Answer **Therefore, the correct answer is 20098.**

Question 4

$217 	imes 217 + 183 	imes 183 = x$

Accepted Answer

### Concept & Formula Just like the previous problem, we are evaluating $a^2 + b^2$. Here, $217$ and $183$ are precisely equidistant from $200$. By exploiting this symmetry, we can collapse the expression using a standard algebraic formula. $$(x + y)^2 + (x - y)^2 = 2(x^2 + y^2)$$ ### Step-by-Step Solution * Define the common base and the offset: The base $x = 200$, and the offset $y = 17$. $$217 = 200 + 17$$ $$183 = 200 - 17$$ * Substitute these into the combined identity: $$(200 + 17)^2 + (200 - 17)^2 = 2 	imes ((200)^2 + (17)^2)$$ * Evaluate the squared terms inside the parenthesis: $$(200)^2 = 40000$$ $$(17)^2 = 289$$ * Add and multiply by 2: $$2 	imes (40000 + 289) = 2 	imes 40289 = 80578$$ ### Exam Strategy & Shortcut Use the "last two digits" shortcut derived from the formula $2(x^2 + y^2)$. Since $x=200$, $x^2$ ends in $0000$. The last digits of the entire sum will entirely depend on $2 	imes y^2$. $$2 	imes (17)^2 = 2 	imes 289 = 578$$ The answer must end in exactly $578$. Scanning the options, only option (b) $80578$ satisfies this rule! You can circle the answer without calculating the front digits. ### Common Pitfall Getting intimidated by the numbers and guessing, or making a calculation error while attempting to find $217 	imes 217$ manually. Always look for a clean base (like 100, 200, 50) hiding between the given numbers. ### Final Answer **Therefore, the correct answer is 80578.**

Question 5

$106 	imes 106 - 94 	imes 94 = x$

Accepted Answer

### Concept & Formula This problem asks for the difference between two perfect squares, represented as $a^2 - b^2$. This is one of the most important factorization identities in mathematics and drastically simplifies the arithmetic. $$a^2 - b^2 = (a + b)(a - b)$$ ### Step-by-Step Solution * Identify $a$ and $b$ from the expression: $$a = 106$$ $$b = 94$$ * Apply the difference of squares identity: $$(106)^2 - (94)^2 = (106 + 94)(106 - 94)$$ * Calculate the sum and the difference inside the brackets: $$106 + 94 = 200$$ $$106 - 94 = 12$$ * Multiply the simplified terms together: $$200 	imes 12 = 2400$$ ### Exam Strategy & Shortcut Whenever you see $a^2 - b^2$ on an aptitude test, instantly calculate the sum $(a+b)$. Here, $106 + 94 = 200$. This means the final answer must be a multiple of $200$. Any multiple of $200$ MUST end in two zeroes ($00$). Looking at the options, $2400$ and $2000$ are the only possibilities. Since the difference $(a-b)$ is $12$, the leading digits come from $12 	imes 2 = 24$. Option (a) is the instant winner. ### Common Pitfall Not recognizing the pattern and manually calculating $106 	imes 106$ and $94 	imes 94$ before subtracting. While it works, it turns a 5-second problem into a 60-second chore, which is fatal in competitive exams. ### Final Answer **Therefore, the correct answer is 2400.**

Question 6

$8796 	imes 223 + 8796 	imes 77 = x$

Accepted Answer

### Concept & Formula This problem is designed to test your knowledge of the distributive property of multiplication. Instead of performing two large multiplications, we can factor out the common term. $$ab + ac = a(b + c)$$ ### Step-by-Step Solution * Identify the common multiplier in both terms: The number 8796 is common to both $8796 	imes 223$ and $8796 	imes 77$. * Apply the distributive property to factor it out: $$8796 	imes (223 + 77)$$ * Perform the addition inside the parentheses: $$223 + 77 = 300$$ * Multiply the factored number by the sum: $$8796 	imes 300 = 2638800$$ ### Exam Strategy & Shortcut When spotting a repeated number in an addition or subtraction string, immediately look for a base-10 sum (10, 100, 300, etc.) hiding in the other multipliers. Calculating $8796 	imes 300$ only requires computing $8796 	imes 3$ and appending two zeros. The unit digit of $6 	imes 3$ is 18, so the answer must end in 800. Since 2638800 is not among the given options (a) through (d), the answer is "None of these". ### Common Pitfall Calculating $8796 	imes 223$ and $8796 	imes 77$ independently. This wastes several minutes and dramatically increases the likelihood of a basic arithmetic mistake during the final addition. ### Final Answer **Therefore, the correct answer is None of these.**

