Question 1

The sum of the digits of a 3-digit number is subtracted from the number. The resulting number is always

Accepted Answer

### Concept & Formula Any generic number can be expanded algebraically into its base-10 place values. This is the key to proving universal divisibility rules for digits. $$ 	ext{A 3-digit number} = 100x + 10y + z $$ $$ 	ext{Where } x, y, 	ext{ and } z 	ext{ are the hundreds, tens, and units digits respectively.} $$ ### Step-by-Step Solution * **Step 1: Set up the original number.** Let our 3-digit number be represented as $100x + 10y + z$. * **Step 2: Set up the sum of the digits.** The literal sum of its digits is simply $(x + y + z)$. * **Step 3: Perform the subtraction.** The problem states to subtract the digit sum from the original number: * Result $= (100x + 10y + z) - (x + y + z)$ * Combine like terms: Result $= 100x - x + 10y - y + z - z$ * Result $= 99x + 9y$ * **Step 4: Factor the result.** * Result $= 9(11x + y)$ * **Conclusion:** Because the final algebraic expression has a global factor of 9, the resulting number will always be a perfect multiple of 9, regardless of what the original digits were. ### Exam Strategy & Shortcut **Pick a real number:** Don't waste time setting up algebra if you can do quick mental math. Pick any random 3-digit number, for example, 123. Sum of its digits $= 1 + 2 + 3 = 6$. Subtract the sum from the number: $123 - 6 = 117$. Check the options: Is 117 divisible by 9? Check the digit sum of 117: $1 + 1 + 7 = 9$. Yes, it is perfectly divisible by 9. ### Common Pitfall Students sometimes randomly guess divisibility by 6 (option d) because they pick a test number like 111 (Sum = 3. $111 - 3 = 108$. 108 is divisible by both 9 and 6). You must pick a number that breaks the false patterns. 123 yields 117, which is NOT divisible by 6, proving option (d) false and leaving option (b) as the universal truth. ### Final Answer Therefore, the correct answer is **divisible by 9**.

Question 2

Find the least 6-digit number which is exactly divisible by 349.

Accepted Answer

### Concept & Logic Similar to finding a 4-digit number, to find the least $6$-digit number divisible by a divisor $D$, we start with the smallest theoretical $6$-digit base number, which is $100000$. We divide this base by the divisor to find the remainder. Because subtracting the remainder would drop us down to a $5$-digit number, we must *add* the difference between the divisor and the remainder to our base. $$ 	ext{Required} = 100000 + (	ext{Divisor} - 	ext{Remainder}) $$ ### Step-by-Step Solution * **Step 1:** Set up the long division of the smallest 6-digit number by 349. $100000 \div 349$ * **Step 2:** Execute the division. $1000 \div 349 \approx 2$. ($349 	imes 2 = 698$) Remainder $= 1000 - 698 = 302$ Bring down a $0$ to make $3020$. $3020 \div 349 \approx 8$. ($349 	imes 8 = 2792$) Remainder $= 3020 - 2792 = 228$ Bring down the final $0$ to make $2280$. $2280 \div 349 \approx 6$. ($349 	imes 6 = 2094$) Remainder $= 2280 - 2094 = 186$ * **Step 3:** Calculate the amount required to reach the next perfect multiple. Number to add $= 	ext{Divisor} - 	ext{Remainder}$ Number to add $= 349 - 186 = 163$ * **Step 4:** Add this to the base number. $100000 + 163 = 100163$ ### Exam Strategy & Shortcut **Unit Digit Check:** Look closely at the options. We are looking for a multiple of $349$ that is just above $100000$. Our quotient was roughly $286$ (giving a remainder). The next multiple must be $349 	imes 287$. Look at the unit digits: $9 	imes 7 = 63$. Therefore, the correct answer MUST end in a $3$. Options (a), (b), and (c) all end in $63$. Let's look at the magnitude. $349 	imes 1000 \approx 349000$. $160000$ (Option c) is way too large. $101000$ (Option b) is also likely too large since $349$ is a relatively small jump from $100000$. $100163$ (Option a) represents a jump of only $163$, which is less than the divisor $349$, perfectly matching the logic of finding the *least* number. ### Common Pitfall Long division with 3-digit divisors is prone to arithmetic errors. A simple mistake in subtraction (e.g., $3020 - 2792$) cascades and guarantees an incorrect remainder, subsequently ruining the final addition. Take an extra 10 seconds to verify your subtraction steps during the long division. ### Final Answer **Therefore, the correct answer is 100163.**

