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how do you get the sum of 25 by using four 4's

by using the ! or the square route or exponents

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Ruth Carver - Four 4's Puzzle - Math Forum


From the Math Forum Ruth Carver's . Math Tips & Tricks
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Four Fours Puzzle - Solution - Math Is Fun


Four Fours Puzzle - Solution. The ... The Puzzle: A popular mathematical pastime: Use exactly four 4's to form every integer from 0 to 50, ... 25 = (4*4!+4)/4 26 = 4 ...
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Puzzles/Arithmetical puzzles/Four 4s Equal.../Solution ...


Puzzles/Arithmetical puzzles/Four 4s Equal.../Solution. From Wikibooks, ... 25 25; 26 26; 27 27; 28 28; 29 29; 30 30; 31 31; 32 32; 33 33; 34 34; 35 35; 36 36; 37 37 ...
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Four fours - Wikipedia


Four fours is a mathematical puzzle. The goal of four fours is to find the simplest mathematical expression for every whole number from 0 to some maximum, using only ...
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The Four Fours Problem - wheels.org


The Four Fours Problem ... Since it is possible to create an infinite number of integers from even a single four (for example, 4! = 24. 4 ... 25: 84: 26: 85: 27: 86 ...
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How do you make the numbers 0-50 using only four 4s


... /.4 22 = 44*sqrt(4)/4 23 = (4*4!-4)/4 24 = 4*4+4+4 25 = (4 ... In words, four squared (16) take 4 (12), … + the sum of 4 ... How do you make the numbers 11-20 ...
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How do you make the numbers 11-20 with four 4's only


Answers.com ® WikiAnswers ® Categories Science Math and Arithmetic Statistics How do you make the numbers 11-20 with four 4's ... 4*4!-4)/4 24 = 4*4+4+4 25 = (4*4 ...
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Suggested Questions And Answer :


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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statistic probability

i. All the possible answers in Quiz 1 by chance can be represented by the terms in the expansion of (0.25+0.75)^8, where 0.25 is the probability of a correct choice of answer for each individual question and 0.75 is the probability of an incorrect choice of answer. The binomial expression has a value of 1, which is a probability of 1, i.e., certainty, made up of the sum of all the terms. Let p=0.25 and 1-p is 0.75. The expansion is p^8+... + 8Cr*p^r(1-p)^(8-r), where r indicates the rth term between r=0 and 8 and 8Cr is the combination function nCr for n=8. The value of 8Cr is given by (8*7*...*(8-r+1))/(1*2*...*r). The coefficients are, in order: 1, 8, 28, 56, 70, 56, 28, 8, 1, the 8th row of Pascal's triangle. The binomial terms start with the probability of 8 correct answers, then 7 correct 1 incorrect, 6 correct 2 incorrect, 5 correct 3 incorrect, and so on. The probability of answering incorrectly 3 question is (3/4)^3=27/64, so the probability of answering at least one question in 3 correctly by chance is 1-27/64=37/64, in other words, this is the probability that at most 3 questions are attempted before obtaining a correct answer. This is better than an evens chance (50%), so is more likely than unlikely. ii. In Quiz 2 p=1/5 or 0.20 and n=15. The probability of less than 4 correct answers is the sum of the last four terms of the binomial series for 15, which give all incorrect, 1, 2, 3 correct. The coefficients are respectively 1, 15, 105, 455. The p terms alone are 0.03518, 008796, 0.002199 and 0.0005498. When we multiply by the coefficients and add the terms together we get 0.6482 or 64.82% probability. iii. The probability of 6 or more correct answers in Quiz 1 is 1-(probability of up to 2 incorrect answers)=1-(0.75^8+8*0.75^7*0.25+28*0.75^6*0.25^2)=1-0.6785=0.3215 or 32.15%. This is less than 1/3, so the odds are 2:1 against Siti getting 6 or more correct answers by chance alone, which makes it reasonably unlikely. iv and v. The problem says "before the 8th question", so there need to be 5 correct answers within the first 7 questions. Using p=0.2, the probability of getting 5 correct answers out of 5 is 0.2^5=0.00032; the probability of getting 5 correct out of 6 is 6*0.2^5*0.8=0.001536; and the probability of 5 out of 7 is 21*0.2^5*0.8^2=0.0043008. Therefore the probability of getting 5 correct before the 8th question is the sum of these=0.0061568 or 0.62%, a most unlikely occurrence, less than 1 in 162.
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With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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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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how do you get the sum of 25 by using four 4's


