Question 1

Unit's digit in $(784)^{126} + (784)^{127}$ is

Accepted Answer

### Concept & Logic For numbers ending in $4$ or $9$, the cyclicity is simply $2$. You don't need to divide by $4$; you only need to look at whether the exponent is odd or even to determine the unit digit. ### Step-by-Step Solution **Given:** Evaluate the unit digit of $(784)^{126} + (784)^{127}$. **Calculation:** * The base for both terms has a unit digit of $4$. We only care about powers of $4$. * The cyclicity rule for $4$ is: * $4^{	ext{odd power}}$ ends in $4$ (like $4^1 = 4$). * $4^{	ext{even power}}$ ends in $6$ (like $4^2 = 16$). * **Term 1:** $(784)^{126}$ has an even exponent ($126$). Thus, its unit digit is $6$. * **Term 2:** $(784)^{127}$ has an odd exponent ($127$). Thus, its unit digit is $4$. * **Addition:** Add the two derived unit digits: $$6 + 4 = 10$$ * The final unit digit of the sum $10$ is $0$. ### Exam Strategy & Shortcut Notice that $a^n + a^{n+1}$ can be factored. $(784)^{126} 	imes (1 + 784) = (784)^{126} 	imes 785$. Any even number (which $(784)^{126}$ definitely is) multiplied by a number ending in $5$ (like $785$) will always produce a trailing zero. This logic completely bypasses cyclicity calculations! ### Common Pitfall Applying the "divide by $4$" cyclicity rule to bases ending in $4$ or $9$ isn't *wrong*, but it is an inefficient use of time. Recognize the odd/even shortcut to save precious seconds on the exam. ### Final Answer **Therefore, the correct answer is 0.**

Question 2

The digit in the unit's place of $[(251)^{98} + (21)^{29} - (106)^{100} + (705)^{35} - 16^4 + 259]$ is

Accepted Answer

### Concept & Logic The numbers $1, 5,$ and $6$ have a cyclicity of $1$. This means that any number ending in $1, 5,$ or $6$ will always retain that same unit digit when raised to any positive integer power. ### Step-by-Step Solution **Given:** Evaluate the unit digit of $[(251)^{98} + (21)^{29} - (106)^{100} + (705)^{35} - 16^4 + 259]$. **Calculation:** * Extract the unit digit for each term based on its base: * $(251)^{98} \rightarrow$ base ends in $1$, so unit digit is $1$. * $(21)^{29} \rightarrow$ base ends in $1$, so unit digit is $1$. * $(106)^{100} \rightarrow$ base ends in $6$, so unit digit is $6$. * $(705)^{35} \rightarrow$ base ends in $5$, so unit digit is $5$. * $16^4 \rightarrow$ base ends in $6$, so unit digit is $6$. * $259 \rightarrow$ unit digit is $9$. * Substitute these unit digits back into the expression: $$1 + 1 - 6 + 5 - 6 + 9$$ * Add the positive values together and the negative values together: * Positive sum: $1 + 1 + 5 + 9 = 16 \rightarrow$ unit digit $6$. * Negative sum: $-6 - 6 = -12 \rightarrow$ unit digit $2$. * Subtract the negative unit sum from the positive unit sum: $$6 - 2 = 4$$ ### Exam Strategy & Shortcut Scan the expression for bases ending in $0, 1, 5,$ or $6$ because their unit digits never change. Since every single term in this expression falls into that category (or is a flat number like $259$), you can instantly write down $1 + 1 - 6 + 5 - 6 + 9$ and solve it mentally in under 15 seconds. ### Common Pitfall A major trap in long polynomial additions/subtractions is ending up with a negative unit digit (e.g., if you evaluated left-to-right as $2 - 6 = -4$). If you ever hit a negative intermediate unit digit, you must "borrow $10$" (so $-4$ becomes $10 - 4 = 6$). Grouping all positives and all negatives first avoids this confusion. ### Final Answer **Therefore, the correct answer is 4.**

Question 3

The digit in the unit's place of the product $(2464)^{1793} 	imes (615)^{317} 	imes (131)^{491}$ is