Question 7

$287 	imes 287 + 269 	imes 269 - 2 	imes 287 	imes 269 = x$

Accepted Answer

### Concept & Formula The given expression strictly follows a well-known algebraic identity for the square of a binomial difference. Recognizing this pattern collapses a complex arithmetic problem into a simple mental math exercise. $$a^2 + b^2 - 2ab = (a - b)^2$$ ### Step-by-Step Solution * Map the expression to the algebraic identity variables: Let $a = 287$ and $b = 269$. * Rewrite the original problem using these variables: $$(287 	imes 287) + (269 	imes 269) - (2 	imes 287 	imes 269) ightarrow a^2 + b^2 - 2ab$$ * Collapse the expression into its factored form: $$(287 - 269)^2$$ * Calculate the difference inside the parentheses: $$287 - 269 = 18$$ * Square the result to find the final value: $$(18)^2 = 324$$ ### Exam Strategy & Shortcut You should have the squares of all numbers up to 30 memorized for competitive exams. Once you spot the $a^2 + b^2 - 2ab$ pattern, instantly subtract the two unique numbers ($287 - 269 = 18$). Then, either recall that $18^2 = 324$, or use unit digits: the unit digit of $(18)^2$ is the unit digit of $8 	imes 8 = 64$, which is 4. Option (d) 324 and (a) 534 end in 4, but $18^2$ is much smaller than 500, leading straight to 324. ### Common Pitfall Brute-forcing the multiplication. Even partial brute-forcing (like only calculating the unit digit of the entire expanded expression) takes longer and is more error-prone than applying the identity. ### Final Answer **Therefore, the correct answer is 324.**

Question 8

$\{(476 + 424)^2 - 4 	imes 476 	imes 424\} = x$

Accepted Answer

### Concept & Formula This expression is a classic application of a derived algebraic identity. It represents the relationship between the square of a sum and the square of a difference. $$(a + b)^2 - 4ab = (a - b)^2$$ ### Step-by-Step Solution * Identify the core variables in the given expression: Let $a = 476$ and $b = 424$. * Substitute these variables into the structural formula: $$(476 + 424)^2 - 4(476)(424) ightarrow (a + b)^2 - 4ab$$ * Simplify the expression into the squared difference: $$(476 - 424)^2$$ * Perform the subtraction: $$476 - 424 = 52$$ * Calculate the final square: $$(52)^2 = 2704$$ ### Exam Strategy & Shortcut To square 52 quickly, use the base-50 squaring method. For a number $(50 + x)^2$, the result is composed of two parts: $(25 + x)$ and $(x^2)$. Here, $x = 2$. First part: $25 + 2 = 27$ Second part: $2^2 = 04$ (must be two digits) Combine them: 2704. You can solve this entirely in your head! ### Common Pitfall Attempting to expand $(476 + 424)^2$ into $900^2 = 810000$ and then trying to manually subtract $4 	imes 476 	imes 424$. While mathematically valid, the sheer size of the numbers almost guarantees a calculation error under time pressure. ### Final Answer **Therefore, the correct answer is 2704.**

Question 9

The value of $112 	imes 5^4$ is

Accepted Answer

### Concept & Strategy Multiplying by powers of 5 can be vastly simplified by converting the 5 into 10/2. This converts a difficult multiplication problem into a simple division problem followed by appending zeros. $$a 	imes 5^n = a 	imes \left(\frac{10}{2}ight)^n = \frac{a 	imes 10^n}{2^n}$$ ### Step-by-Step Solution * Recognize that $5^4$ is equal to $(10/2)^4$: $$5^4 = \frac{10^4}{2^4} = \frac{10000}{16}$$ * Substitute this fraction back into the original expression: $$112 	imes \frac{10000}{16}$$ * Rearrange to perform the division first: $$\left(\frac{112}{16}ight) 	imes 10000$$ * Divide 112 by 16: $$112 \div 16 = 7$$ * Multiply by 10000: $$7 	imes 10000 = 70000$$ ### Exam Strategy & Shortcut You should memorize the powers of 2 up to $2^{10}$ and basic multiples of 16. When you see $	imes 5^4$, instantly think "divide by 16, add four zeros". $112 \div 16 = 7$. Appending four zeros gives 70000. The entire problem takes less than 5 seconds without writing anything down. ### Common Pitfall Manually calculating $5^4 = 625$ and then multiplying $112 	imes 625$. While 625 is not extremely large, multiplying a three-digit number by a three-digit number manually wastes valuable time. ### Final Answer **Therefore, the correct answer is 70000.**

Question 10

Multiply $5746320819$ by $125$.

