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what is the sum of 5 + (-100)

Apparently, my daughter''s teacher has explained to her that the answer is -105. However, everything I was ever taught leds me to believe the answer is -95. Can someone please let me know? Thank you.

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Sum of numbers and/or digits 1-100 - The Math Forum at NCTM


Sum of numbers and/or digits 1-100 Date: 03/29/97 at 02:45:58 From: Adam Cooper Subject: Sum of digits 1-100? How would one go about adding up all the digits 1-100?
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What is the sum of the first 100 whole numbers?


Hi Jo, The question you asked relates back to a famous mathematician, Gauss. In elementary school in the late 1700’s, Gauss was asked to find the sum of the numbers ...
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Techniques for Adding the Numbers 1 to 100 – BetterExplained


Sum for 5 + 6 + 7 + 8 + … n = [n * (n + 1) / 2] – 10 And for any starting number a: ... 296 Comments on "Techniques for Adding the Numbers 1 to 100"
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Adding fractions (denominators 10 & 100) (video ... - Khan ...


Adding fractions (denominators 10 & 100) Google Classroom Facebook Twitter Email. Fractions with denominators of 10 and 100. Visually converting tenths and hundredths.
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What is the sum of 2+4+6+8+…+98+100? - Quora


Some people have used a general equation for calculating sums with equidistant terms. Here's how to arrive at that equation. We are going to look at a sum of...
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What is the sum of 1 2 3 4 5 . . . . . 100 - The Most Trusted ...


There is a formula, called the formula for an arithmetic series, which you can use to calculate this: n/2 * [2a + (n - 1)d] Where: n is the number of terms (100) a is ...
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if 28% of the sum is $100.80 what is the sum - Brainly.com


if 28% of the sum is $100.80 what is the sum - 453543
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What is the sum of 1 2 3 4 5 . . . . . 100? - Weknowtheanswer


What is the sum of 1 2 3 4 5 . . . . . 100? Find answers now! No. 1 Questions & Answers Place. More questions about Education, School Subjects, Math and Arithmetic, what
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What is the value of the sum 5 + 10 + 15 + ... + 95 + 100 ...


Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than ...
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Sum | Define Sum at Dictionary.com


Sum definition, the aggregate of two or more numbers, magnitudes, quantities, or particulars as determined by or as if by the mathematical process of addition: The ...
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Suggested Questions And Answer :


use the integers 0, -2, -3, -4, -5, -6, -7, -8, and -10 to fill in a 3x3 magic square

First, find out what the sum is. Consider 3 rows of 3 numbers: each row has sum S. The rows contain all the numbers, so since the sum of the numbers is -45, 3S=-45 and S=-15. Take each number in turn and work out what other numbers can go with it to make -15: 0: (-10,-5),(-8,-7) -2: (-10,-3),(-8,-5),(-7,-6) -3: (-10,-2),(-8,-4),(-7,-5) -4: (-9,-2),(-8,-3),(-6,-5) -5: (-10,0),(-8,-2),(-7,-3),(-6,-4) -6: (-7,-2),(-5,-4) -7: (-8,0),(-6,-2),(-5,-3) -8: (-7,0),(-5,-2),(-4,-3) -10: (-5,0),(-3,-2) Now, consider the square: C1 C2 C3 C4 C5 C6 C7 C8 C9 Each cell participates in various sums. C1, for example, is in the first row and column and on a diagonal, so it is involved in 3 sums. C3, C7 and C9 are also similarly involved. Here's a complete list: C2,C4,C6,C8: 2 sums C1,C3,C7,C9: 3 sums C5: 4 sums Now we match the number of sums with the numbers that can be involved in at least the same number of sums. C5=-5. 2 sums: {C2 C4 C6 C8}={0 -6 -10} (exact) < {-2 -3 -4 -7 -8} (3 sums) 3 sums: {C1 C3 C7 C9}={-2 -3 -4 -7 -8}, one of these can be used to satisfy 2 sums, leaving 4 to match the cells. So {C1 C3 C7 C9}={-2 -3 -4 -7)|{-2 -3 -4 -8}|{-2 -3 -7 -8}|-2 -4 -7 -8}|{-3 -4 -7 -8} and {C2 C4 C6 C8}={0 -6 -8 -10}|{0 -6 -7 -10}|{0 -4 -6 -10}|{0 -3 -6 -10}|{0 -2 -6 -10} Because of symmetry we can assign 0 to C2. So C8=-10: C1 0 C3 C4 -5 C6 C7 -10 C9 C7+C9=-5, which means only -2 and -3. Because of symmetry we can assign either number to C7: C1 0 C3 C4 -5 C6 -2 -10 -3 C1-8=-15 so C1=-7; C3-7=-15, so C3=-8: -7 0 -8 C4 -5 C6 -2 -10 -3 C4=-6 and C6=-4: -7 0 -8 -6 -5 -4 -2 -10 -3
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addition using vedic maths