Accepted Answer

### Concept & Strategy The unit digit of a product is entirely dependent on the unit digits of its multipliers. Furthermore, multiplying any even number by a number ending in $5$ will always yield a product that ends in $0$. ### Step-by-Step Solution **Given:** Find the unit digit of $(2464)^{1793} 	imes (615)^{317} 	imes (131)^{491}$. **Calculation:** * Look at the base of the middle term: $615$. Since it ends in $5$, any power of it will also end in $5$. * $(615)^{317} ightarrow$ unit digit is $5$. * Look at the base of the first term: $2464$. Since it ends in an even number ($4$), any integer power of it will result in an even unit digit. * $(2464)^{1793} ightarrow 4^{	ext{odd}} ightarrow$ unit digit is $4$ (an even number). * Look at the base of the last term: $131$. It ends in $1$, so its power will end in $1$. * Multiply the resulting unit digits: $$4 	imes 5 	imes 1 = 20$$ * The unit digit of $20$ is $0$. ### Exam Strategy & Shortcut Do not waste time calculating the exact unit digit for every term using cyclicity rules. The moment you spot a factor that will end in $5$ (like $615^{317}$) and *any* factor that will be even (like $2464^{1793}$), the unit digit of the entire product instantly becomes $0$. ### Common Pitfall Students often go into "autopilot" mode, meticulously calculating $1793 \pmod 4$ to find the power of $4$, taking unnecessary time. Always do a quick holistic scan of the expression before applying brute-force rules. ### Final Answer **Therefore, the correct answer is 0.**

Question 4

If $x$ is an even number, then $x^{4n}$, where $n$ is a positive integer, will always have

Accepted Answer

### Concept & Logic Every non-zero even digit ($2, 4, 6, 8$) has a power cyclicity that results in the unit digit $6$ when raised to the $4$th power (or any multiple of $4$). The only exception is $0$, which always remains $0$. ### Step-by-Step Solution **Given:** $x$ is an even number. We need the unit digit of $x^{4n}$ for any positive integer $n$. **Deduction:** * Since $x$ is an even number, its unit digit can be $0, 2, 4, 6,$ or $8$. * Let's test the $4$th power (setting $n=1$) for each possible even unit digit: * If unit digit is $0$: $0^4 = 0 \rightarrow$ ends in $0$. * If unit digit is $2$: $2^4 = 16 \rightarrow$ ends in $6$. * If unit digit is $4$: $4^4 = 256 \rightarrow$ ends in $6$. * If unit digit is $6$: $6^4 = 1296 \rightarrow$ ends in $6$. * If unit digit is $8$: $8^4 = 4096 \rightarrow$ ends in $6$. * Therefore, the resulting unit digit will always be either $0$ (if $x$ ends in $0$) or $6$ (if $x$ ends in $2, 4, 6,$ or $8$). ### Exam Strategy & Shortcut Instead of writing out all the powers, you can quickly test the two extremes of even numbers. Let $n=1$. Try $x=2$ (gives $2^4 = 16 \rightarrow 6$) and try $x=10$ (gives $10^4 = 10000 \rightarrow 0$). This immediately proves the answer must encompass both $0$ and $6$. ### Common Pitfall The most frequent mistake is forgetting that $0$ is considered an even digit (or that numbers ending in $0$, like $10$, are even). If you only test $2, 4, 6,$ and $8$, you will confidently but incorrectly choose "6 in the unit's place". ### Final Answer **Therefore, the correct answer is either 0 or 6 in the unit's place.**

Question 5

If $m$ and $n$ are positive integers, then the digit in the unit's place of $5^n + 6^m$ is always

Accepted Answer

### Concept & Rule The numbers $5$ and $6$ are special because they are invariant under exponentiation. Any positive integer power of $5$ will end in $5$, and any positive integer power of $6$ will end in $6$. ### Step-by-Step Solution **Given:** Find the unit digit of $5^n + 6^m$ where $m, n > 0$. **Calculation:** * Analyze the first term, $5^n$: * $5^1 = 5$, $5^2 = 25$, $5^3 = 125\dots$ The unit digit is always $5$. * Analyze the second term, $6^m$: * $6^1 = 6$, $6^2 = 36$, $6^3 = 216\dots$ The unit digit is always $6$. * Add the unit digits together: $$5 + 6 = 11$$ * The unit digit of this sum ($11$) is $1$. ### Exam Strategy & Shortcut This is a pure "knowledge check" question that requires zero arithmetic. Commit the static digits ($0, 1, 5, 6$) to memory. Seeing $5^n + 6^m$ should instantly trigger $5 + 6 = 11 \rightarrow 1$. This should take no more than 3 seconds on an exam. ### Common Pitfall Students sometimes get intimidated by abstract variables like $m$ and $n$ with unknown values, thinking they need to set up algebraic equations or that the answer depends on whether $m$ and $n$ are odd or even. Recognize when variables are irrelevant decoys. ### Final Answer **Therefore, the correct answer is 1.**