Accepted Answer

### Concept & Strategy Multiplying any enormous number by 125 is a prime candidate for the fraction substitution technique. Since $125 = 1000 / 8$, we can divide the large number by 8 and append three zeros. $$x 	imes 125 = x 	imes \frac{1000}{8}$$ ### Step-by-Step Solution * Rewrite 125 as a fraction of 1000: $$5746320819 	imes \frac{1000}{8}$$ * Set up the division of the large number by 8: $$\frac{5746320819}{8} 	imes 1000$$ * Perform the step-by-step division: $57 \div 8 = 7$ (remainder 1) $14 \div 8 = 1$ (remainder 6) $66 \div 8 = 8$ (remainder 2) ... continuing this process yields 718290102. * Handle the decimal remainder for the final digits: The remainder at the end is 3. So we have 718290102 and 3/8. $3/8 	imes 1000 = 375$. * Combine the whole and fractional parts: $$718290102375$$ ### Exam Strategy & Shortcut Look at the options before you divide! You only need to divide the first few digits to find the correct answer. $57 \div 8 = 7$ (Starts with 7). $14 \div 8 = 1$ (Second digit is 1). The prefix is 71. Look at the options: (a) 718,290... (b) 728,490... (c) 748,290... (d) 798,290... Option (a) is the only one that starts with 718. You can confidently select it after calculating just the first two digits! ### Common Pitfall Attempting a standard multi-line multiplication with 125. Even if you are fast at multiplying, adding up three offset rows of a 10-digit number virtually guarantees an addition or alignment error. ### Final Answer **Therefore, the correct answer is 718,290,102,375.**

Question 11

$935421 	imes 625 = x$

Accepted Answer

### Concept & Formula Multiplying a large number by 625 can be vastly simplified by expressing 625 as a power of 10 divided by a power of 2. $$625 = \frac{10000}{16}$$ ### Step-by-Step Solution * Rewrite the mathematical expression by replacing 625 with its fractional equivalent. $$935421 	imes \frac{10000}{16}$$ * This means we must divide $9354210000$ by 16. * First digits: $935 \div 16 = 58$ with a remainder of 7. * Carry down the 4 to get 74. $74 \div 16 = 4$ with a remainder of 10. * Continue this division sequence: $9354210000 \div 16 = 584638125$. ### Exam Strategy & Shortcut Use the **Digital Root (Digit Sum)** method for immediate elimination. Sum of digits of $935421 = 9+3+5+4+2+1 = 24 ightarrow 6$. Sum of digits of $625 = 6+2+5 = 13 ightarrow 4$. Product digital root = $6 	imes 4 = 24 ightarrow 6$. Check the digital roots of the options: (a) 5+7+5+6+4+8+1+2+5 $ightarrow$ 7 (b) 5+8+4+6+3+8+1+2+5 $ightarrow$ 6 (Matches) (c) 5+8+4+6+4+9+1+2+5 $ightarrow$ 8 (d) 5+8+5+6+2+8+1+2+5 $ightarrow$ 6 Options (b) and (d) remain. Use basic estimation: $935 	imes 625$ will be slightly more than $900 	imes 600 = 540000$. Thus, $584$ is the correct starting sequence over $585$. ### Common Pitfall Attempting brute-force multiplication without converting 625 to a fraction. This consumes massive amounts of time and increases the risk of simple arithmetic errors in an exam setting. ### Final Answer Therefore, the correct answer is **584638125**.

Question 12

$(999)^2 - (998)^2 = x$

Accepted Answer

### Concept & Formula This problem is a direct, straightforward application of the algebraic identity for the difference of two perfect squares. $$a^2 - b^2 = (a - b)(a + b)$$ ### Step-by-Step Solution * Identify the terms from the equation, where $a = 999$ and $b = 998$. * Substitute these values directly into the algebraic identity: $$(999)^2 - (998)^2 = (999 - 998)(999 + 998)$$ * Simplify the mathematical terms inside the parentheses: $$(1)(1997)$$ * Multiply the final results: $$1997$$ ### Exam Strategy & Shortcut Whenever you are subtracting the squares of two consecutive numbers (meaning $a - b = 1$), the final answer is always simply the sum of those two numbers. Instantly perform $999 + 998 = 1997$ in your head to save time. ### Common Pitfall Students often attempt to manually calculate the massive square of 999 and 998 individually. This is a massive time sink and highly prone to calculation errors. ### Final Answer Therefore, the correct answer is **1997**.

