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what set of numbers do I need that will get me 7,214 from 5,335?

I need help please!!

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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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Interpolate the data set (1, 150), (3, 175), (4, 185), (6, 200), (8, 300) to estimate the amount of money Gracie may earn if she displays her items for 7 hours

Since 7 is halfway between 6 and 8, Gracie should earn an amount about halfway between 200 and 300, that is, 250 (triangular interpolation). This is the simplest interpolation, but see later. Interpolating for 2 and 5 we get 162.5 and 192.5, that is, respectively, a difference of 12.5 (162.5-150 or 175-162.5) and 7.5 (192.5-185 or 200-192.5), while 250 is a difference of 50 from 200 and 300. So the interpolated figures fluctuate. For a more sophisticated approach, we need to take the whole dataset and look for a formula that best fits. One way to do this is to fit a polynomial F(x)=ax^4+bx^3+cx^2+dx+e into the five given points. This polynomial has 5 unknown coefficients, so with 5 simultaneous equations we should be able to find them. The process can be simplified slightly by taking the lowest "x" coord and using that as the zero starting point. In this case the lowest coord is 1 (hour) so we subtract 1 from the first coord of each pair to get: (0,150), (2,175), etc. F(0)=e=150. So we have the constant 150. The next step is to subtract 150 from each of the other "y" coords so we arrive at the following set of equations: (1) F(2)=16a+8b+4c+2d=25 (2) F(3)=81a+27b+9c+3d=35 (3) F(5)=625a+125b+25c+5d=50 (4) F(7)=2401a+343b+49c+7d=150 and we already have F(0)=150=e. We can now eliminate d from (1) and (2): 2F(3)-3F(2): (162-48)a+(54-24)b+(18-12)c=70-75=-5; (5) 114a+30b+6c=-5. and we can eliminate d from (3) and (4): 5F(7)-7F(5): (12005-4375)a+(1715-875)b+(245-175)c=750-350; 7630a+840b+70c=400 which simplifies to (6) 763a+84b+7c=40 or 109a+12b+c=40/7 We can eliminate c between (5) and (6): 6(6)-(5): (654-114)a+(72-30)b=240/7+5=275/7; (7) 540a+42b=275/7 or 90a+7b=275/42. So b=(275/42-90a)/7. From (6) we have: 109a+12b+c=109a+12(275/42-90a)/7+c=40/7, so c=40/7-109a-12(275/42-90a)/7; c=40/7-550/7+(1080/7-109)a=-510/7+317a/7=(317a-510)/7. We now have b and c in terms of a. We can continue to find d in terms of a. From (1) d=(25-16a-8b-4c)/2=25-16a-8(275/42-90a)/7-4(317a-510)/7; d=25-1100/147+2040/7+(-16+720/7-1268/7)a= (3675-1100+42840)/147+(-112+720-1268)a/7; d=45415/147-660a/7. We have b, c and d in terms of a, so we can find a by substituting into an equation containing all four coefficients (but not (1), because we used it to find d). Let's pick (2) and hope we get a sensible result! 81a+27(275/42-90a)/7+9(317a-510)/7+3(45415/147-660a/7)=35. From this a=5143/2772=1.855. Therefore b=-22.919, c=78.510, d=-67.687, e=150. And F(x)=1.855x^4-22.919x^3+78.510x^2-67.687x+150. This results need to be checked before we use F to find an interpolated value. Unfortunately, this polynomial approach produces inconsistent results, and needs to be discarded. Lagrange's method seems the obvious choice, even if it is tedious to do. We have 5 x values which we'll symbolise as x0, x1, x2, x3, x4 and 5 function values f0, f1, f2, f3, f4. If the function we're looking for is f(x) then: f(x)=(x-x1)(x-x2)(x-x3)(x-x4)f0/((x0-x1)(x0-x2)(x0-x3)(x0-x4))+         (x-x0)(x-x2)(x-x3)(x-x4)f1/((x1-x0)(x1-x2)(x1-x3)(x1-x4))+         (x-x0)(x-x1)(x-x3)(x-x4)f2/((x2-x0)(x2-x1)(x2-x3)(x2-x4))+... x0=1, x1=3, x2=4, x3=6, x4=8; f0=150, f1=175, f2=185, f3=200, f4=300. We want x=7, so f(7) is given by: 4.3.1.-1.150/(-2.-3.-5.-7)+6.3.1.-1.175/(2.-1.-3.-5)+ 6.4.1.-1.185/(3.1.-2.-4)+6.4.3.-1.200/(5.3.2.-2)+ 6.4.3.1.300/(7.5.4.2) This comes to: -60/7+105-185+240+540/7=1600/7=228.57 (229 to the nearest whole number) compared with 250 from the simple interpolation. 
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if we owe 7 dollars and are paying it back at the rate of 2 dollars a day what would be an equation for that problem