Question 6

The number formed from the last two digits (ones and tens) of the expression $2^{12n} - 6^{4n}$, where $n$ is any positive integer is

Accepted Answer

### Concept & Formula The problem requires finding the last two digits of the algebraic expression $x^m - y^m$. We use the algebraic identity that $a^k - b^k$ is always perfectly divisible by $(a - b)$ for all positive integers $k$. If the difference $(a - b)$ ends in two zeros, the entire expression will end in two zeros. ### Step-by-Step Solution * **Given Expression:** $2^{12n} - 6^{4n}$ * Rewrite the bases to have a common exponent $n$: $$2^{12n} = (2^{12})^n = 4096^n$$ $$6^{4n} = (6^4)^n = 1296^n$$ * Now the expression is in the form of $a^n - b^n$, where $a = 4096$ and $b = 1296$. * According to the algebraic property, $a^n - b^n$ is divisible by $(a - b)$. * Calculate the difference: $$4096 - 1296 = 2800$$ * Since the base difference $2800$ is a multiple of $100$ (it ends in $00$), any integer power expansion or multiple of this base difference will also end in $00$. ### Exam Strategy & Shortcut Instead of using algebraic expansion, use the **Substitution Method**. Assume the simplest positive integer value for $n$, which is $n = 1$. Expression becomes $2^{12} - 6^4 = 4096 - 1296 = 2800$. The last two digits of $2800$ are $00$. This must hold true for all $n$. ### Common Pitfall Students often try to find the cyclicity of the last two digits of $2$ and $6$ separately, which becomes highly complex and time-consuming. Grouping the exponents to form a single power $n$ is the key. ### Final Answer Therefore, the correct answer is **00**.

Question 7

The last digit in the decimal representation of $\left(\frac{1}{5}\right)^{2000}$ is

Accepted Answer

### Concept & Logic To find the last non-zero digit in a decimal representation of a fraction, it helps to convert the denominator to a power of $10$. This transforms the problem into a simple unit digit cyclicity question. ### Step-by-Step Solution * **Given:** $\left(\frac{1}{5}ight)^{2000}$ * Convert the fraction to a decimal-friendly format by multiplying the numerator and denominator by $2$: $$\left(\frac{1 	imes 2}{5 	imes 2}ight)^{2000} = \left(\frac{2}{10}ight)^{2000}$$ * This can be written as: $$\frac{2^{2000}}{10^{2000}}$$ * The denominator $10^{2000}$ simply defines the decimal point position (creating $2000$ decimal places). * The last digit of the decimal expansion will therefore be exactly the unit digit of the numerator, $2^{2000}$. * Find the unit digit of $2^{2000}$ using cyclicity. The powers of $2$ repeat their unit digits in a cycle of $4$: $(2, 4, 8, 6)$. * Divide the power $2000$ by the cycle length $4$. The remainder is $0$, meaning it lands on the 4th value in the cycle. * The 4th value in the cyclicity of $2$ is $6$. ### Exam Strategy & Shortcut Recognize immediately that $1/5 = 0.2$. The problem is asking for the last digit of $(0.2)^{2000}$, which is governed purely by the unit digit of $2^{2000}$. Since $2000$ is a multiple of $4$, the unit digit is $2^4 ightarrow 16 ightarrow 6$. Solved in 10 seconds. ### Common Pitfall A common mistake is assuming the last digit of a decimal expansion involving fractions like $1/5$ always ends in $5$, confusing division rules with exponent rules. ### Final Answer Therefore, the correct answer is **6**.

