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does the zero mean anything in a set of numbers when calculating mean?

more info well heres the problem 0,10,10,20 and now i have to find the mean well i add it up and now i have 40 but do i divide by 4 or 5? because of the extra zero and is it just really a zero and doesnt count?

Research, Knowledge and Information :

How to Find the Mean - Math is Fun - Maths Resources

How to Find the Mean. ... Negative Numbers. How do you handle negative numbers? Adding a negative number is the same as subtracting the number (without the negative).
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How to Calculate the Mean: 4 Steps (with Pictures) - wikiHow

How to Calculate the Mean. ... the 0 in the amount of numbers at the end when you divide your big ... different results than the mean from the same set of numbers.
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probability - Why are mean 0 and standard deviation 1 ...

Why are mean 0 and standard deviation 1 distributions ... number of standard deviations that score is from the mean. The equation is calculating the (score - mean) ...
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If zero is in your number set do you count it when figuring ...

If zero is in your number set do you count it when figuring the mean? ... is the number zero a member of the set of natural numbers? ie does 0 ∈ ℕ The answer is: ...
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How To Analyze Data Using the Average – BetterExplained

You take a set of numbers, ... 1/3 — 1 over anything, really.) The harmonic mean helps us calculate ... How To Analyze Data Using the Average @ BetterExplainedThe ...
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Calculating the Mean and Standard Deviation with Excel ...

Finding the Mean Enter the scores in one of the columns on the ... Calculating the Mean and Standard Deviation with ... If a data set had more than one ...
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Mean Absolute Deviation - MathBitsNotebook(A1 - CCSS Math)

... (average) of all deviations in a set equals zero ... of any value on a number line from zero. ... The mean absolute deviation is the "average" of the ...
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Mean absolute deviation (MAD) (video) | Khan Academy

Mean absolute deviation is a way to describe variation in a data set. Mean absolute ... Well, how do I ... "Is there a number that can ... is zero, one , two, three ...
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Learn How to Calculate the Mean or Average (Your GPA)

Here's how to calculate the mean or average of a set of numbers. ... Here's how to calculate the mean or average of a set of ... Worked Examples of Calculating the Mean.
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Suggested Questions And Answer :

does the zero mean anything in a set of numbers when calculating mean?

0,10,10, 20 is 4 numbers tu get the averaej...suma 4 numbers & divide bi 4 =40/4=10 (krats oft kall this the "arithmatik meen")
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how can I divide a sextillion by eight million?

I may need stronger glasses; however, isn't the number you put up there 1 septillion (as in million, billion, trillion, quadrillion, quintillion, sextillion, septillion, octillion, nonillion, decillion, undecillion, duodecillion, tredecillion and my all time favorite--Quattrodecillion and etc, etc)?  In any event, I would just do it old school and start striking out 3 zero's at a time.  Thus, after getting rid of 2 sets of 3 zeros from 8 million leaves you with 8.  Getting rid of the equivalent 2 sets of 3 zeros from the other number leave you with 6 sets of 3 zeros with the '1' in front of them which should be 1 quintillion (or 1 quadrillion if you really wanted to start with 1 sextillion).  Still too many freakin zeros so then I guess I would start taking away 3 sets of zeros with the first set giving you 0.008 and either 1 quadrillion or 1 trillion--still too many zeros so I'm going to live dangerously and take away 2 more sets (i.e., 6 total zeros) just to get this over with.  Now your down to 0.000000008 and either just 1 billion or 1 million.  Now your down to something you can at least input into your calculator, but if it doesn't have scientific notation you still won't have your answer and buy a better calculator.  I was able to get an answer on my IPhone if I held it hortizonally and because it will answer using scientific notation.  So, if it is the number you typed in (1 septillion), the answer I got was 1.25 x 1017 (that 17 is supposed to be a superscript) or 125,000,000,000,000,000 a/k/a 125 quadrillion.
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how do you solve quadratic equations by factoring