First, we make all the numbers the same length by prefixing zeroes. The longest number has 8 digits, so we use leading zeroes for numbers with fewer than 8 digits.  0 8 7 6 5 4 3 1  0 6 5 6 6 6 8 7  0 0 8 7 6 5 4 3  7 6 5 4 3 4 3 3  0 0 0 8 7 6 5 5 A gap has been left before each digit to leave space for a marker. Starting on the bottom left we add together the digits in the same column and we insert a marker when this sum exceeds 10, and we keep in mind the residue. We're essentially adding digits a pair at a time. The first column presents no problem so we simply write down 7 (temporarily) as the sum of the first column, but the second column illustrates the method better. Starting at the bottom, we have the digits 0 6 0 6 8. The zeroes we can ignore. So 6+6=12. We place a marker in front of the second 6 and keep in mind the residue 2. We add 2 to 8=10 and we insert another marker to the left of 8 and record the sum as 0. So far we have:  0•8 7 6 5 4 3 1  0•6 5 6 6 6 8 7  0 0 8 7 6 5 4 3  7 6 5 4 3 4 3 3  0 0 0 8 7 6 5 5  7 0 next column: 0 5 8 5 7 we proceed the same. Ignore the zero; 5+8=13; mark the 8 and keep 3 in mind. Add 3 to 5=8, then add 7 to 8=15, and insert another marker before 7. Put 5 in the answer.  0•8•7 6 5 4 3 1  0•6 5 6 6 6 8 7  0 0•8 7 6 5 4 3  7 6 5 4 3 4 3 3  0 0 0 8 7 6 5 5  7 0 5 We continue this way up to the last column:  0•8•7•6 5 4 3 1  0•6 5•6•6•6•8 7  0 0•8 7 6 5•4•3  7 6 5•4•3•4 3 3  0 0 0 8 7 6 5 5  7 0 5 1 7 5 3 9 +2 2 3 2 2 2 1  (sum of markers)   9 2 8 3 9 7 4 9 Now we count the markers. Between the first and second columns we have 2 markers so we add 2 to the result of the first column to give us 9 replacing the 7. Then 0+2=2, 5+3=8, and so on. We do this up to the last column. I've shown the sum of the markers for each column. The Vedic arithmetic only requires you to be able to add only the numbers 1 to 9 together. The faster you can do this the faster you can do Vedic addition (drop tens method).
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How many different distributions can the manager make if every employee receives at least one voucher?