Question 8

Let $x$ be the product of two numbers $3,659,893,456,789,325,678$ and $342,973,489,379,256$. The number of digits in $x$ is

Accepted Answer

### Concept & Strategy The number of digits in the product of two large numbers can be determined by representing them in standard scientific notation (powers of 10). If number $A$ has $m$ digits and $B$ has $n$ digits, their product $A 	imes B$ will have either $(m+n-1)$ or $(m+n)$ digits, depending on whether the product of their leading digits exceeds $10$. ### Step-by-Step Solution * Count the digits in the first number: $A = 3,659,893,456,789,325,678$. This is a $19$-digit number. It can be approximated as $3.66 	imes 10^{18}$. * Count the digits in the second number: $B = 342,973,489,379,256$. This is a $15$-digit number. It can be approximated as $3.43 	imes 10^{14}$. * Find the product $x = A 	imes B$: $$x \approx (3.66 	imes 10^{18}) 	imes (3.43 	imes 10^{14})$$ $$x \approx (3.66 	imes 3.43) 	imes 10^{32}$$ * Calculate the approximate product of the leading coefficients: $3.66 	imes 3.43 \approx 12.55$ * So, $x \approx 12.55 	imes 10^{32} = 1.255 	imes 10^{33}$. * A number written as $C 	imes 10^k$ (where $1 \le C < 10$) has $k + 1$ digits. * Number of digits = $33 + 1 = 34$. ### Exam Strategy & Shortcut Use the digit sum rule: Number of digits = $m + n$ (if the product of the first digits of each number $\ge 10$) or $m + n - 1$ (if $< 10$). Here, $m = 19$ and $n = 15$. The leading digits are $3.6\dots$ and $3.4\dots$. Their product is $3.6 	imes 3.4 > 10$. Therefore, digits = $19 + 15 = 34$. ### Common Pitfall Students often simply add the number of digits $(19 + 15 = 34)$ without checking the product of the leading digits. While it works here, if the numbers started with $1$ and $2$ (e.g., $1.5 	imes 2.5 = 3.75$), the answer would be $19 + 15 - 1 = 33$. Always verify the carry-over. ### Final Answer Therefore, the correct answer is **34**.

Question 9

Let a number of three digits have for its middle digit the sum of the other two digits. Then it is a multiple of

Accepted Answer

### Concept & Logic A standard three-digit number with digits $a$ (hundreds), $b$ (tens), and $c$ (units) can be expressed algebraically as $100a + 10b + c$. Substituting given digit constraints into this algebraic form reveals its divisibility properties. ### Step-by-Step Solution * Let the three-digit number be represented as $100a + 10b + c$, where $a, b,$ and $c$ are the digits. * **Given constraint:** The middle digit is the sum of the other two digits. $$b = a + c$$ * Substitute $b$ into the original algebraic expansion of the number: $$100a + 10(a + c) + c$$ * Expand and simplify the expression: $$100a + 10a + 10c + c$$ $$110a + 11c$$ * Factor out the common term: $$11(10a + c)$$ * Since the entire number can be expressed as $11$ multiplied by an integer $(10a + c)$, the number must be perfectly divisible by $11$. ### Exam Strategy & Shortcut **Option Elimination via Examples:** Pick any three-digit number satisfying the condition. Example: Let first digit = 1, last digit = 2. Middle digit = $1+2 = 3$. The number is $132$. Test $132$ against options: Not a multiple of $10$ or $50$. $132 / 18$ is not an integer. $132 / 11 = 12$. Option (b) is the only valid choice. ### Common Pitfall Trying to memorize divisibility rules without understanding algebraic expansion. While you might know the rule for $11$ (alternating digit sum), mapping it inversely from the algebraic equation is more robust for abstract questions. ### Final Answer Therefore, the correct answer is **11**.

Question 10

What least value must be given to $n$ so that the number $6135n2$ becomes divisible by 9?

Accepted Answer

### Concept & Formula The divisibility rule for $9$ states that a number is perfectly divisible by $9$ if and only if the sum of all its digits is a multiple of $9$. ### Step-by-Step Solution * **Given Number:** $6135n2$, where $n$ is a single-digit integer ($0 \le n \le 9$). * Calculate the sum of the known digits: $$	ext{Sum} = 6 + 1 + 3 + 5 + n + 2$$ $$	ext{Sum} = 17 + n$$ * For the entire number to be divisible by $9$, the expression $(17 + n)$ must equal a multiple of $9$. * The multiples of $9$ are $9, 18, 27, 36, \dots$ * Since $n$ must be a positive, single-digit value, the closest multiple of $9$ greater than or equal to $17$ is $18$. * Set up the equation to find the least value: $$17 + n = 18$$ $$n = 18 - 17 = 1$$ ### Exam Strategy & Shortcut **Casting out Nines:** Ignore any digits that sum to $9$ while adding. In $6135n2$: $(6+3)=9$ (cast out). Remaining digits: $1 + 5 + 2 = 8$. We need $8 + n$ to equal $9$. Therefore, $n = 1$. This saves precious calculation time. ### Common Pitfall A common mistake is skipping over zero when checking for least values. Always remember $0$ is a valid digit. If the sum had been exactly $18$ without $n$, the least value for $n$ would be $0$, not $9$. However, here $17+n=18$, so $n=1$. ### Final Answer Therefore, the correct answer is **1**.