Question 3

The smallest 6-digit number exactly divisible by 111 is

Accepted Answer

### Concept & Formula To find the smallest $n$-digit number perfectly divisible by a divisor $D$, start with the smallest possible $n$-digit base number. Find the remainder upon division, and add the difference between the divisor and the remainder to bridge the gap to the next multiple. $$ 	ext{Required Number} = 	ext{Base} + (	ext{Divisor} - 	ext{Remainder}) $$ ### Step-by-Step Solution * **Step 1:** Identify the absolute smallest 6-digit number. The base number is $100000$. * **Step 2:** Execute long division to find the remainder. $100000 \div 111$ We know $111 	imes 9 = 999$. So, $111 	imes 900 = 99900$. Subtracting this from the base: $100000 - 99900 = 100$. The remainder is $100$. * **Step 3:** Calculate the amount required to reach the next perfect multiple. Number to add $= 	ext{Divisor} - 	ext{Remainder}$ Number to add $= 111 - 100 = 11$. * **Step 4:** Add this value to the original base number. $100000 + 11 = 100011$. ### Exam Strategy & Shortcut **Option Elimination:** Look at the options and apply the divisibility rule for 3 (since 111 is divisible by 3, the answer must also be). (a) $111111 ightarrow$ Sum is 6 (Divisible, but is it the smallest?) (b) $110011 ightarrow$ Sum is 4 (Not divisible) (c) $100011 ightarrow$ Sum is 3 (Divisible) (d) $110101 ightarrow$ Sum is 4 (Not divisible) Comparing the valid candidates, $100011$ is significantly smaller than $111111$. A quick mathematical check ($100011 = 99900 + 111$) confirms it is built from exact multiples. ### Common Pitfall Students often incorrectly assume a number like $111111$ is the smallest simply because it repeats the digits of the divisor and feels conceptually "clean". Never assume visual patterns bypass mathematical rules; always start your logic from the fundamental numerical base $100000$. ### Final Answer **Therefore, the correct answer is 100011.**

Question 4

$(7^{19} + 2)$ is divided by $6$. The remainder is

Accepted Answer

### Concept & Logic This problem uses the **Additive Property of Remainders**. If you are dividing a sum of two terms by a divisor, you can find the remainder of each term individually and then add those remainders together. $$ (A + B) \pmod D = [A \pmod D + B \pmod D] \pmod D $$ ### Step-by-Step Solution * **Given:** * Expression: $7^{19} + 2$ * Divisor: $6$ * **Calculation:** * Break the expression into two parts and divide each by $6$. * **Part 1:** Find the remainder of $7^{19} \div 6$. * Since $7$ is exactly $1$ more than the divisor $6$, we can write $7 = 6 + 1$. * $7^{19} \pmod 6 = (6 + 1)^{19} \pmod 6$. * As per the remainder theorem, this simply becomes $1^{19} = 1$. * **Part 2:** Find the remainder of $2 \div 6$. * Since $2$ is smaller than $6$, the remainder is simply $2$. * **Combine:** Add the individual remainders together. $$ 	ext{Total Remainder} = 1 + 2 = 3 $$ ### Exam Strategy & Shortcut Whenever the base is exactly $1$ more than the divisor (like $7$ and $6$), the remainder of that term to *any* power is always exactly $1$. You can mentally skip all the math: $1 + 2 = 3$. This should take no more than 3 seconds to solve on test day. ### Common Pitfall A careless mistake is trying to evaluate the power or add the $2$ to the base before dealing with the exponent (e.g., falsely thinking it acts like $9^{19}$). Always isolate terms separated by a plus or minus sign when applying the remainder theorem. ### Final Answer **Therefore, the correct answer is 3.**