There must be 100 vouchers because each is worth RM5 and the total value is 500 ringgits or RM500. (i) Each employee receives at least 1 voucher, so that means there are 95 vouchers left to distribute. We can write each distribution as {A,B,C,D,E}: starting with {0,0,0,0,95}, then {0,0,0,1,94}, {0,0,0,2,93}, ..., {0,0,0,95,0}, {0,0,1,0,94}, ..., {0,0,95,0,0}, ..., ..., {95,0,0,0,0}. So when A=B=C=0, {D,E} range from {0,95}, {1,94}, ..., to {95,0}, 96 ways. When A=B=0 and C=1, {D,E} range from {0,94} to {94,0}, 95 ways. Finally when C=95 so {C,D,E}={95,0,0} we will have covered 96+95+...+1=96*97/2=4656 ways. (The sum of the whole numbers from 1 to n is given by S=n(n+1)/2.) That was for B=0; when B=1, {C,D,E} ranges from {0,0,94} to {94,0,0} to cover 95*96/2=4560 ways. When B=2 it's 94*95/2=4465 ways. So for A=0 we have 4656+4560+4465+...+3+1 = 96^2+94^2+...+2^2 = 4(48^2+47^2+46^2+...+2^2+1^2)=4*48*49*97/6=152096 (the sum of the squares of the whole numbers from 1 to n is given by S=n(n+1)(2n+1)/6. Also, the sum of whole numbers between 1 and n taken in pairs gives us: n(n+1)/2+(n-1)n/2 for each pair. This is n^2/2+n/2+n^2/2-n/2=n^2. For the next pair we get (n-2)^2 and so on.) That was just for A=0! For A=1 we have {1,0,0,0,94} to {1,94,0,0,0}. This will give us 4560+4465+...+10+6+3+1 = 95^2+93^2+...+5^2+3^2+1. There is a formula for this sum. It is S=(n+1)(2n+1)(2n+3)/3, where 2n+1=95, so n=47. So for A=1, the number of ways is 48*95*97/3=147440. For A=2 we have {2,0,0,0,93} to {2,93,0,0,0} which produces 94^2+92^2+...+4^2+2^2 = 4(47^2+46^2+...+1) = 4*47*48*95/6 = 142880. So we alternate between two formulae as A continues to go from 3 to 95.  More to follow...
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find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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how many ways are there to add and get the sum of 180

There are an infinite number of ways to get 180 from two numbers, if we count decimals and fractions as well as other real numbers; but if we are limited to positive integers greater than zero and just the sum of two of them, we are limited to x and 180-x. If we also exclude 90+90 because the numbers are the same, then we have 1 to 89 combined with 179 to 91, which is 89 pairs. Moving on to the sum of three different numbers, let's make 1 plus another two different numbers adding up to 179. So we have 2+177, 3+176, ..., 87+92, 88+91, 89+90, which is 88 groups combined with 1. Move on to 2 plus another two different numbers adding up to 178: 3+175, ..., 87+91, 88+90, which is 86 groups. Then we move on to 3 plus 177: 4+173, ..., 86+91, 87+90, 88+89, 85 groups. And so on, with reducing numbers, until we get to 59, 60 and 61. Let's divide the numbers into two groups A and B. In A we start with 1 and in B we put 2 and (180-A-B)=177 as a pair (2,177). Then we put the next pair in group B: (3,176), then (4,175) and keep going till we have used up all the numbers, ending up with (88,90). Then we count how many pairs there are in group B and pair it up with the number in group A, so we start with (1,88) which covers all the combinations of numbers in group B. Now we move to 2 in group A, put all the pairs adding up to 178 in group B, and finally put the count of these pairs with 2 in group A: (2,86). We then move on to 3, and so on, putting in the counts to make up the number pair in group A. When we've finished by putting the last count in group A, which is (59,1), we can forget about group B and look at the pattern in group A. What we see is this: (1,88), (2,86), (3,85), (4,83), (5,82), (6,80), (7,79), ... See how the counts come in pairs with a gap? All the multiples of 3 are missing in the counts sequence (e.g., 87, 84, 81). We find there are 29 pairs and one odd count, 88, which is unpaired. Number the pairs 0 to 28 and refer to the pair number as N. Add the counts in the pairs together so we start with pair 0 as 86+85=171, pair 1 as 165, pair 2 as 159, and so on. The sequence 171, 165, 159, ..., 3 is an arithmetic sequence with a start of 171 and a difference of 6 between each term in the sequence. [Note also that the terms in the series are all multiples of 3: 3*57, 3*55, 3*55, ...] The rule for the Nth term is 171-6N. When N=0 we have the first term 171 and when N=28 the last term is 3. There is one more term at the end which is unpaired made up of the numbers 59, 60 and 61. We can combine this with the unpaired (1,88). We can find the sum of the terms in the series, which will tell us how many ways there are of adding three different integers so that their sum is 180 (like the sum of the angles of a triangle).  To find the sum of the terms of the series we note that there are 29 terms (0 to 28) and they all contain 171, so that's 171*29=4959. We also have to subtract 6(0+1+2+3+...+28)=6*28*29/2=2436. So 4959-2436=2523. [The sum of the series is also 3(57+55+53+...+5+3+1)=2523.] To this we add the "odd couple" 88+1=89 and 2523+89=2612. Add also the 89 which is the number of pairs of integers adding up to 180 we calculated at the beginning. The total so far is 2612+89=2701 ways of adding 2 or 3 positive integers so that their sum is 180. If you want to go further, please feel free to do so!
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what is this pattern? 2, 5, 9, 19, 40, 77, 137

