answer：To find the original average of the sequence, let's denote the first number of the original sequence as ( x ). Since the numbers are consecutive integers, the sequence can be represented as:( x, x+1, x+2, x+3, x+4, x+5, x+6, x+7, x+8, x+9 )The sum of these 10 numbers is:( S = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7) + (x+8) + (x+9) )Simplifying, we get:( S = 10x + (1+2+3+4+5+6+7+8+9) )( S = 10x + 45 )The average of the original sequence is ( S/10 ), which is ( x + 4.5 ).Next, consider the new sequence after the deductions:( x-9, x+1-8, x+2-7, x+3-6, x+4-5, x+5-4, x+6-3, x+7-2, x+8-1, x+9 )The sum of these numbers is:( S' = (x-9) + (x-7) + (x-5) + (x-3) + (x-1) + (x+1) + (x+3) + (x+5) + (x+7) + (x+9) )Notice that the negative and positive numbers cancel each other out, so we are left with:( S' = 10x )The new average is given as 15.5, so:( frac{10x}{10} = 15.5 )( x = 15.5 )Now, we can find the original average by substituting ( x ) back into the original average formula:Original average = ( x + 4.5 )Original average = ( 15.5 + 4.5 )Original average = ( 20 )Therefore, the original average of the integers was boxed{20}.

answer：Okay, so I need to find the sum of the digits of the integer k, where k is equal to 10^100 minus 250. Hmm, that sounds a bit tricky, but let's break it down step by step.First, I know that 10^100 is a 1 followed by 100 zeros. That's a really big number! So, 10^100 looks like 100000...000 with 100 zeros in total. Now, I need to subtract 250 from this number. Subtracting 250 from such a large number might seem simple, but I have to be careful with the borrowing that will occur during the subtraction.Let me visualize this. If I have 10^100, which is 1 followed by 100 zeros, and I subtract 250, I'm essentially subtracting 250 from the last three digits of this number. But since the last three digits are all zeros, I'll need to borrow from the preceding digits. However, all the digits before the last three are also zeros, so I'll have to borrow from the very first digit, which is 1.This borrowing process will turn the first digit into 0 (but since we're dealing with a 100-digit number, the leading zero doesn't count, so effectively, the number becomes 999...99750). Wait, let me think about that again. If I have 10^100, which is 1 followed by 100 zeros, and I subtract 250, the result should be a number that has 99 nines, followed by 750. Is that right?Let me check with a smaller exponent to see if this pattern holds. For example, if I take 10^4, which is 10000, and subtract 250, I get 9750. That's 9750, which has two nines at the beginning. Wait, that's not 99 nines. Hmm, maybe my initial thought wasn't entirely accurate.Actually, when I subtract 250 from 10^100, I should consider how the borrowing affects each digit. Starting from the rightmost digit, I need to subtract 250, which means I'll have to borrow from the fourth digit from the end. But since all the digits in between are zeros, each borrow will turn a zero into a 9, except for the digit from which I'm borrowing, which will become 9 as well. This process will continue all the way to the first digit, which is 1. After borrowing, the first digit becomes 0, but since we don't write leading zeros, the number effectively becomes 999...99750, with 99 nines followed by 750.Now, to find the sum of the digits of this number, I need to add up all the digits. There are 99 nines, each contributing 9 to the sum, and then the last three digits are 7, 5, and 0. So, the total sum would be 99 times 9 plus 7 plus 5 plus 0.Calculating that, 99 times 9 is 891. Adding 7 gives 898, then adding 5 gives 903. The zero doesn't add anything. So, the sum of the digits is 903.Wait, let me double-check that. If I have 99 nines, that's 99 times 9, which is indeed 891. Then adding 7 and 5 gives 891 + 7 = 898, and 898 + 5 = 903. Yeah, that seems correct.I think I've got it. The sum of the digits of k, where k is 10^100 minus 250, is 903.