If you know that the quadratic is going to factorise, then you need to look at the coefficient of the x^2 term. The number of x^2 you have can be split into factors. For example, if the coefficient is 6, then we have pairs of factors (1,6),(2,3). Then we look at the constant term and split that into possible pairs of factors. For example, if the constant is 24 then the pairs are (1,24),(2,12),(3,8),(4,6). Then we look at the sign of the constant. If it's + then in the next step we are going to ADD, otherwise we're going to SUBTRACT.  The next bit is like a game of tennis doubles. We'll call the game Quadrattack. Instead of the players being in a tennis court, they're in brackets, where the contents of the brackets consist of a team of two players. There's a pair of brackets, one for each team of two. Here's what the brackets look like as factors for the quadratic: (ax+b)(cx+d). We have two teams a and b versus c and d. The team members each represent a number (which may be printed on their T-shirts). The tables below show all possible "games" (fixtures) for the two teams a and b versus c and d, the second table being the "swapped partner" set where b and d change teams. The tables allow you to solve for a, b, c, d where: AX^2+BX+C=(aX+b)(cX+d) and ac=A and bd=C. The specific example of 6X^2+...+24 is given so ac=6 and bd=24. The tables show all possible values for B. X is any variable (representing x, y, t, etc.). The products ad and bc are calculated by multiplying the numbers on their T-shirts. Only players with the right numbers on their T-shirts can take part in the series of games. Plus or minus (+) in front of C tells you whether to use the Sum column (+) or Diff column (-). On with the game!   Game fixtures Game a b c (=A/a) d (=C/b) ad bc Sum Diff 1 1 1 6 24 24 6 30 18 2 1 2 6 12 12 12 24 0 3 1 3 6 8 8 18 26 10 4 1 4 6 6 6 24 30 18 5 2 1 3 24 48 3 51 45 6 2 2 3 12 24 6 30 18 7 2 3 3 8 16 9 25 7 8 2 4 3 6 12 12 24 0 Swap partners (b and d) Game a b c (=A/a) d (=C/b) ad  bc Sum Diff 9 1 24 6 1 1 144 145 143 10 1 12 6 2 2 72 74 70 11 1 8 6 3 3 48 51 3 12 1 6 6 4 4 36 40 4 13 2 24 3 1 2 72 74 2 14 2 12 3 2 4 36 40 4 15 2 8 3 3 6 24 30 6 16 2 6 3 4 8 18 26 8 In an actual question to solve or factorise a quadratic when you know that it can be factorised into rational real factors, you would set up the tables of fixtures for the games according to the given values of A and C. I've just selected A=6 and C=24 as an example. The repeated values of Sum and Diff effectively tell you that the games are in fact identical, so in the example Games 1, 4 and 6 have the same Sum, and it would not be necessary to play more than one of them if you have +C. Whoever decides the games (devises the quadratics) should pick values of A, B and C so that no two games are the same. In Games 2 and 8 Diff is zero, which means the quadratic contains no X term (B=0). The purpose of these games is to play each one until the Sum (where we have +C) or Diff (where we have -C) matches the given value of B. Both teams receive the prize or points, or the individual players receive points. There's just one more thing to do after the winning game: what is the sign in each bracket? For +C the sign is minus for -B and plus for +B; for -C and -B the sign is minus for the b or d player with bc or ad as the larger number and plus for the other b or d player; for -C and +B the sign is plus for the b or d player with bc or ad as the larger number and minus for the other b or d player. For example, if the quadratic to be solved was 6x^2-10x-24, then Game 3 is the winning game (Diff=10) and the factors are (x-3)(6x+8). Here we have -C (-24) and -B (-10) and bc (3*6) is larger than ad (1*8), so b (3) is minus and d (8) is plus.  
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show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem

f(x) = sqrt(x^2 + 1) ; [(0, sqrt(8)] Okay, so for the Mean Value Theorem, two things have to be true: f(x) has to be continuous on the interval [0, sqrt(8)] and f(x) has to be differentiable on the interval (0, sqrt(8)). First find where sqrt(x^2 + 1) is continous on. We know that for square roots, the number has to be greater than or equal to zero (definitely no negative numbers). So set the inside greater than or equal to zero and solve for x. You'll get an imaginary number because when you move 1 to the other side, it'll be negative. So, this means that the number inside the square root will always be positive, which makes sense because the x is squared and you're adding 1 to it, not subtracting. There would be no way to get a negative number under the square root in this situation. Therefore, since f(x) is continuos everywhere, (-infinity, infinity), then f(x) is continuous on [0, sqrt(8)]. Now you have to check if it is differentiable on that interval. To check this, you basically do the same but with the derivative of the function. f'(x) = (1/2)(x^2+1)^(-1/2)x2x which equals to f'(x) = x/sqrt(x^2+1). So for the derivative of f, you have a square root on the bottom, but notice that the denominator is exactly the same as the original function. Since we can't have the denominator equal to zero, we set the denominator equal to zero and solve to find the value of x that will make it equal to zero. However, just like in the first one, it will never reach zero because of the x^2 and +1. Now you know that f'(x) is continous everywhere so f(x) is differentiable everywhere. Therefore, since f(x) is differentiable everywhere (-infinity, infinity), then it is differentiable on (0, sqrt(8)). So the function satisfies the two hypotheses of the Mean Value Theorem. You definitely wouldn't have to write this long for a test or homework; its probably one or two lines of explanation at most. But I hope this is understandable enough to apply to other similar questions!
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Difference between mean, average and median?