the equation in y=mx+b or slope intercept form is y=2x-7 x represents the number of days it will take to pay the money back. -7 represent the amount of money you own in at the start and y represents the money you owe at a given time. to graph this we can ask the question how much will I owe on day 0 If we set x to 0 and solve for y we get y=2*0-7 y=-7 So our first point on the line is (0,-7) which is the y intercept find the next point by asking, how much you will owe on day 1 set x equal to 1 and solve for y y=2*1-7 y=-5 So our next point is 1, the number of days and -5 the amount of money still owed (1,-5) Here is the graph of the equation: The x intercept represents the day when all the money has been paid back. (about 3.5 days) the slope we know is 2 or 2/1. We get that from the y=mx+b equation. M=the slope and B=y intercept our m value is 2 and our b value is -7
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I need help, I really need help pleeease.

I assume that (3) is a graph, because it doesn't display on my tablet. Question 1 a) 1 because the daily operating cost is stated specifically. b) 3 because the profit of $500 will be a point on the graph corresponding to a value of t; 2 because if we put P=500 and solve for t we get t=1525/7.5=203.3 so the graph would show t=203-204. c) 2 because we set P=0 and solve for t=1025/7.5=136.7, so 137 tickets sold would give a profit of $2.5, but 136 tickets would make a loss of $5; alternatively, if (3) is a graph then P=0 is the t-axis so it's where the line cuts the axis between 136 and 137. d) The rate of change is 7.5 from (2).  e) 2, because the format shows it to be a linear relationship; a straight line graph for (3) also shows linearity. Question 2 a) profit=sales- operating costs so 1025 in (2) is the negative value representing these costs. If (3) is a graph, it's the intercept on the P axis at P=-1025; (4) is the P value when t=0. b) For (1) you would need to find out how many tickets at $7.50 you would need to cover the operating costs of $1025 plus the profit of $500. That is, how many tickets make $1525? Divide 1525 by 7.5; 2 and 3 have already been dealt with; to use (4) you would note that P=$500 somewhere between t=200 and 250 in the table. c) For (1), work out how many tickets cover the operating costs. 1025/7.5=136.7, so pick 137 which gives the smallest profit to break even; 2 and 3 already given; in (4) it's where P goes from negative to positive, between 100 and 150. d) For (1) the only changing factor is the number of tickets sold. The rate is simply the price of the ticket, $7.50; if (3) is a graph, the rate of change is the slope of the graph, to find it make a right-angled triangle using part of the line as the hypotenuse, then the ratio of the vertical side (P range) and the horizontal side (t range) is the slope=rate of change; for (4) take two P values and subtract the smallest from the biggest, then take the corresponding t values and subtract them, and finally divide the two differences to give the rate of change: example: (475-(-275))/(200-100)=750/100=7.5. e) For (1) it's clear that the profit increases (or loss decreases) with the sale of each ticket by the same amount as the price of a ticket, so there is a linear relationship; 2 and 3 already dealt with; in the table in (4) the profit changes by the fixed value of 50 tickets=$375, showing that a linear relationship applies between P and t: for every 50 tickets we just add $375 to the profit.
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how we count the subsets of any set? what is the formula of counting the subsets of any set?

Let's say there are N objects in a set. Now you need a rule for the subset. Whatever obeys the rule stays in the set, whatever doesn't you put to one side. Those that obey the rule are in the subset and they can be counted just as the original objects were. You may have to apply the rule to every individual object in the original set, or there may be a way of separating out the whole subset or build up from groups instead of counting each individual object. For example, all the numbers from 1 to 100 form the original set, where N=100. The subset may consist of just the odd numbers. You can count them individually: 1, 3, 5, etc., or you can find a formula: 2n-1, that generates them. Then you need to find the range of n. When n=1 we get 0, the first in the subset, and when n=50 we get the last, 99. So n ranges from 1 to 50, which is 50 objects. So the size of the subset is 50.
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find all reconstuctions of the sum: ABC+DEF+GHI=2014 if all letters are single digits