answer：First, I need to understand what it means for a number to be a square of a perfect square. A perfect square is a number that is the square of an integer, such as 1, 4, 9, 16, and so on. Therefore, a square of a perfect square would be the square of one of these perfect squares.Mathematically, if ( n ) is a perfect square, then ( n = k^2 ) for some integer ( k ). Consequently, the square of this perfect square would be ( (k^2)^2 = k^4 ). So, the problem is essentially asking for the sum of all positive integers less than 1000 that are fourth powers of integers.Next, I need to identify all integers ( k ) such that ( k^4 < 1000 ). I'll start by testing consecutive integers:- For ( k = 1 ): ( 1^4 = 1 ), which is less than 1000.- For ( k = 2 ): ( 2^4 = 16 ), which is less than 1000.- For ( k = 3 ): ( 3^4 = 81 ), which is less than 1000.- For ( k = 4 ): ( 4^4 = 256 ), which is less than 1000.- For ( k = 5 ): ( 5^4 = 625 ), which is less than 1000.- For ( k = 6 ): ( 6^4 = 1296 ), which is greater than 1000.Since ( 6^4 ) exceeds 1000, I stop here. The valid integers ( k ) are 1, 2, 3, 4, and 5.Now, I'll calculate the fourth powers of these integers:- ( 1^4 = 1 )- ( 2^4 = 16 )- ( 3^4 = 81 )- ( 4^4 = 256 )- ( 5^4 = 625 )Finally, I'll sum these values to find the total:[1 + 16 + 81 + 256 + 625 = 979]Therefore, the sum of all positive integers less than 1000 that are squares of perfect squares is 979.

answer：Okay, so Masha has six pies in total—three apple pies and three cherry pies. They got mixed up while she was carrying them home, and now she needs to separate them into two groups, each containing only one type of pie. She has this machine that can help her determine if there are different kinds of pies in a group by turning on a green light. The challenge is to figure out how to divide all the pies into two groups of the same kind using the machine only four times.First, I need to understand how the machine works. If Masha puts several pies into the machine and there are pies of different kinds among them, the green light will turn on. If all the pies are of the same kind, the light stays off. So, the machine essentially tells her whether a group has mixed pies or not. Since there are six pies, and she needs to split them into two groups of three each, both groups should be homogeneous—either all apple or all cherry. The key is to use the machine strategically to identify which pies are of the same kind without having to check every possible combination, which would take too many uses.Let me think about how to approach this. Maybe she can start by testing small groups and using the information from each test to inform the next one. For example, if she takes two pies and puts them in the machine, and the light doesn't turn on, she knows they're the same kind. If the light does turn on, she knows they're different. But since she has six pies, she needs a systematic way to test and eliminate possibilities. Maybe she can use a process of elimination, testing pairs or triplets and using the results to narrow down the types of pies.Wait, if she takes three pies and puts them in the machine, and the light doesn't turn on, that means all three are the same kind. If the light does turn on, there's a mix. But she wants to end up with two groups of three, so maybe she can use this method to identify one group of three first.Let me outline a possible strategy:1. **First Test:** Take any three pies and put them in the machine. - If the light is off, all three are the same kind. Let's say they're apple pies. Then the remaining three must be cherry pies. She's done in one use. - If the light is on, there's a mix. So, among these three, there are both apple and cherry pies.But the problem is that she might not be so lucky with the first test. If the first test results in a mix, she needs to figure out how to separate them further.2. **Second Test:** If the first test had a mix, take two pies from the first group and test them. - If the light is off, they're the same kind. Let's say both are apple. Then the third pie in the first group must be cherry. - If the light is on, they're different. So one is apple and one is cherry.But this might not be efficient because she could end up with more tests than needed. Maybe a better approach is to use a reference pie.3. **Reference Pie Method:** - Choose one pie as a reference. Let's say she picks Pie A. - Compare Pie A with another pie, Pie B. - If the light is off, they're the same kind. Then she knows both are either apple or cherry. - If the light is on, they're different kinds.But she needs to do this in four tests, so she needs to be careful with how she uses each test.4. **Detailed Strategy:** - **Test 1:** Compare Pie 1 and Pie 2. - If the light is off, they're the same