First, it is usually a good idea to arrange your data in numerical order from lowest to highest.  We can arrange this data set as follows: 2,4,8,9,12,14,14,16,17,19 Now that the data is properly arranged, we can easily calculate the median. The median is the middle number in the data set. For this set, since there are 10 numbers there is no clear middle. In this case we take the two middle values, add them together and divide by 2 12+14 = 26 20/2 = 13 The median value for this data set is 10. mean and average are interchangeable terms. They are calculated by taking the sum of all data points and dividing by the number of data points.  For this data set we have: 2+4+8+9+12+14+14+16+17+19 = 115 115/10 = 11.5 Both mean and median can be useful in different situations. In this situation for example the mean is much lower than the median because data points 2 and 4 are weighed less heavily. When you have data which has a few extremes that may not fairly represent the majority of the data it may be useful to use the median instead of the mean.
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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:
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integral from 0 to infinity of (cos x * cos(x^2)) dx

The behaviour of this function f(x)=cos(x)cos(x^2) is interesting. The integral is the area between the curve and the x axis. If the functions cos(x) and -cos(x) are plotted on the same graph, the latter form an envelope for f(x). Between x=(pi)/2 and 3(pi)/2, the curve has 4 maxima and 3 minima; between x=3(pi)/2 and 5(pi)/2 there are 7 maxima and 6 minima; between x=(2n-1)(pi)/2 and (2n+1)(pi)/2 there are 3n+1 maxima and 3n minima (integer n>0), a total of 6n+1. (These are based purely on observation, and need to be supported by sound mathematical deduction.) As n becomes large the envelope appears to fill as the extrema become closer together. As x tends to infinity n also tends to infinity. The envelope apparently has as much area above (positive) as below (negative) the x axis so the total area will be zero as the positive and negative areas cancel out. The question is: do the areas cancel out exactly? As x gets larger, the curve starts to develop irregularities and patterns, but it stays within the envelope, and positive irregularities appear to be balanced by negative irregularities, so the overall symmetry appears to be preserved. f(x)=0 when cos(x)=0 or cos(x^2)=0, which means that x=(2n-1)(pi)/2 or sqrt((2n-1)(pi)/2), where n>0. Between 3(pi)/2 and 5(pi)/2, for example, we have sqrt(3(pi)/2), sqrt(5(pi)/2), ..., sqrt(13(pi)/2), because sqrt(13(pi)/2)<3(pi)/20, and this lies between (2n-1) and (2n+1); so n is defined by 2n-1<(2m+1)^2(pi)/2<2n+1.  For m-1 we have 2(n-z)-1<(2m-1)^2(pi)/2<2(n-z)+1, where z is related to the number of zeroes in the current "batch". For example, take m=3: 2n-1<49(pi)/2<2n+1; 49(pi)/2=76.97 approx., so 2n-1=75, and n=38. Also 2(n-z)-1<25(pi)/2<2(n-z)+1 so, because 25(pi)/2=39.27 approx., 2(n-z)+1=41, n-z=20, and z=18. When m=2, 2n-1=39, n=20; 2(20-z)-1<9(pi)/2<2(20-z)+1; 2(20-z)-1=13, 20-z=7, z=13. The actual number of zeroes, Z, including the end points is 2 more than this: Z=z+2. Now we have an exact way to calculate the number of zeroes in each batch. So Z and n are both related to m. The number of extrema, E=Z-1=z+1. In fact, E=int(2(pi)(m-1)+1), where int(a) means the integer part of a, so as m increases, there are proportionately more extrema over the range (2m-1)(pi)/2 to (2m+1)(pi)/2. The figure of 6n+1 deduced earlier by observation approximates to the mathematical findings, because 2(pi) is approximately equal to 6. But we still need to show, or disprove, that the areas above and below the x axis are equal and therefore cancel out. Unfortunately, if we consider the area under the first maximum (between x=(pi)/2 and sqrt(3(pi)/2)), and the area above the first minimum (between x=sqrt(3(pi)/2) and sqrt(5(pi)/2)), they are not the same, so do not cancel out. More...
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Finding a polynomial of a given degree with given zeros: Complex zeros