Compound Arithmetic Sequence a0          a1          a2          a3          a4          a5          a6         b0          b1         b2          b3          b4           b5  ------- 1st differences               c0          c1          c2          c3          c4  -------------- 2nd differences                        d            d            d             d   ------------------- 3rd differences = constant We have here an irregular sequence (a_n) = (a_1, a_2, a_3, ..., a_k, ...). The differences between the elements of (a_n), the 1st differences, are non-constant. So also are the differences between the elements of (b_n), the 2nd differences. However, the differences between the elements of (c_n) are constant. This makes (c_n) a regular arithmetic sequence, and so we can write, cn = c0 + nd,  n = 0,1,2,3,... The sequence (b_n) is non-regular, but we can write, b_(n+1) = b_n + c_n Using the expression for the c-sequence, b_(n+1) = b_n + c_0 + nd Solve the recurrence relation for b_n Develop the terms of the sequence. b1 = b0 + c0 b2 = b1 + c0 + 1.d = b0 + 2.c0 + 1.d b3 = b2 + c0 + 2.d = b0 + 3.c0 + (1 + 2).d b4 = b3 + c0 + 3.d = b0 + 4.c0 + (1 + 2 + 3).d ‘ ‘ ‘ b(n+1) = b_0 + (n+1).c0 + sum[k=1..n](k) * d b_(n+1) = b_0 + (n+1).c0 + (nd/2)(n + 1) b_(n+1) = b_0 + (n+1)(c_0 + nd/2) We could also write c0 = b1 – b0 in the above expression. Now that we have a rule/expression for the general term of the b-sequence, let us analyse the a-sequence. Since the sequence (a_n) is non-regular like (b_n), we can write, a_(n+1) = a_n + b_n Substituting for the general term of the b-sequence, a_(n+1) = a_n + b0 + n(c0 + (n-1)d/2)      (co can be replaced later with c0 = b1 – b0) Solve the recurrence relation for a_n Develop the terms of the sequence. a1 = a0 + b0 a2 = a1 + b0 + 1.c0 + 1.0.d/2 = a0 + 2.b0 + 1.c0 + 1.0.d/2 a3 = a2 + b0 + 2.c0 + 2.1.d/2 = a0 + 3.b0 + (1 + 2).c0 + (1.0 + 2.1).d/2   (N.B. here, 1.0 = 1*0, 2.1 = 2*1) a4 = a3 + b0 + 3.c0 + 3.2.d/2 = a0 + 4.b0 + (1 + 2 + 3).c0 + (1.0 + 2.1 + 3.2).d/2 . . . a_n = a0 + n.b0 + sum[k=1..n-1] k * c0 + sum[k=1..n-1](k(k-1)) * (d/2) a_n = a0 + n.b0 + (c0/2).n(n-1) + sum[k=1..n-1](k^2) * (d/2) – sum[k=1..n-1](k) * (d/2) a_n = a0 + n.b0 + (c0.n/2)(n-1) + (d/2){(n/6)(n-1)(2n-1) – (n/2)(n-1)} a_n = a0 + n.b0 + (c0.n/2)(n-1) + (nd/6)(n^2 – 3n + 2) Substituting for c0 =  b1 – b0, b1 = a2 – a1, b0 = a1 – a0 Then, c0 = a0 – 2a1 + a2. These substitutions give, a_n = (1 + (n/2)(n – 3)).a0 – n(n – 2).a1 – (n/2)(1 – n).a2 + (nd/6)(n^2 – 3n + 2) Now substituting for a0 = 2, a1 = 5, a2 = 9 and d = 5, the rule is, a_n = 2 + n(n – 3) – 5n(n – 2) – (9/2).n.(1 – n) + (5/6).n.(n^2 – 3n + 2)  
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arrange 0,7,5,1,-1,-4,-5,-2,2 in circular order such that every three digit summation is 0