Question 5

$n$ being any odd number greater than 1, $n^{65} - n$ is always divisible by

Accepted Answer

### Concept & Logic This problem utilizes properties of consecutive numbers and factorization. The key insight is that for any odd number $n$, $(n-1)$ and $(n+1)$ are consecutive even numbers, guaranteeing divisibility by higher powers of 2. $$n^{65} - n = n(n^{64} - 1)$$ ### Step-by-Step Solution * **Calculation:** First, factor out $n$ from the expression: $$n^{65} - n = n(n^{64} - 1)$$ * Using the difference of squares, expand $(n^{64} - 1)$ recursively: $$n(n^{32} + 1)(n^{16} + 1)(n^8 + 1)(n^4 + 1)(n^2 + 1)(n - 1)(n + 1)$$ * Look closely at the terms $n(n - 1)(n + 1)$. This can be rearranged as $(n - 1)n(n + 1)$, which is the product of three consecutive integers. * The product of any three consecutive integers is always divisible by $3! = 6$. * Since $n$ is explicitly given as an **odd number**, $(n - 1)$ and $(n + 1)$ are consecutive **even numbers**. * Among any two consecutive even numbers, one is a multiple of 2 and the other is a multiple of 4. Therefore, their product $(n - 1)(n + 1)$ is always divisible by $2 	imes 4 = 8$. * Combining these facts: the term $(n-1)n(n+1)$ is divisible by 3 (from being 3 consecutive integers) and divisible by 8 (from the even numbers property). * Since the lowest common multiple of 3 and 8 is 24, the entire expression is always divisible by 24. ### Exam Strategy & Shortcut The fastest method for "always divisible by" questions with variables is **Value Substitution**. * Pick the smallest valid odd number greater than 1: let $n = 3$. * Evaluate the expression for the first few terms: $n(n-1)(n+1) = 3(2)(4) = 24$. * Since 24 must be a factor of the full expression $3^{65}-3$, check the options. 24 is directly present. (Note: While 5 is also a factor due to Fermat's Little Theorem, 24 is the intended answer because it directly relies on the specific "odd number" constraint provided in the prompt). ### Common Pitfall A major pitfall is picking $n=1$ to test, which gives $1^{65} - 1 = 0$, leading to confusion since 0 is divisible by everything. Always read constraints carefully: the question specifies $n > 1$. ### Final Answer Therefore, the correct answer is **24**.

Question 6

The least number of five digit is exactly divisible by 88 is

Accepted Answer

## Concept & Logic This problem is based on the concept of finding a number within a specific range that is completely divisible by a given divisor. To find the least $n$-digit number divisible by a divisor $d$: 1. Identify the smallest $n$-digit number. 2. Divide it by $d$ to find the remainder. 3. Add the difference of the divisor and the remainder $(d - r)$ to the smallest $n$-digit number. ## Step-by-Step Solution * **Given:** We need the least 5-digit number divisible by $88$. * The smallest 5-digit number is $10000$. * **Calculation:** Divide $10000$ by $88$. $$10000 \div 88$$ $88 	imes 113 = 9944$ * The remainder ($r$) is: $$10000 - 9944 = 56$$ * **Deduction:** If we subtract $56$ from $10000$, the number becomes a 4-digit number ($9944$), which is divisible by $88$. * To get the next multiple of $88$ (which will be the smallest 5-digit multiple), we must add the divisor $88$ to $9944$: $$9944 + 88 = 10032$$ * Alternatively, use the standard formula: Least Number + (Divisor - Remainder) $$10000 + (88 - 56) = 10000 + 32 = 10032$$ ## Exam Strategy & Shortcut Use the **Divisibility Rule Option Elimination** method. A number is divisible by $88$ if it is divisible by both $8$ and $11$. * **Rule of 8:** The last 3 digits must be divisible by 8. * (a) $032 \div 8 = 4$ (Divisible) * (b) $132 \div 8 = 16.5$ (Not divisible) * (c) $088 \div 8 = 11$ (Divisible) * (d) $023$ (Not even, so not divisible) * We are left with (a) $10032$ and (c) $10088$. * The question asks for the *least* number. $10032$ is smaller than $10088$. * Verify $10032$ with the **Rule of 11** (Difference of alternating sums is 0 or 11): $(1+0+2) - (0+3) = 3 - 3 = 0$. It works! ## Common Pitfall A very common mistake is simply subtracting the remainder ($56$) from $10000$ and arriving at $9944$, then forgetting that $9944$ is a 4-digit number. Always remember to add the divisor back or use the $(d - r)$ addition rule. ## Final Answer Therefore, the correct answer is 10032.

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