If the polynomial has an even degree, meaning that the highest power of the variable (for example: x) is an even number (for example: x^4), then all the zeroes could be complex. If odd, there must be at least one real zero. A complex zero is given by the complex expression: a+ib, where a and b are real, and i=sqrt(-1). If the polynomial has real coefficients (all the numbers in at are, or represent, real numbers), as is usually the case, then the imaginary part of the complex zeroes must combine to make a real number. For example, if one complex zero is identified as a+ib, then there will be another zero a-ib such that (x-a-ib)(x-a+ib)=x^2-2ax+a^2+b^2. (Note that, although a and b are real, they are not necessarily rational numbers.) There are no imaginary components in this expansion. Example: x^4+x^3-x-1 is degree 4 polynomial. It has real zeroes 1 and -1 so that it factorises to (x-1)(x+1)(x^2+x+1). The zeroes of x^2+x+1 are found by setting the quadratic to zero and solving using the formula; so x=(-1+sqrt(1-4))/2=(-1+sqrt(-3))/2=(-1+isqrt(3))/2. The two complex roots are -1/2-isqrt(3)/2 and -1/2+isqrt(3)/2. In this case a=-1/2 and b=sqrt(3)/2. Note that a^2+b^2=1. Example: you're told that a degree 5 polynomial has a real zero 1, and two complex zeroes 1+i and -2-i. Find all zeroes and the polynomial. The two missing zeroes will be 1-i and -2+i. The polynomial is (x-1)(x-1-i)(x-1+i)(x+2+i)(x+2-i)=(x-1)(x^2-2x+2)(x^2+4x+5)=x^5+x^4-3x^3-x^2+12x-10. Strictly speaking, the polynomial is a(x^5+x^4-3x^3-x^2+12x-10), where a is a real number, because multiplying a polynomial by a constant doesn't affect its zeroes.
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complex, rational and real roots

The quintic function should have 5 roots.  The changes of sign (through Descartes) tell us the maximum number of positive roots. Since there are two changes of sign there is a maximum of 2 positive roots. To find the number of negative roots we negate the terms with odd powers and check for sign changes: that means that there is at most one negative root, because there's only one sign change between the first and second terms. Complex roots always come in matching or complementary pairs, so that means in this case 2 or 4.  Put x=-1: function is positive; for x=-2 function is negative, so there is a real root between -1 and -2, because the x axis must be intercepted between -1 and -2. This fulfils the maximum for negative roots. That leaves 4 more roots. They could be all complex; there could be two complex and two positive roots. Therefore there are at least two complex roots. To go deeper we can look at calculus and a graph of the function: The gradient of the function is 10x^4-5. When this is zero there is a turning point: x^4=1/2. If we differentiate again we get 40x^3. When x is negative this value is negative so the turning point is a maximum when we take the negative fourth root of 1/2; at the positive fourth root of 1/2 the turning point is a minimum, and these are irrational numbers. The value of the function at these turning points is positive. The fourth root of 1/2 is about 0.84. and once the graph has crossed the x axis between -1 and -2, it stays positive, so all other roots must be complex. The function is 12 when x=0, so we can now see the behaviour: from negative values of x, the function intersects the x axis between -1 and -2 (real root); at x=4th root of 1/2 (-0.84) it reaches a local maximum, intercepts the vertical axis at 12 until it reaches 4th root of 1/2 (0.84) and a local minimum, after which it ascends rapidly at a steep gradient.  [Incidentally, one way to find the real root is to rearrange the equation: x^5=2.5x-6=-6(1-5x/12); x=-(6(1-5x/12))^(1/5)=-6^(1/5)(1-5x/12)^(1/5) We can now use an iterative process to find x. We start with x=0, so x0, the first approximation of the solution, is the fifth root of 6 negated=-1.430969 approx. To find the next solution x1, we put x=x0 and repeat the process to get -1.571266. We keep repeating the process until we get the accuracy we need, or the calculator reaches a fixed value. After just a few iterations, my calculator gave me -1.583590704.]
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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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