Question 13

$(80)^2 - (65)^2 + 81 = x$

Accepted Answer

### Concept & Formula Simplify the squared terms first by applying the difference of two squares identity before adding the final constant. $$a^2 - b^2 = (a - b)(a + b)$$ ### Step-by-Step Solution * First, isolate the squared terms to process them efficiently: $(80)^2 - (65)^2$. * Apply the identity where $a = 80$ and $b = 65$: $$(80 - 65)(80 + 65) = (15)(145)$$ * Calculate the resulting product by splitting the terms logically: $$15 	imes 145 = 15 	imes (100 + 45) = 1500 + 675 = 2175$$ * Add the remaining constant of 81 to this intermediate result: $$2175 + 81 = 2256$$ ### Exam Strategy & Shortcut You can utilize the **Unit Digit Method** to eliminate options rapidly. The number $(80)^2$ ends in 0, and $(65)^2$ ends in 5. So, their difference ends in $10 - 5 = 5$. Then, add the unit digit of 81 (which is 1). $5 + 1 = 6$. The correct answer must end in 6, leaving only options (a) and (d). Since $80^2 = 6400$ and $65^2 = 4225$, the difference is roughly 2200. Option (d) matches this perfectly. ### Common Pitfall Calculating $65^2$ manually can take unnecessary time if you don't know the standard ending-in-5 shortcut. Utilizing the $a^2 - b^2$ formula avoids large manual squaring equations entirely. ### Final Answer Therefore, the correct answer is **2256**.

Question 14

$(24 + 25 + 26)^2 - (10 + 20 + 25)^2 = x$

Accepted Answer

### Concept & Formula First, you must evaluate the arithmetic sums inside the parentheses, and then apply the difference of squares identity to conclude the equation efficiently. $$a^2 - b^2 = (a - b)(a + b)$$ ### Step-by-Step Solution * Calculate the sum found in the first bracket: $$24 + 25 + 26 = 75$$ * Calculate the sum found in the second bracket: $$10 + 20 + 25 = 55$$ * Substitute these evaluated sums back into the original expression: $$(75)^2 - (55)^2$$ * Apply the difference of squares formula where $a = 75$ and $b = 55$: $$(75 - 55)(75 + 55) = (20)(130)$$ * Multiply the final simplified terms: $$20 	imes 130 = 2600$$ ### Exam Strategy & Shortcut Notice that $24 + 25 + 26$ is a simple arithmetic progression. Its sum is just the middle term multiplied by the number of terms ($25 	imes 3 = 75$). This quickly gives you the first squared term mentally. Then, using the standard $(a-b)(a+b)$ algebraic rule transforms a complex squaring problem into basic mental math. ### Common Pitfall Squaring 75 and 55 manually ($5625 - 3025$) mathematically works but leaves immense room for basic subtraction errors under pressure. Always default to $(a-b)(a+b)$ when you naturally spot a difference of squares. ### Final Answer Therefore, the correct answer is **2600**.

Question 15

$(65)^2 - (55)^2 = x$

Accepted Answer

### Concept & Formula This is another direct application of the fundamental algebraic identity for evaluating the difference of two perfect squares without manually multiplying. $$a^2 - b^2 = (a - b)(a + b)$$ ### Step-by-Step Solution * Identify the components of the given expression where $a = 65$ and $b = 55$. * Apply the identity substitution directly: $$(65)^2 - (55)^2 = (65 - 55)(65 + 55)$$ * Solve the basic arithmetic expressions inside the parentheses: $$65 - 55 = 10$$ $$65 + 55 = 120$$ * Multiply the isolated results together: $$10 	imes 120 = 1200$$ ### Exam Strategy & Shortcut For any $a^2 - b^2$ problem where the difference between $a$ and $b$ is exactly 10, the answer is simply $10 	imes (a + b)$. You can mentally add $65 + 55 = 120$ and immediately append a zero to get 1200 instantly without writing anything down. ### Common Pitfall Squaring the numbers ending in 5 manually ($4225 - 3025$) is a valid mathematical trick, but it takes significantly more cognitive load and exam time than simply applying the standardized $a^2 - b^2$ formula. ### Final Answer Therefore, the correct answer is **1200**.