Let's look at groups of 3 digits that sum to zero (5 groups): (0,5,-5), (0,1,-1), (0,-2,2), (7,-5,-2), (5,-1,-4) 0 appears in 3 groups; 7 appears in 1 group; 5 appears in 2 groups; 1 appears in 1 group; -1 appears in 2 groups; -4 appears in 1 group; -5 appears in 2 groups; -2 appears in 2 groups; 2 appears in 1 group. Three numbers also appear with negative counterparts: (1,-1), (2,-2), (5,-5). We have 5 intersecting regions: Region A contains the numbers 7 -5 -2 Region B contains the numbers 5 -1 -4 Region C includes (overlaps) -1 from Region B, 0 and 1 Region D includes (overlaps) -2 from Region A and 0 from Region C and 2 Region E includes (overlaps) -5 from Region A, 5 from Region B and 0 from Region C Each region contains numbers whose sum is zero. The singletons 7, 1, -4 and 2 lie outside the intersecting parts of their respective regions (A, C, B and D). 0 is part of overlapping Regions C, D and E and occupies the central position in the regions. The picture below shows 7 circular or elliptical shapes. The central circle containing 0 is enclosed by three ellipses but sits in between two other circles, the contents of which sum to zero, Region A being the right-hand circle and Region B the left. The contents of each ellipse also sum to zero. Region C is the ellipse leaning to the right, Region D the ellipse leaning to the left and Region E is the horizontal ellipse.    
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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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make two magical square with single digit

3 x 3 MAGIC SQUARE SOLUTIONS Represent square using letters: A B C D E F G H I A+B+C=S=D+E+F=G+H+I; A+B+C+D+E+F+G+H+I=3S. A+E+I=B+E+H=C+E+G=D+E+F=S (A+B+C+D+E+F+G+H+I)+3E=4S; 3S+3E=4S, E=S/3. A+E+I=S, I=S-E-A, I=2S/3-A. H=S-E-B, H=2S/3-B. C=S-(A+B). G=2S/3-C=2S/3-S+(A+B), G=A+B-S/3. D+G=B+C=B+S-(A+B)=S-A; D=S-A-G=S-A-A-B+S/3, D=4S/3-(2A+B). F=2S/3-D=2S/3-4S/3+2A+B, F=2A+B-2S/3. Completed square:           A            B             S-(A+B) 4S/3-(2A+B)    S/3    2A+B-2S/3    A+B-S/3    2S/3-B      2S/3-A So A and B are arbitrary; S must be a multiple of 3 if square is to be whole numbers only. EXAMPLE: A=1, B=5, S=18:   1  5  12 17  6  -5   0  7  11 Single digits can be 1 to 9 (sum=45) or 0 to 8 (sum=36). The common sum is 45/3=15 or 36/3=12. In one case the middle digit is 5 (15/3)  and in the other it's 4 (12/3). In the first case, 5 must be in the middle of the square, and we need to see where 9 fits in. The common sum is 15 so 15-9=6 and the other two numbers must be (1,5) or (2,4). This tells us that 9 can only participate in two sums and therefore it must be in the middle of a side with 2 and 4 on either side of it. So B=9 and A=2. 2 9 4 7 5 3 6 1 8 is a solution. In the case for 0-8 we simply subtract 1 from each square: 1 8 3 6 4 2 5 0 7 and we can reorientate this: 7 2 3 0 4 8 5 6 1 There we have it: two solutions. 
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What is the sum of all even numbers between 0 and 398?

sum even numbers, start=0, end maebee=398 if inklude 0 & 398, hav 200 numbers first=0, last=398 sum=n*averaej, n=200 averaej=(0+398)/2=199 sum=200*199=39,800
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