Question 11

Find the multiple of 11 in the following numbers.

Accepted Answer

### Concept & Formula The rule for divisibility by $11$ states that the difference between the sum of digits at odd positions and the sum of digits at even positions must be either $0$ or a multiple of $11$. ### Step-by-Step Solution Let us systematically test each option by applying the divisibility rule: * **Option (a) 112144:** Sum of digits at odd places: $1 + 2 + 4 = 7$ Sum of digits at even places: $1 + 1 + 4 = 6$ Difference: $7 - 6 = 1$ (Not divisible by $11$) * **Option (b) 447355:** Sum of digits at odd places: $4 + 7 + 5 = 16$ Sum of digits at even places: $4 + 3 + 5 = 12$ Difference: $16 - 12 = 4$ (Not divisible by $11$) * **Option (c) 869756:** Sum of digits at odd places: $8 + 9 + 5 = 22$ Sum of digits at even places: $6 + 7 + 6 = 19$ Difference: $22 - 19 = 3$ (Not divisible by $11$) * **Option (d) 978626:** Sum of digits at odd places: $9 + 8 + 2 = 19$ Sum of digits at even places: $7 + 6 + 6 = 19$ Difference: $19 - 19 = 0$ (Divisible by $11$) ### Exam Strategy & Shortcut Do not write out the sums during the exam. Mentally track a running total by adding and subtracting alternating digits from left to right. For option (d): $9 - 7 + 8 - 6 + 2 - 6 = 0$. Since the final result is $0$, it is perfectly divisible by $11$. ### Common Pitfall A common error is grouping adjacent pairs of digits instead of correctly identifying strictly alternating positions, which leads to incorrect sums and false conclusions. ### Final Answer Therefore, the correct answer is **978626**.

Question 12

111,111,111,111 is divisible by

Accepted Answer

### Concept & Logic Numbers formed by repeating identical digits (repunits) have specific divisibility properties based on their length. A $6$-digit repunit (like $111,111$) is always a multiple of $7$, $11$, and $13$, because it equals $111 	imes 1001$. ### Step-by-Step Solution * **Divisibility by 3:** The sum of the digits is $12 	imes 1 = 12$, which is divisible by $3$. * **Divisibility by 11:** The difference between the sum of digits in odd places ($6$) and even places ($6$) is $0$, which is divisible by $11$. * **Divisibility by 37 and 111:** Any $3$-digit repunit ($111$) is divisible by $37$ ($37 	imes 3 = 111$). Since our $12$-digit number is constructed from blocks of $111$, it is inherently divisible by both $37$ and $111$. * **Divisibility by 1001:** The number $1001$ equals $7 	imes 11 	imes 13$. A $6$-digit block of $1$s ($111,111$) is mathematically $111 	imes 1001$, meaning it is divisible by $1001$. Because $12$ is a multiple of $6$, a $12$-digit block of $1$s is also divisible by $1001$. Since all conditions hold, the number is divisible by all listed factors. ### Exam Strategy & Shortcut Memorize the core factorization rule: $1001 = 7 	imes 11 	imes 13$. Any number that repeats a $3$-digit block twice (like $abcabc$) is divisible by $1001$. Since $111,111$ repeats a $3$-digit block, it is divisible by $1001$. The $12$-digit version simply scales this up. This immediately confirms option (d). ### Common Pitfall Students often stop testing after finding that the number is divisible by $3$, $11$, and $37$, mistakenly selecting an incomplete option like (b) or (c) because they fail to test for the larger $1001$ block property. ### Final Answer Therefore, the correct answer is **3, 11, 37, 111 and 1001**.

Question 13

Which of the following numbers is not divisible by 18?

