answer:To find all natural numbers ( n geq 3 ) such that [ d(n-1) + d(n) + d(n+1) leq 8, ] we need to examine the properties of the divisor function ( d ) closely. We'll check even and odd ( n ) separately and consider numbers divisible by 3. 1. **Considering even ( n geq 6 )**: - For any even number ( k geq 6 ), there are at least four divisors: 1, 2, ( frac{k}{2} ), and ( k ) itself. Hence, ( d(k) geq 4 ). 2. **Checking small values of ( n )**: - For ( n = 3 ): [ d(2) = 2,, d(3) = 2,, d(4) = 3 ] [ d(2) + d(3) + d(4) = 2 + 2 + 3 = 7 ] which satisfies the condition since ( 7 leq 8 ). Hence, ( n = 3 ) is a solution. - For ( n = 5 ): [ d(4) = 3,, d(5) = 2,, d(6) = 4 ] [ d(4) + d(5) + d(6) = 3 + 2 + 4 = 9 ] which does not satisfy the condition since ( 9 notleq 8 ). Hence, ( n = 5 ) is not a solution. 3. **For odd ( n geq 7 )**: - Given ( n-1 ) is even, ( d(n-1) geq 4 ). - Both even numbers ( n-1 ) and ( n+1 ) provide at least 4 divisors each. - Combining these, ( d(n-1) + d(n) + d(n+1) geq 4 + d(n) + 4 ). - This results in [ d(n-1) + d(n) + d(n+1) geq 4 + d(n) + 4 > 8 ] for any ( d(n) geq 1 ). Hence, no odd ( n geq 7 ) satisfies the condition. 4. **Considering even ( n geq 6 )**: - When ( n geq 8 ): - If ( n-1 ) is divisible by 3, then ( d(n-1) geq 3 ). - Hence, [ d(n-1) + d(n) + d(n+1) geq 3 + 4 + d(n+1) > 8 ] - Similarly, if ( n+1 ) is divisible by 3, [ d(n-1) + d(n) + d(n+1) geq d(n-1) + 4 + 3 > 8 ] 5. **Only remaining case when ( n ) is divisible by 6**: - Let ( n = 6k geq 12 ): - For ( n geq 12 ), which is divisible by 6, the divisors include 1, 2, 3, ( frac{n}{3} ), ( frac{n}{2} ), and ( n ), giving at least 6 divisors: ( d(n) geq 6 ). - Hence, [ d(n-1) + d(n) + d(n+1) geq d(n-1) + 6 + d(n+1) > 8 ] Therefore, there are no more possible solutions for ( n geq 12 ). # Conclusion: Summarizing the computations, we find the only solutions for ( n geq 3 ) are ( n = 3, 4, ) and ( 6 ). [ boxed{n = 3, 4, 6} ]

answer:1. **Define Variables and Given Information:** - Let ( n ) be the number of coins initially in the purse. - The average value of the coins is given as 17 cents. - Therefore, the total value of the ( n ) coins is ( 17n ) cents. 2. **Formulate the Value After Removing a Coin:** - When a penny (1 cent) is removed, the number of coins becomes ( n-1 ). - The new average value becomes 18 cents. - Thus, the new total value of coins is ( 18(n-1) ) cents. 3. **Set Up the Equation:** - The total value of the coins before removing the penny is ( 17n ) cents. - After removing one penny, the total value decreases by 1 cent, making it: [ 17n - 1 = 18(n - 1) ] 4. **Solve for ( n ):** - Expand and simplify the equation: [ 17n - 1 = 18n - 18 ] [ -1 + 18 = 18n - 17n ] [ 17 = n ] - So, ( n = 17 ). Therefore, the original collection of coins has 17 coins. 5. **Determine the Total Value:** - The total value of these 17 coins is: [ 17 times 17 = 289 text{ cents} ] 6. **Analyze the Composition of Coins:** - We know the coin values end in a 4 cents difference after removing a penny. Therefore, aside from pennies, the total value without pennies must be divisible by 5. - The removed pennies must leave a remainder divisible by 5. Considering the properties, we conclude there are at least 4 pennies (i.e., ( 4 text{ pennies} )): 7. **Recalculate without Pennies:** - This leaves ( 17 - 4 = 13 ) coins. - The remaining value is ( 289 text{ cents} - 4 text{ cents} = 285 text{ cents} ). 8. **Investigate Number of Quarters:** - ( 13 ) coins: Considering the higher valued quarters because they fit our reduced total in more multiples: - ( 11 text{ quarters} ), ( 1 text{ dime} ), and ( 1 text{ nickel} ). 9. **Verify our Counts:** - ( 11 text{ quarters} = 11 times 25 = 275 text{ cents} ). - ( 1 times 10 = 10 text{ cents} ). - ( 2 times 5 = 10 text{ cents}). - Therefore, (boxed{(11 times 25) + (2times 5) + (4times1)} = 275+10+4 = 289.) # Conclusion: Thus, there must be (boxed{2}) nickels in the purse. [ boxed{A} ]