There is no solution if A to I each uniquely represent a digit 1 to 9, because the 9's remainder of ABC+DEF+GHI (0) cannot equal the 9's remainder of 2014 (=7). Let me explain. The 9's remainder, or digital root (DR), is obtained by adding the digits of a number, adding the digits of the result, and so on till a single digit results. If the result is 9, the DR is zero and it's the result of dividing the number by 9 and noting the remainder only. E.g., 2014 has a DR of 7. When an arithmetic operation is performed, the DR is preserved in the result. So the DRs of individual numbers in a sum give a result whose DR matches. If we add the numbers 1 to 9 we get 45 with a DR of zero, but 2014 has a DR of 7, so no arrangement can add up to 2014. If 2 is replaced by 0 in the set of available digits, the DR becomes 7 (sum of digits drops to 43, which has a DR of 7). 410+735+869=2014 is just one of many results of applying the following method. Look at the number 2014 and consider its construction. The last digit is the result of adding C, F and I. The result of addition can produce 4, 14 or 24, so a carryover may apply when we add the digits in the tens column, B, E and H. When these are added together, we may have a carryover into the hundreds of 0, 1 or 2. These alternative outcomes can be shown as a tree. The tree: 04 >> 11, >> 19: ............21 >> 18; 14 >> 10, >> 19: ............20 >> 18; 24 >> 09, >> 19: ............19 >> 18. The chevrons separate the units (left), tens (middle) and hundreds (right). The carryover digit is the first digit of a pair. For example, 20 means that 2 is the carryover to the next column. Each pair of digits in the units column is C+F+I; B+E+H in the tens; A+D+G in the hundreds. Accompanying the tree is a table of possible digit summations appearing in the tree. Here's the table: {04 (CFI): 013} {09 (BEH): 018 036 045 135} {10 (BEH): 019 037 046 136 145} {11 (BEH): 038 047 137 056 146} {14 (CFI): 059 149 068 158 167 347 086 176 356} {18 (ADG): 189 369 459 378 468 567} {19 (BEH/ADG): 379 469 478 568} {20 (BEH): 389 479 569 578} {21 (BEH): 489 579 678} {24 (CFI): 789} METHOD: We use trial and error to find suitable digits. Start with units and sum of C, F and I, which can add up to 4, 14 or 24. The table says we can only use 0, 1 and 3 to make 4 with no carryover. The tree says if we go for 04, we must follow with a sum of 11 or 21 in the tens. The table gives all the combinations of digits that sum to 11 or 21. If we go for 11 the tree says we need 19 next so that we get 20 with the carryover to give us the first two digits of 2014. See how it works? Now the fun bit. After picking 013 to start, scan 11 in the table for a trio that doesn't contain 0, 1 or 3. There isn't one, so try 21. We can pick any, because they're all suitable, so try 489. The tree says go for 18 next. Bingo! 567 is there and so we have all the digits: 013489576. We have a result for CFIBEHADG=013489576, so ABCDEFGHI=540781693. There are 27 arrangements of these because we can rotate the units, tens and hundreds independently like the wheels of an arcade jackpot machine. For example: 541+783+690=2014. Every solution leads to 27 arrangements. See how many you can find using the tree and table!
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How to determine the number of unique subsets of 2,3,5 in a set of 7 or 8?

SET OF SEVEN C(7,5)=C(7,2) or 7C5 or 7C2=21 sets of 2 or 5 (=7*6/2=7*6*5*4*3/(1*2*3*4*5)). C(7,3)=7C3=35 sets of 3 (=7*6*5/(1*2*3)). SETS OF EIGHT C(8,2)=28 sets of 2; C(8,3)=C(8,5)=56 sets of 3 or 5. The palindromic pattern follows Pascal's Triangle for row 7 and row 8; ROW 7: 1 7 21 35 35 21 7 1 = C(7,0) C(7,1) C(7,2),..., C(7,6) C(7,7), and the sum is 2^7=128 ROW 8: 1 8 28 56 70 56 28 8 1 I don't think { 1 2 3 } can be considered a subset of { 1 2 3 } but the subsets will include the null set { }; the set itself is included in the power set, and the total number of sets in the power set is 2^3=8 in this case and 2^n in the general case where n is the set size.
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magic square 16 boxes each box has to give a number up to16 but when added any 4 boxes equal 34

