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# the square of the sum of a number and 11

what is the square of the sum of a number and 11

## Research, Knowledge and Information :

### Square number - Wikipedia

... a square number or perfect square is an integer that is the square of an ... 11 2 = 121 12 2 = 144 13 2 ... The sum of two consecutive square numbers is a ...

### The sum of a number and its square is 110. Find the numbers ...

The sum of a number and its square is 110. ... the sum of -11 and its square, ... Enter your number and we’ll text you a download link.

### square of sum | planetmath.org

square of sum. Primary tabs. View ... The square of a sum is equal to the sum of the squares of all the summands plus the sum of ... This is true for any number of ...

### The sum of a number and its square is 72. What ... - jiskha.com

In a number square, the sum of the numbers in each row, column, and main diagonal is the same. ... CCA Unit 4 Lesson 11 3. If a number is not a rational number, ...

### The square of the sum of a number and 13 - Answers.com

The square of the sum of the number and 13 is (N+13)2 N2 + 26 N + 169. ... The sum of a number and its square is 6 Find the number? There are two solution: ...

### Sum of the Squares of the first n Natural Numbers

... Sum of the Squares of the First n Natural Numbers The sum of the squares ... [3.11 ] Simplifying: [3.12] ... coefficients of the square and the cubic Ken Ward's ...

### The sum of a number squared and three less than ... - Socratic

Let's call the number x "A number squared" =x^2 "Three less than twice the number" =2x-3 "The sum ... If we take out the square, ... SOCRATIC ...

### Translating Verbal Expressions – Terms - Del Mar College

Translating Verbal Expressions – Terms . ADDITION: ... square of the square of a number x2. ... 11. The sum of two consecutive odd ...

### Mathematical Phrase Translation - Cerritos College

Mathematical Phrase Translation The sum of x and 5 x + 5 ... The sum of the product of 8 and a number, and 3. 11. ... The square of the differen ce of the opposite of ...

### Find the numbers such that the square of the sum of the ...

Find the numbers such that the square of the sum of the number and 6 is 169. ... 11/10/2015 ... 3. "the square of the sum of the number and 6 is 169" = (x + 6) ...

## Suggested Questions And Answer :

### 40 sum from 3 by 3 magic squares

Label the squares A to I: A B C D E F G H I  If each row and column is to sum to 40 then the three rows added together must sum to 120: (A+B+C)+(D+E+F)+(G+H+I)=40+40+40=120, which is the sum of all numbers used. The consecutive numbers 1 to 9 add up to 45, which is 75 short of 120. If we use the numbers 2, 4, 6, ..., 18 the sum would be 90, 30 short of 120. If we bias the numbers by starting from a different number, we can make up the difference. Let the numbers be in order, nx+a where x is between 1 and 9, and a is a bias (a number to be added to each nx) and n is to be an integer, then the sum of all the numbers is n+a+2n+a+3n+a+...+9n+a=45n+9a=120. That is, 5n+a=120/9=40/3. So one solution is n=2 and a=40/3-10=10/3. For numbers 1 to 9, a magic square is typically: 8 1 6 3 5 7 4 9 2 We use the conversion nx+a: 1→16/3, 2→22/3, 3→28/3, 4→34/3, 5→40/3, 6→46/3, 7→52/3, 8→58/3, 9→64/3: 58/3 16/3 46/3 28/3 40/3 52/3 34/3 64/3 22/3  In this square the fractions sum to 40. If we write a=40/3-5n=5(8-3n)/3 or n=(40/3-a)/5=(40-3a)/15 or 8/3-a/5, we can find other values for a and n. Then we can convert the numbers 1 to 9 using the formula x→nx+a. GENERAL 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, where S is the common sum. 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 Since squares can be rotated and reflected, A and B could have been 1 and 17, 12 and 5, 11 and -5, etc., to produce the same square effectively.

### 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

### 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.

### answer for a magic square where the sum is 35

If the sum is 35 for each row, column and diagonal, the numbers must total 3*35=105. Now we need to find out if the numbers are consecutive. The numbers 1-9, for example, add up to 45. If the numbers form an arithmetic progression (AP): a, a+d, a+2d, ..., a+8d, the sum is 9a+d(0+1+2+...+8)=9a+d*8*9/2=9a+36d=105; a+4d=11+2/3. Since a and d are supposed to be integers, this equation cannot be satisfied because of the fraction 2/3. The numbers 1 4 6 9 12 14 17 20 22 could be used to construct a 3x3 magic square (almost). 20  1 14  6 12 17 all but one sum is 35  9 22   4 The following numbers form a true magic square: 1 11/3 19/3 9 35/3 43/3 17 59/3 67/3: 59/3      1 43/3 19/3 35/3    17      9 67/3 11/3 The numbers 5 20/3 25/3 10 35/3 40/3 15 50/3 55/3 similarly form a true magic square.

### the sum of the consecutive numbers is 35. what is the square of the largest of these numbers?

???????? HOW MANY ??????? sum=35 me assume "number" is INTEGER if yu sum 1 number, biggest number=35 & square=35*35=1225 if 2 numbers, yu hav 17 &18, biggest=18 18*18=324

### the magic square which has both positive and negative numbers and the sum of numbers is same horizontal, vertical, and diagonal

For odd-numbered Magic Squares, we start with the lowest number in the top centre square. If we have a 3X3 square, we start with -4 and end with 4. Let's refer to the square using a grid with rows A, B, C and columns 1, 2, 3. Using the up-1 right-1 rule: Square A2=-4; C3=-3; B1=-2; C1=-1; B2=0; A3=1; B3=2; A1=3; C2=4. 3 -4 1 -2 0 2 -1 4 -3 5X5 square: 4 11 -12 -5 2 10 -8 -6 1 3 -9 -7 0 7 9 -3 -1 6 8 -10 -2 5 12 -11 -4 Both of these squares sum to zero horizontally, vertically and diagonally.. Using the diagonal rule for 4X4: -7 7 6 -4 4 -2 -1 1 0 2 3 -3 5 -5 -6 8 The total is 2.

### square numbers with the sum of 13

The problem below is from my son's third grade math homework. There is a cube with 9 squares and the instructions are below. Fill in the boxes with the numbers 1,3,5,7,9,11,13,15,and17 such that the sum of the numbers in the horizontal,vertical and diagonal positions are the same. Each number can only be used once.

### I really need help with these optimization problems. Thanks.

Well the difference between them is even though they are the same numbers, you said negative and positive. I see -25 and +25 are different because -25 is below zero and +25 is above zero. That is what is different between them.

### What is the value of SS (sum of squared deviations) for the following sample?

data=2,3,4, 7 num numbers=4, su sum=16, averaej=4, median=3.5, biggest =7 sum av squares=14 varians = (sum av squares) / (num numbers)=3.5 std deviashun=1.87083