Accepted Answer

### Concept & Strategy For a number to be divisible by a composite number like $18$, it must be divisible by its co-prime factors, which are $2$ and $9$. $$18 = 2 	imes 9$$ Therefore, the number must be even (divisible by $2$) and the sum of its digits must be a multiple of $9$. ### Step-by-Step Solution We are looking for the number that is **not** divisible by $18$. The easiest way to break this is to find a number that fails the simplest divisibility rule first. * **Test for divisibility by 2:** A number is divisible by $2$ if its unit digit is $0, 2, 4, 6, 	ext{or} 8$. * **Examine the options:** (a) $34056$ ends in $6$ (Even - Divisible by $2$) (b) $50436$ ends in $6$ (Even - Divisible by $2$) (c) $54036$ ends in $6$ (Even - Divisible by $2$) (d) $65043$ ends in $3$ (Odd - **Not divisible by 2**) Since $65043$ is an odd number, it cannot possibly be a multiple of an even number like $18$. We do not even need to check the rule for $9$. ### Exam Strategy & Shortcut Always check divisibility by $2$ or $5$ first before checking sums ($3$ or $9$) because they only require looking at the final digit. This visual scan takes less than two seconds, allowing you to bypass tedious addition. ### Common Pitfall The most common trap is ignoring the unit digit and immediately adding up the sum of the digits for all four options. While $65043$ ($6+5+0+4+3=18$) is technically divisible by $9$, doing this calculation wastes precious exam time. ### Final Answer Therefore, the correct answer is **65043**.

Question 14

The number 89715938* is divisible by 4. The unknown non-zero digit marked as * will be

Accepted Answer

### Concept & Formula The divisibility rule for $4$ dictates that a number is divisible by $4$ if and only if the number formed by its last two digits is divisible by $4$. ### Step-by-Step Solution * **Given:** The number is $89715938*$. The last two digits form the number $8*$, where $*$ represents a single digit. * **Calculation:** We need to find which option can replace $*$ such that $8*$ is a multiple of $4$. Let us evaluate the given choices: If $* = 2$, the number is $82$. ($82 / 4 = 20.5 ightarrow$ Not divisible) If $* = 3$, the number is $83$. (Odd number $ightarrow$ Not divisible) If $* = 4$, the number is $84$. ($84 / 4 = 21 ightarrow$ **Divisible**) If $* = 6$, the number is $86$. ($86 / 4 = 21.5 ightarrow$ Not divisible) Only substituting $4$ creates a valid multiple. ### Exam Strategy & Shortcut You can eliminate odd numbers instantly. Since $4$ is even, any multiple of $4$ must also be even, eliminating option (b). Furthermore, if the tens digit is even (like $8$), the unit digit MUST be $0, 4, 	ext{or} 8$ to be a multiple of $4$. This immediately singles out $4$ as the only mathematically sound option. ### Common Pitfall Some test-takers confuse the rule for $4$ with the rule for $8$ (last three digits) or $3$ (sum of all digits). Trying to sum the entire string of $89715938*$ will waste immense time and lead to a completely incorrect framework. ### Final Answer Therefore, the correct answer is **4**.

Question 15

Which one of the following numbers is divisible by 3?

Accepted Answer

### Concept & Strategy The fundamental rule for divisibility by $3$ states that a number is divisible by $3$ if the sum of its individual digits is a multiple of $3$. ### Step-by-Step Solution Let us test the sum of the digits for each provided option to find the valid multiple of $3$: * **Option (a) 4006020:** Sum: $4 + 0 + 0 + 6 + 0 + 2 + 0 = 12$ Since $12$ is perfectly divisible by $3$ ($12 / 3 = 4$), this entire number is divisible by $3$. * **Option (b) 2345678:** Sum: $2 + 3 + 4 + 5 + 6 + 7 + 8 = 35$ $35$ is not a multiple of $3$. * **Option (c) 2876423:** Sum: $2 + 8 + 7 + 6 + 4 + 2 + 3 = 32$ $32$ is not a multiple of $3$. * **Option (d) 9566003:** Sum: $9 + 5 + 6 + 6 + 0 + 0 + 3 = 29$ $29$ is not a multiple of $3$. ### Exam Strategy & Shortcut Use the "digit cancellation" or "casting out 3s" method to calculate faster. Instead of adding all digits, cross out any $3, 6, 9$, or combinations of digits that add to a multiple of $3$. For option (a): $4006020$. Cross out $6$. You are left with $4$ and $2$. $4 + 2 = 6$, which is a multiple of $3$. This guarantees the number is divisible without performing long addition. ### Common Pitfall The main risk here is simple arithmetic errors when adding up long strings of digits (like in option b or c). Using the cancellation shortcut prevents these basic addition mistakes. ### Final Answer Therefore, the correct answer is **4006020**.