The magic square consists of 4X4 boxes. We don't yet know whether all the numbers from 1 to 16 are there. Each row adds up to 34, so since there are 4 rows the sum of the numbers must be 4*34=136. When we add the numbers from 1 to 16 we get 136, so we now know that the square consists of all the numbers between 1 and 16. We can arrange the numbers 1 to 16 into four groups of four such that within the group there are 2 pairs of numbers {x y 17-x 17-y}. These add up to 34. We need to find 8 numbers represented by A, B, ..., H, so that all the rows, columns and two diagonals add up to 34. A+B+17-A+17-B=34, C+D+17-C+17-D=34, ... (rows) A+C+17-A+17-C=34, ... (columns) Now there's a problem, because the complement of A, for example, appears in the first row and the first column, which would imply duplication and mean that we would not be able to use up all the numbers between 1 and 16. To avoid this problem we need to consider other ways of making up the sum 34 in the columns. Let's use an example. The complement of 1 is 16 anewd its accompanying pair in the row is 2+15; but if we have the sum 1+14 we need another pair that adds up to 19 so that the sum 34 is preserved. We would perhaps need 3+16. We can't use 16 and 15 because they're being used in a row, and we can't use 14, because it's being used in a column, so we would have to use 6+13 as the next available pair. So the row pairs would be 1+16 and 2+15; the column pairs 1+14 and 6+13; and the diagonal pairs 1+12 and 10+11. Note that we've used up 10 of the 16 numbers so far. This type of logic applies to every box, apart from the middle two boxes of each side of the square (these are part of a row and column only), because they appear in a row, a column and diagonal. There are only 10 equations but 16 numbers. In the following sets, in which each of the 16 boxes is represented by a letter of the alphabet between A and P, the sum of the members of each set is 34: {A B C D} {A E I M} {A F K P} {B F J N} {C G K O} {D H L P} {D G J M} {E F G H} {I J K L} {M N O P} The sums of the numbers in the following sets in rows satisfy the magic square requirement of equalling 34. This is a list of all possibilities. However, there are also 24 ways of arranging the numbers in order. We've also seen that we need 10 out of the 16 numbers to satisfy the 34 requirement for numbers that are part of a row, column and diagonal; but the remaining numbers (the central pair of numbers on each side of the square) only use 7 out of 16. The next problem is to find out how to combine the arrangements. The vertical line divides pairs of numbers that could replace the second pair of the set. So 1 16 paired with 2 15 can also be paired with 3 14, 4 13, etc. The number of alternative pairs decreases as we move down the list. 17 X 17: 1 16 2 15 | 3 14 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 2 15 3 14 | 1 16 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 3 14 4 13 | 1 16 | 2 15 | 5 12 | 6 11 | 7 10 | 8 9 4 13 5 12 | 1 16 | 2 15 | 3 14 | 6 11 | 7 10 | 8 9 5 12 6 11 | 1 16 | 2 15 | 3 14 | 4 13 | 7 10 | 8 9 6 11 7 10 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 8 9 7 10 8 9 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 6 11 16 X 18: 1 15 2 16 | 4 14 | 5 13 | 6 12 | 7 11 | 8 10 2 14 3 15 | 5 13 | 6 12 | 7 11 | 8 10 3 13 4 14 | 2 16 | 6 12 | 7 11 | 8 10 4 12 5 13 | 2 16 | 3 15 | 7 11 | 8 10 5 11 6 12 | 2 16 | 3 15 | 4 14 | 8 10 6 10 7 11 | 2 16 | 3 15 | 4 14 | 5 13 7 9 8 10 | 2 16 | 3 15 | 4 14 | 5 13 | 6 12 15 X 19: 1 14 3 16 | 4 15 | 6 13 | 7 12 | 8 11 2 13 4 15 | 3 16 | 5 14 | 7 12 | 8 11 3 12 5 14 | 4 15 | 6 13 | 8 11 4 11 6 13 | 3 16 | 5 14 | 7 12 5 10 7 12 | 3 16 | 4 15 | 6 13 | 8 11 6 9 8 11 | 3 16 | 4 15 | 5 14 | 7 12  7 8 9 10 | 3 16 | 4 15 | 5 14 | 6 13 14 X 20: 1 13 4 16 | 5 15 | 6 14 | 8 12 | 9 11 2 12 5 15 | 4 16 | 6 14 | 7 13 | 9 11 3 11 6 14 | 4 16 | 5 15 | 7 13 | 8 12 4 10 7 13 | 5 15 | 6 14 | 8 12 | 9 11 5 9 8 12 | 4 16 | 6 14 | 7 13 | 9 11 6 8 9 11 | 4 16 | 5 15 | 7 13 13 X 21: 1 12 5 16 | 6 15 | 7 14 | 8 13 | 10 11 2 11 6 15 | 5 16 | 7 14 | 8 13 | 9 12 3 10 7 14 | 5 16 | 6 15 | 8 13 | 9 12 4 9 8 13 | 5 16 | 6 15 | 7 14 | 10 11 5 8 9 12 | 6 15 | 7 14 | 10 11 6 7 10 11 | 5 16 | 8 13 | 9 12 12 X 22: 1 11 6 16 | 7 15 | 8 14 | 9 13 | 10 12 2 10 7 15 | 6 16 | 8 14 | 9 13 | 10 12 3 9 8 14 | 6 16  4 8 9 13 | 6 16 | 7 15  5 7 10 12 | 6 16 11 X 23: 1 10 7 16 | 8 15 | 9 14 2 9 8 15 | 7 16  3 8 9 14 | 7 16  10 X 24: 1 9 8 16 To give an example of how the list could be used, let's take row 3 in 17 X 17 {3 14 4 13}. If 3 is the number in the row column and diagonal, then we need to inspect the list to find another row in a different group containing 3 that doesn't duplicate any other numbers. So we move on to 15 X 19 and we find {3 12 8 11} and another row in 13 X 21 with {3 10 5 16}. So we've used the numbers 3, 4, 5, 8, 10, 11, 12, 13, 14, 16. That leaves 1, 2, 6, 7, 9, 15. In fact, one answer is: 07 12 01 14 (see 15x19) 02 13 08 11 16 03 10 05 09 06 15 04
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Prove, by contradiction, that if w,z∈C.... (elementary algebra)