kind. Let's say they're both apple. Then she knows at least two apple pies. - If the light is on, they're different. So one is apple and one is cherry. - **Test 2:** Compare Pie 1 and Pie 3. - If the light is off, Pie 3 is the same as Pie 1. - If the light is on, Pie 3 is different. - **Test 3:** Compare Pie 1 and Pie 4. - Similarly, determine if Pie 4 is the same as Pie 1. - **Test 4:** Compare Pie 1 and Pie 5. - Determine if Pie 5 is the same as Pie 1. After these four tests, she can categorize Pies 1-5 relative to Pie 1. Then, Pie 6 can be determined by comparing it with any known pie. For example: - If Pie 1 is apple, and Pies 2, 3, 4, 5 are also apple, then Pie 6 must be cherry. - If some of Pies 2-5 are different, she can group accordingly.But this might not always work because she could end up with more than three pies of one kind. Wait, no, since there are exactly three of each, she can count how many are the same as Pie 1 and adjust accordingly.Alternatively, she can use a different approach by dividing the pies into groups and using the machine to check for homogeneity.5. **Group Division Method:** - **Test 1:** Divide the six pies into two groups of three and test one group. - If the light is off, that group is homogeneous. The other group is also homogeneous. - If the light is on, there's a mix. Then she needs to further test. - **Test 2:** If the first group was mixed, take two pies from that group and test them. - If the light is off, they're the same. The third pie is different. - If the light is on, they're different. - **Test 3 and 4:** Use the information from the previous tests to reassign pies into homogeneous groups.This seems a bit vague. Maybe a better way is to use the machine to identify pairs and then build up from there.6. **Pair Testing:** - **Test 1:** Test Pies 1 and 2. - **Test 2:** Test Pies 3 and 4. - **Test 3:** Test Pies 5 and 6. From these three tests, she can determine which pairs are homogeneous and which are mixed. - If a pair lights up, they're different. - If not, they're the same. Then, in the fourth test, she can use the information to combine pairs or singles to form homogeneous groups.But this might not always lead to the correct grouping in four tests. She needs a more reliable method.7. **Using a Reference Pie and Elimination:** - Choose Pie 1 as a reference. - **Test 1:** Compare Pie 1 with Pie 2. - If same, proceed. - If different, note that Pie 2 is different. - **Test 2:** Compare Pie 1 with Pie 3. - Similarly, determine if Pie 3 is same or different. - **Test 3:** Compare Pie 1 with Pie 4. - Determine Pie 4's type. - **Test 4:** Compare Pie 1 with Pie 5. - Determine Pie 5's type. Now, she knows the types of Pies 1-5 relative to Pie 1. Then, Pie 6 can be determined by comparing it with any known pie. For example, if Pie 1 is apple, and Pies 2-5 include two apples and two cherries, then Pie 6 must be the remaining type to make three of each.This seems promising. Let's see:- Suppose Pie 1 is apple.- Test 1: Pie 1 & 2: If same, both apple; if different, Pie 2 is cherry.- Test 2: Pie 1 & 3: Similarly.- Test 3: Pie 1 & 4: Similarly.- Test 4: Pie 1 & 5: Similarly.After these four tests, she knows how many apples and cherries are in Pies 1-5. Since there are exactly three of each, she can deduce Pie 6's type without testing.For example, if in Pies 1-5, there are three apples and two cherries, then Pie 6 must be cherry. If there are two apples and three cherries, Pie 6 must be apple.This method uses four tests and ensures that she can separate the pies into two groups of three each, all of the same kind.So, the strategy is:1. Choose one pie as a reference.2. Test this reference pie against four others, one by one.3. Based on the results, determine the types of the tested pies.4. The sixth pie can be deduced without testing, as there are exactly three of each type.This should work because after four tests, she has enough information to categorize all six pies correctly.

answer：First, I recognize that Ram is half as efficient as Krish, which means if Krish's efficiency is ( K ), then Ram's efficiency is ( frac{1}{2}K ).Next, I know that Ram alone takes 30 days to complete the task. Using the formula ( text{Work} = text{Efficiency} times text{Time} ), the total work can be expressed as ( W = frac{1}{2}K times 30 ).When Ram and Krish work together, their combined efficiency is ( frac{1}{2}K + K = frac{3}{2}K ). They complete the task in 10 days, so the total work is also ( W = frac{3}{2}K times 10 ).Setting the two expressions for work equal to each other:[frac{1}{2}K times 30 = frac{3}{2}K times 10]Simplifying both sides:[15K = 15K]This confirms the relationship between their efficiencies. Therefore, the ratio of Ram's efficiency to Krish's efficiency is ( 1:2 ).