Question 16

A number is divisible by 11 if the difference between the sums of the digits in odd and even places respectively is

Accepted Answer

### Concept & Logic The divisibility rule for 11 relies on the alternating sum of its digits. For any integer to be divisible by 11, the difference between the sum of the digits in its odd positions and the sum of the digits in its even positions must be exactly 0 or a multiple of 11 (such as 11, 22, 33). ### Step-by-Step Solution This is a direct theoretical question testing knowledge of foundational divisibility rules. * The question states: "A number is divisible by 11 if the difference between the sums of the digits in odd and even places respectively is..." * According to the standard mathematical rule for divisibility by 11, this difference must equate to 0, 11, 22, etc. * Looking at the options, only option (d) correctly identifies this requirement. ### Exam Strategy & Shortcut There is no calculation required here. Memorizing the fundamental divisibility rules for numbers 2 through 11 is mandatory for competitive exams, as it allows you to answer definitive theoretical questions in under five seconds. ### Common Pitfall A common mistake is confusing this rule with the rule for 3 or 9, where the sum of *all* digits is considered. Test-takers might wrongly guess options related to 3 if they misremember the rule structure. ### Final Answer Therefore, the correct answer is **zero or a multiple of 11**.

Question 17

Which one of the following numbers is divisible by 11?

Accepted Answer

### Concept & Strategy To determine if a number is divisible by 11, we use the alternating sum method. Find the sum of digits in the odd positions (1st, 3rd, 5th, etc.) and the sum of digits in the even positions (2nd, 4th, 6th, etc.). The number is divisible by 11 if the difference between these two sums is 0 or a multiple of 11. ### Step-by-Step Solution Let us test the options systematically: * **Option (a) 4823718:** Odd places sum: $8 + 7 + 2 + 4 = 21$ Even places sum: $1 + 3 + 8 = 12$ Difference: $21 - 12 = 9$ (Not a multiple of 11) * **Option (b) 4832718:** Odd places sum: $8 + 7 + 3 + 4 = 22$ Even places sum: $1 + 2 + 8 = 11$ Difference: $22 - 11 = 11$ (This is a multiple of 11) Since we found the correct answer, there is no need to fully calculate the rest during an exam, but for completion: * Option (c) yields a difference of 17. * Option (d) yields a difference of 19. ### Exam Strategy & Shortcut Instead of writing out separate addition equations, use a mental "running difference" technique moving from right to left (or left to right). For option (b): $8 - 1 + 7 - 2 + 3 - 8 + 4 = 7 + 5 - 5 + 4 = 11$. This keeps numbers small and prevents arithmetic errors when dealing with large totals. ### Common Pitfall The most frequent error is misaligning the odd and even positions, especially in numbers with repeating digits or similar sequences. Always double-check your initial grouping before adding. ### Final Answer Therefore, the correct answer is **4832718**.

Question 18

Which one of the following numbers is divisible by 15?

Accepted Answer

### Concept & Strategy For a number to be divisible by a composite number like 15, it must be divisible by its co-prime factors, which are 3 and 5. $$15 = 3 	imes 5$$ Therefore, the number must end in 0 or 5 (divisibility by 5), and the sum of its digits must be a multiple of 3 (divisibility by 3). ### Step-by-Step Solution * **Check divisibility by 5:** All four options (17325, 23755, 29515, 30560) end in either 5 or 0. Thus, all are divisible by 5. We must rely on the rule for 3. * **Check divisibility by 3 (Sum of digits):** * Option (a) 17325: $1 + 7 + 3 + 2 + 5 = 18$. (18 is divisible by 3). * Option (b) 23755: $2 + 3 + 7 + 5 + 5 = 22$. (Not divisible by 3). * Option (c) 29515: $2 + 9 + 5 + 1 + 5 = 22$. (Not divisible by 3). * Option (d) 30560: $3 + 0 + 5 + 6 + 0 = 14$. (Not divisible by 3). Only option (a) satisfies both conditions. ### Exam Strategy & Shortcut Use the "digit cancellation" method for the rule of 3 to save time. In option (a) 17325, cancel out the 3 immediately. You are left with 1, 7, 2, and 5. $7 + 2 = 9$ (cancel it). $1 + 5 = 6$ (cancel it). Everything cancels out, meaning it is a perfect multiple of 3 without doing a single large addition. ### Common Pitfall A common mistake is trying to divide large numbers directly by 15. This is highly time-consuming and prone to long-division arithmetic errors. Always break composite numbers down into co-prime rules. ### Final Answer Therefore, the correct answer is **17325**.