Question 16

If $a$ and $b$ be positive integers such that $a^2 - b^2 = 19$, then the value of $a$ is

Accepted Answer

### Concept & Formula When a difference of two perfect squares equals a prime number, factor the squares and equate the factors to the prime number and 1. $$a^2 - b^2 = (a - b)(a + b)$$ ### Step-by-Step Solution * **Given:** $a^2 - b^2 = 19$, where $a$ and $b$ are positive integers. * Expand the left side using the standard algebraic identity: $$(a - b)(a + b) = 19$$ * Since 19 is a prime number, its only positive integer factors are 1 and 19. * Because $a$ and $b$ are positive integers, the sum $(a + b)$ must be greater than the difference $(a - b)$. * Therefore, we can set up a system of linear equations: $$a - b = 1$$ $$a + b = 19$$ * Add the two equations together to eliminate $b$: $$(a - b) + (a + b) = 1 + 19$$ $$2a = 20$$ $$a = 10$$ ### Exam Strategy & Shortcut Whenever $a^2 - b^2 = P$ (where $P$ is a prime number), the larger number $a$ is always $\frac{P + 1}{2}$ and the smaller number $b$ is always $\frac{P - 1}{2}$. Mentally calculate $\frac{19 + 1}{2} = 10$ to find the answer instantly without writing anything down. ### Common Pitfall Students often try to blindly guess perfect squares whose difference is 19 (e.g., trying $25-6$, $36-17$, $100-81$) which wastes valuable exam time. Always rely on prime factorization properties instead of trial and error. ### Final Answer Therefore, the correct answer is **10**.

Question 17

If $a$ and $b$ are positive integers, $a > b$ and $(a + b)^2 - (a - b)^2 > 29$, then the smallest value of $a$ is

Accepted Answer

### Concept & Formula The difference between the square of a sum and the square of a difference of two variables simplifies directly to four times their product. $$(a + b)^2 - (a - b)^2 = 4ab$$ ### Step-by-Step Solution * **Given:** $(a + b)^2 - (a - b)^2 > 29$, where $a > b$, and both are positive integers. * Substitute the simplified algebraic identity into the given inequality: $$4ab > 29$$ * Divide both sides by 4 to isolate the product $ab$: $$ab > \frac{29}{4}$$ $$ab > 7.25$$ * Since $a$ and $b$ are positive integers, their product $ab$ must be an integer of at least 8. * We need the smallest possible value for $a$, keeping in mind the strict condition that $a > b$. * Let's test the options starting from the smallest integer: * If $a = 3$, the maximum integer value for $b$ (since $a > b$) is 2. The product $ab = 6$. This is not greater than 7.25. * If $a = 4$, the possible integer values for $b$ are 1, 2, or 3. If $b = 2$, $ab = 8$. If $b = 3$, $ab = 12$. Both results are greater than 7.25. * Thus, the smallest valid integer for $a$ that satisfies all conditions is 4. ### Exam Strategy & Shortcut Recognize the $4ab$ expansion immediately. Knowing $4ab > 29$ means $ab \ge 8$. Since $a > b$, $a$ must be strictly greater than $\sqrt{8}$. The square root of 8 is approximately 2.8, so $a$ must be at least 3. Test $a=3 ightarrow 	ext{max } b=2 ightarrow ab=6$ (Fail). Move to $a=4 ightarrow 	ext{max } b=3 ightarrow ab=12$ (Pass). The answer is 4. ### Common Pitfall A frequent mistake is ignoring the condition $a > b$ and assuming $b$ can be any large integer, which might lead one to believe $a$ could be 1 or 2. Always strictly enforce all inequality constraints while testing options. ### Final Answer Therefore, the correct answer is **4**.