answer：Okay, so I have this problem where four students, A, B, C, and D, applied for a high school mathematics teaching position. Their scores have some relationships, and I need to figure out who has the highest score. Let me try to break this down step by step.First, let's list out the given information:1. The sum of A and B's scores is the same as the sum of C and D's scores. So, mathematically, that would be: [ a + b = c + d ] 2. The sum of B and D's scores is greater than the sum of A and C's scores: [ b + d > a + c ] 3. A's score is greater than the sum of B and C's scores: [ a > b + c ]Alright, so I have these three equations. My goal is to determine who among A, B, C, or D has the highest score.Let me start by looking at the first equation: ( a + b = c + d ). This tells me that the total score of A and B together is equal to the total score of C and D together. So, if I can find relationships between the individual scores, that might help.The third equation says that A's score is greater than the sum of B and C's scores: ( a > b + c ). That seems significant because it tells me that A is not just higher than B or C individually, but higher than their combined scores. That suggests that A might have a very high score.But before jumping to conclusions, let's see what the second equation tells us: ( b + d > a + c ). So, the sum of B and D is greater than the sum of A and C. Hmm, interesting. Since we know from the first equation that ( a + b = c + d ), maybe we can substitute or manipulate these equations to find more information.Let me try to express ( d ) in terms of the other variables from the first equation. From ( a + b = c + d ), we can solve for ( d ):[ d = a + b - c ]Now, let's substitute this expression for ( d ) into the second equation ( b + d > a + c ):[ b + (a + b - c) > a + c ]Simplify the left side:[ b + a + b - c > a + c ]Combine like terms:[ a + 2b - c > a + c ]Subtract ( a ) from both sides:[ 2b - c > c ]Add ( c ) to both sides:[ 2b > 2c ]Divide both sides by 2:[ b > c ]Okay, so from this, we've deduced that B's score is greater than C's score. That's useful information.Now, going back to the third equation: ( a > b + c ). Since we know ( b > c ), maybe we can find a relationship between A and D.From the first equation, we have ( d = a + b - c ). Let's see if we can express ( d ) in terms of ( a ) and ( b ) since we know ( a > b + c ).But wait, since ( a > b + c ), and ( d = a + b - c ), let's substitute ( c ) from the inequality ( a > b + c ). If ( a > b + c ), then ( c < a - b ). Plugging this into the expression for ( d ):[ d = a + b - c ]Since ( c < a - b ), replacing ( c ) with something smaller would make ( d ) larger. So:[ d > a + b - (a - b) ]Simplify:[ d > a + b - a + b ][ d > 2b ]So, D's score is greater than twice B's score. That's a significant finding.Now, let's see if we can compare A and D. We know from the first equation that ( a + b = c + d ). Since we've established that ( d > 2b ), let's substitute that into the equation:[ a + b = c + d > c + 2b ]So:[ a + b > c + 2b ]Subtract ( b ) from both sides:[ a > c + b ]But wait, that's exactly what the third equation says: ( a > b + c ). So, this doesn't give us new information.Let me try another approach. Since ( d = a + b - c ) and ( a > b + c ), let's see what that implies for ( d ).From ( a > b + c ), we can write ( a - b > c ). Plugging this into the expression for ( d ):[ d = a + b - c ]Since ( a - b > c ), then ( c < a - b ). Therefore:[ d = a + b - c > a + b - (a - b) ]Simplify:[ d > a + b - a + b ][ d > 2b ]Which is the same conclusion as before.So, D's score is more than twice B's score. Now, let's see how A and D compare.From the first equation, ( a + b = c + d ), and since ( d > 2b ), we can write:[ a + b = c + d > c + 2b ]So:[ a + b > c + 2b ]Subtract ( b ) from both sides:[ a > c + b ]Which is again the third equation, so no new information.Let me try to express A in terms of D. From ( d = a + b - c ), we can solve for ( a ):[ a = d - b + c ]But from the third equation, ( a > b + c ), so:[ d - b + c > b + c ]Subtract ( c ) from both sides:[ d - b > b ]Add ( b ) to both sides:[ d > 2b ]Again, the same result.So, D is greater than twice B. Now, let's