Given the validity of the comment attached to the question, and it's resolution, we can take a look at the implication of the question. To be true for w and z generally, we can take a specific case to demonstrate the argument. Let w=0.8+0.6i, which satisfies the requirement |w|=1, a special case of |w|<1. For any value of n, the binomial expansion of (a+b)^n=1 only when a+b=1, i.e., b=1-a. The binomial expansion is (a+b)^n=a^n+nba^(n-1)+(n(n-1)/2)b^2a^(n-2)+...+b^n=1. The general term is (n!/(r!(n-r)!))a^(n-r)b^r, where 01/2. Let w=0.03+0.04i, |w|=0.05, z=1-(0.03+0.04i)=0.97-0.04i, |z|=0.9708 approx>1/2. As |w| gets smaller, |z| gets larger, so moving further away from 1/2 and towards 1. What is w+z? In all of the above cases w+z=1, by definition, and so the binomial expansion also =1. Just seen your comment. Let w=0.96+0.28i, |w|=1, z=0.04-0.28i, |z|=0.2828 approx, which is <1/2. So the proposition would not be true for the binomial expansion of (w+z)^n=1 for all n. That means we require a different expression than w+z. However, we do know that (...)^n must be 1 for all n and we need to find an alternative expression similar to the one given. Let w=a+isqrt(p^2-a^2), |w|=sqrt(a^2+p^2-a^2)=p<1. This is the general form for w. In the above, z=1-a-isqrt(p^2-a^2) and |z|=sqrt((1-a)^2+p^2-a^2)=sqrt(1-2a+a^2+p^2-a^2)=sqrt(1-2a+p^2). When p=1, this is sqrt(2(1-a)) which can be less than 1/2 (7/81/2) and wz/(w+z)=1 so ((w+z)/wz)^n=((1/z)+(1/w))^n=(wz/(w+z))^n=1. ((1/z)+(1/w))^n=(w+z)^n=1. Therefore, w and z are particular complex numbers for which the proposition is true, and the binomial expansion applies.
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what is the sum of the remainders if the numbers 100, 101, 102...998, 999,1000 are divided by nine?

all integers from 102 thru 1000 divide bi 9 & add the leftovers yu hav 899 numbers thats 99.9 sets av 9 numbers eech set av 10 numbers hav leftover=0, 1, 2, 3, 4, 5, 6, 7, 8 1 set sum=36 100 sets=3600 then chek a fyu numbes at both ends 102/9=11 with leftover=3 1000/9=111 with leftover=1 lak leftover=2, su anser=3600-2=3598
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