Question 18

$397 	imes 397 + 104 	imes 104 + 2 	imes 397 	imes 104 = x$

Accepted Answer

### Concept & Formula This expression exactly matches the expanded form of a perfect square trinomial. Recognizing this pattern eliminates the need for any manual multiplication. $$a^2 + b^2 + 2ab = (a + b)^2$$ ### Step-by-Step Solution * Identify the repeating terms in the given expression and assign them to variables: * Let $a = 397$ * Let $b = 104$ * Rewrite the numerical expression using these assigned variables: $$a 	imes a + b 	imes b + 2 	imes a 	imes b$$ $$a^2 + b^2 + 2ab$$ * Apply the perfect square trinomial identity to condense the expression: $$(a + b)^2$$ * Substitute the original numerical values back into the condensed formula: $$(397 + 104)^2$$ * Perform the basic addition inside the parentheses first: $$(501)^2$$ * Square the resulting number rapidly by splitting it into $(500 + 1)^2$: $$500^2 + 1^2 + 2(500)(1) = 250000 + 1 + 1000 = 251001$$ ### Exam Strategy & Shortcut Use the **Unit Digit Method** to cross-check. The expression condenses to $(397 + 104)^2 = (501)^2$. The unit digit of 501 is 1, so the square must end in 1. All options end in 1, so unit digit alone isn't enough. However, $500^2$ is exactly 250,000. Because 501 is slightly larger than 500, the answer must be slightly larger than 250,000. Option (b), 251,001, fits perfectly without doing the full expansion. ### Common Pitfall Attempting to multiply $397 	imes 397$ and $104 	imes 104$ manually is a massive trap. It consumes excessive time and highly increases the likelihood of arithmetic errors. Always scan for algebraic patterns before calculating. ### Final Answer Therefore, the correct answer is **251001**.

Question 19

If $(64)^2 - (36)^2 = 20 	imes x$, then $x =$

Accepted Answer

### Concept & Formula Apply the difference of two perfect squares formula to the left side of the equation to simplify the large numbers without resorting to manual squaring. $$a^2 - b^2 = (a - b)(a + b)$$ ### Step-by-Step Solution * **Given:** $(64)^2 - (36)^2 = 20x$ * Focus on the left side of the equation and apply the difference of squares identity where $a = 64$ and $b = 36$: $$(64 - 36)(64 + 36) = 20x$$ * Perform the basic subtraction and addition inside the parentheses: $$(28)(100) = 20x$$ * Multiply the terms to simplify the left side completely: $$2800 = 20x$$ * Divide both sides by 20 to isolate and solve for $x$: $$x = \frac{2800}{20}$$ $$x = 140$$ ### Exam Strategy & Shortcut You can skip full multiplication by manipulating the factors directly. Once you reach $(28)(100) = 20x$, simply divide 100 by 20 in your head to get 5. Then, calculate $28 	imes 5 = 140$. This eliminates the intermediate step of dealing with 2800, minimizing trailing zeros and saving a few extra seconds. ### Common Pitfall Manually squaring 64 ($4096$) and 36 ($1296$) before subtracting is the most common error. While mathematically sound, it introduces massive potential for calculation mistakes and consumes precious time compared to algebraic factoring. ### Final Answer Therefore, the correct answer is **140**.

Question 20

$$\frac{(489 + 375)^2 - (489 - 375)^2}{(489 	imes 375)} = x$$

Accepted Answer

### Concept & Formula This fraction follows a universal algebraic template. The difference between the square of a sum and the square of a difference always simplifies to exactly four times the product of the two numbers. $$(a + b)^2 - (a - b)^2 = 4ab$$ ### Step-by-Step Solution * Analyze the structure of the numerator and assign variables to the recurring numbers to reveal the underlying algebra: * Let $a = 489$ * Let $b = 375$ * Substitute these variables back into the entire original fractional expression: $$\frac{(a + b)^2 - (a - b)^2}{ab}$$ * Apply the standard algebraic identity to simplify the numerator: $$(a + b)^2 - (a - b)^2 = 4ab$$ * Replace the numerator in our fraction with this highly simplified form: $$\frac{4ab}{ab}$$ * Cancel out the common $ab$ terms from both the numerator and the denominator: $$4$$ * Because the variables completely cancel out, the actual numerical values of $a$ (489) and $b$ (375) are entirely irrelevant to the final outcome. ### Exam Strategy & Shortcut Memorize the identity $\frac{(a+b)^2 - (a-b)^2}{ab} = 4$. Whenever you see this exact structural format in a competitive exam, immediately mark 4 as the answer. Do not write anything down or look at the actual numbers provided, as they are merely distractors. ### Common Pitfall A critical mistake is attempting to add and subtract the numbers inside the parentheses, squaring the massive results, and then trying to perform long division. This will drain minutes of exam time and almost guarantee an arithmetic error. ### Final Answer Therefore, the correct answer is **4**.

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