see if we can find a relationship between A and D.From ( a = d - b + c ), and since ( d > 2b ), let's see:[ a = d - b + c > 2b - b + c ][ a > b + c ]Which is consistent with the third equation.Hmm, seems like we're going in circles. Maybe I need to assign some hypothetical values to see the relationships better.Let me assume some numbers that satisfy the given conditions.Let's say B = 10. Then, from ( b > c ), let's say C = 5.From the third equation, ( a > b + c = 10 + 5 = 15 ). So, A must be greater than 15. Let's say A = 20.From the first equation, ( a + b = c + d ):[ 20 + 10 = 5 + d ][ 30 = 5 + d ][ d = 25 ]Now, let's check the second equation: ( b + d > a + c )[ 10 + 25 > 20 + 5 ][ 35 > 25 ]Which is true.So, in this case, A = 20, B = 10, C = 5, D = 25. So, D has the highest score.Let me try another set of numbers to see if this holds.Let B = 20, so C must be less than 20. Let's say C = 10.From the third equation, ( a > 20 + 10 = 30 ). Let's take A = 40.From the first equation:[ 40 + 20 = 10 + d ][ 60 = 10 + d ][ d = 50 ]Check the second equation:[ 20 + 50 > 40 + 10 ][ 70 > 50 ]True.Again, D = 50 is the highest.Another example: B = 5, C = 3.Then, A > 5 + 3 = 8. Let A = 10.From the first equation:[ 10 + 5 = 3 + d ][ 15 = 3 + d ][ d = 12 ]Check the second equation:[ 5 + 12 > 10 + 3 ][ 17 > 13 ]True.Again, D = 12 is higher than A = 10.Wait, in this case, D is higher than A. So, in all these examples, D ends up being higher than A.Is this always the case? Let me see.From the first equation, ( a + b = c + d ), so ( d = a + b - c ).From the third equation, ( a > b + c ), so ( a - b > c ).Plugging into ( d ):[ d = a + b - c > a + b - (a - b) ][ d > a + b - a + b ][ d > 2b ]So, D is greater than twice B.From the second equation, ( b + d > a + c ), and since ( d > 2b ), we have:[ b + d > a + c ]But ( d = a + b - c ), so:[ b + (a + b - c) > a + c ][ a + 2b - c > a + c ][ 2b - c > c ][ 2b > 2c ][ b > c ]Which we already knew.So, D is greater than twice B, and B is greater than C.Now, let's compare A and D.From ( d = a + b - c ), and since ( a > b + c ), let's see:[ d = a + b - c > (b + c) + b - c ][ d > 2b ]Which we've established.But how does D compare to A?From ( d = a + b - c ), and since ( a > b + c ), let's see:[ d = a + b - c > (b + c) + b - c ][ d > 2b ]But we need to compare D and A.Let me express A in terms of D:[ a = d - b + c ]From the third equation, ( a > b + c ), so:[ d - b + c > b + c ][ d - b > b ][ d > 2b ]Which is consistent.But does this mean D is greater than A?From ( a = d - b + c ), and since ( d > 2b ), let's see:[ a = d - b + c > 2b - b + c ][ a > b + c ]Which is the third equation.But to compare A and D, let's see:From ( d = a + b - c ), and since ( a > b + c ), let's see:[ d = a + b - c > (b + c) + b - c ][ d > 2b ]But we need to see if D is greater than A.Let me rearrange ( d = a + b - c ):[ a = d - b + c ]So, A is equal to D minus B plus C.But since ( a > b + c ), we have:[ d - b + c > b + c ][ d - b > b ][ d > 2b ]Again, same conclusion.But to see if D > A, let's see:From ( a = d - b + c ), and since ( d > 2b ), let's see:[ a = d - b + c > 2b - b + c ][ a > b + c ]Which is consistent, but doesn't directly compare A and D.Wait, maybe I can express D in terms of A.From ( d = a + b - c ), and since ( a > b + c ), let's see:[ d = a + b - c > a + b - (a - b) ][ d > a + b - a + b ][ d > 2b ]Again, same result.But to compare A and D, let's see:From ( d = a + b - c ), and since ( a > b + c ), let's see:[ d = a + b - c > (b + c) + b - c ][ d > 2b ]But we need to see if D > A.Let me think differently. Suppose A is the highest, then D would be less than A. But from the examples I tried earlier, D ended up being higher than A. So, maybe D is always higher than A.Let me try to see if A can be higher than D.Suppose A is higher than D. Then, from ( d = a + b - c ), since A > D, we have:[ a > a + b - c ]Subtract A from both sides:[ 0 > b - c ][ c > b ]But from earlier, we have ( b > c ). So, this leads to a contradiction. Therefore, A cannot be greater than D. Hence, D must be greater than A.So, putting it all together:From the given conditions, we've deduced that:1. ( b > c )2. ( d > 2b )3. ( d > a )Therefore, D has the highest score.