Question 19

7386038 is divisible by

Accepted Answer

### Concept & Strategy To find the correct divisor for a large number, apply basic divisibility rules one by one based on the given options, starting with the simplest rules (like 4) and moving to sums (like 3 and 9) or alternating sums (like 11). ### Step-by-Step Solution Let us test the number 7386038 against each option: * **Option (b) Check for 4:** A number is divisible by 4 if its last two digits are divisible by 4. The last two digits are 38. $38 / 4 = 9.5$. Not divisible by 4. * **Option (a) Check for 3 & Option (c) Check for 9:** A number is divisible by 3 or 9 if the sum of its digits is a multiple of 3 or 9. Sum: $7 + 3 + 8 + 6 + 0 + 3 + 8 = 35$. 35 is neither a multiple of 3 nor 9. * **Option (d) Check for 11:** Difference between sum of digits in odd and even positions. Odd places sum (right to left): $8 + 0 + 8 + 7 = 23$. Even places sum (right to left): $3 + 6 + 3 = 12$. Difference: $23 - 12 = 11$. Since the difference is 11, the number is divisible by 11. ### Exam Strategy & Shortcut Always check the rule for 4 first because it only requires looking at the last two digits, taking less than two seconds. If you check 3 and it fails (sum is 35), you can instantly eliminate 9 as well, because any number divisible by 9 MUST be divisible by 3. This dual-elimination saves significant time. ### Common Pitfall Test-takers often recalculate the digit sum for 9 after already finding it for 3. Remember the hierarchical nature of multiples: if it fails the 3 test, do not waste time testing the 9 test. ### Final Answer Therefore, the correct answer is **11**.

Question 20

Consider the following statements:  The numbers 24984, 26784 and 28584 are  (1) divisible by 3  (2) divisible by 4  (3) divisible by 9  Which of these are correct?

Accepted Answer

### Concept & Strategy This question requires verifying three separate divisibility rules across three different numbers. * Rule for 4: The last two digits form a number divisible by 4. * Rule for 3: The sum of digits is divisible by 3. * Rule for 9: The sum of digits is divisible by 9. ### Step-by-Step Solution Let us evaluate the statements systematically. * **Test Statement (2): Divisible by 4** Look at the last two digits of all three numbers: 24984, 26784, 28584. They all end in 84. Since 84 is a multiple of 4 ($84 / 4 = 21$), all three numbers are perfectly divisible by 4. Statement (2) is correct. * **Test Statements (1) and (3): Divisible by 3 and 9** We need to find the sum of the digits for each number. * For 24984: $2 + 4 + 9 + 8 + 4 = 27$ * For 26784: $2 + 6 + 7 + 8 + 4 = 27$ * For 28584: $2 + 8 + 5 + 8 + 4 = 27$ The sum for all three numbers is exactly 27. Since 27 is divisible by 3, statement (1) is correct. Since 27 is divisible by 9, statement (3) is correct. Because statements (1), (2), and (3) are all valid, the correct option encompasses all of them. ### Exam Strategy & Shortcut Notice the pattern in the given numbers: 24984, 26784, 28584. The only digit changing is the thousands digit, which increases by 2 each time (4 to 6 to 8), while the hundreds digit decreases by 2 each time (9 to 7 to 5). This means the sum of the digits remains absolutely constant. You only need to calculate the sum for the very first number (27) to know the sum for all three, saving you from doing three separate additions. ### Common Pitfall Students often check only one number (e.g., just 24984) and assume the rest follow suit without verifying if the properties hold true for all items in the list. While it worked out here due to the hidden arithmetic pattern, blindly assuming can lead to trap answers. ### Final Answer Therefore, the correct answer is **1, 2 and 3**.

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