answer:Let's denote the weight of the empty bag as ( E ) and the weight of the marbles as ( M ). We know that the total weight of the bag with the marbles is 3.4 kg, so we can write the following equation: [ E + M = 3.4 ] When 1/5 of the marbles are taken out, the weight of the marbles in the bag is reduced by 1/5 of ( M ). The new weight of the marbles in the bag is ( M - frac{1}{5}M ) or ( frac{4}{5}M ). The new total weight of the bag with the remaining marbles is 2.98 kg, so we can write another equation: [ E + frac{4}{5}M = 2.98 ] Now we have a system of two equations: 1. ( E + M = 3.4 ) 2. ( E + frac{4}{5}M = 2.98 ) We can solve this system to find the value of ( E ). Let's first express ( M ) from the first equation: [ M = 3.4 - E ] Now we can substitute ( M ) in the second equation: [ E + frac{4}{5}(3.4 - E) = 2.98 ] Expanding the equation, we get: [ E + frac{4}{5} times 3.4 - frac{4}{5}E = 2.98 ] [ E + 2.72 - frac{4}{5}E = 2.98 ] Now, let's combine like terms by subtracting ( frac{4}{5}E ) from ( E ): [ frac{5}{5}E - frac{4}{5}E = 2.98 - 2.72 ] [ frac{1}{5}E = 0.26 ] To find ( E ), we multiply both sides by 5: [ E = 0.26 times 5 ] [ E = 1.3 ] So, the weight of the empty bag is boxed{1.3} kg.

answer:1. **Calculate the percentage of students scoring 98 points**: Given that 15% + 30% + 25% + 10% = 80% of participants have mentioned scores, the remaining percentage who scored 98 points is 100% - 80% = 20%. 2. **Assuming 100 students**: The distribution based on 100 students would be: - 15% of 100 students = 15 students scored 72 points - 30% of 100 students = 30 students scored 84 points - 25% of 100 students = 25 students scored 86 points - 10% of 100 students = 10 students scored 92 points - 20% of 100 students = 20 students scored 98 points 3. **Determine the median score**: Organizing the scores in order, since there are 100 students, the median is the average of the 50^{text{th}} and 51^{text{st}} scores: - Ordered scores include: 15 scores of 72, 30 scores of 84, 25 scores of 86, 10 scores of 92, 20 scores of 98 - Thus, both the 50^{text{th}} and 51^{text{st}} scores are 86 (as these encompass scores from 46th position to 70th). Therefore, the median score is 86. 4. **Calculate the mean score**: [ text{Mean} = frac{72 times 15 + 84 times 30 + 86 times 25 + 92 times 10 + 98 times 20}{100} = frac{1080 + 2520 + 2150 + 920 + 1960}{100} = frac{8630}{100} = 86.3 ] 5. **Difference between the mean and the median**: [ 86.3 - 86 = 0.3 ] Therefore, the difference between the mean and median score on this competition is 0.3. The final answer is boxed{B. 0.3}