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four identical looking objects weigh 3,5,8 and 11 oz

S=SET, 1=A~B+C, 2=A~B+D, 3=A~C+D, 4=B~C+D, 5=A+B~C+D, 6=A+C~B+D (1 to 6 are weighing configurations). ~ (twiddles) means "is balanced against"; < means "lighter than"; > means "heavier than"; = means "balances". S   A B C D     1    2    3    4    5    6 1    3 5 8 11    <    <    <    <    <    < 1    5 8 11 3    <    <    <    <    <    > 1    8 11 3 5    <    <    =    >    >    < 1    11 3 5 8    >    =    <    <    >    > 2    3 5 11 8    <    <    <    <    <    > 2    5 11 8 3    <    <    <    =    >    < 2    11 8 3 5    =    <    >    =    >    > 2    8 3 5 11    =    <    <    <    <    < 3    3 8 5 11    <    <    <    <    <    < 3    8 5 11 3    <    =    <    <    <    > 3    5 11 3 8    <    <    <    =    >    < 3    11 3 8 5    =    >    <    <    >    > 4    3 8 11 5    <    <    <    <    <    > 4    8 11 5 3    <    <    =    >    >    < 4    11 5 3 8    >    <    =    <    >    > 4    5 3 8 11    <    <    <    <    <    < 5    3 11 5 8    <    <    <    <    >    < 5    11 5 8 3    <    >    =    <    >    > 5    5 8 3 11    <    <    <    <    <    < 5    8 3 11 5    <    =    <    <    <    > 6    3 11 8 5    <    <    <    <    >    < 6    11 8 5 3    <    =    >    =    >    > 6    8 5 3 11    =    <    <    <    <    < 6    5 3 11 8    <    <    <    <    <    > The table shows all possible arrangements of the four weighted objects, together with various weighing results. Before carrying out any weighings, we first label the four objects so that we can't get them mixed up. The labels are weightless. We carry out weighings 5 and 6 first. We take the four objects and put two in one scale and two in the other and note the results of the two different weighings. There are four possible results. Note how the table is arranged. Each set is ordered and the order is rotated ABCD, BCDA, CDAB and DABC to make up the set. Note that each set is circular: DABC is followed by ABCD, and ABCD is preceded by DABC. This is important for later. After weighing the pairs, we will have identified one arrangement in each set that would produce the weighing results we've just obtained. We have two weighings left. Using these we should arrive at the unique set of weights. Let's use an example. Suppose the first two weighings gave the result < and < (the pair in the left scale pan was lighter than the pair in the right scale pan in each case). That gives us 6 possible values for A, B, C and D, one group from each of the six sets: 3 5 8 11, 8 3 5 11, 3 8 5 11, 5 3 8 11, 5 8 3 11 and 8 5 3 11. Note that D has already been found to weigh 11 ounces, because it's common to all. If we look at the table at the configuration immediately preceding each of the six groups we've identified in each set and look at the weighing results 2 and 3, we can see that the results are unique. For example, the arrangement preceding 3 5 8 11 is 11 3 5 8 and the weighing results are  = and <; the arrangement preceding 5 8 3 11 is 11 5 8 3 and the results are > and =. All we have to do is carry out the weighings 2 and 3 on this preceding arrangement and note the results. Let's continue with the example: let's suppose that the weighing results are actually > and <. The only ABCD line satisfying this is 3 8 5 11. So these are the weights of the labelled objects in our example. (The arrangement for weighings 2 and 3 is 11 3 8 5.) Another example: weighings 5 and 6 are < and >. This implies 5 8 11 3, 3 5 11 8, 8 5 11 3, 3 8 11 5, 8 3 11 5, 5 3 11 8. This time C=11. However, we can't use the weighings of the previous arrangements, because the results aren't unique. But if we move down or up two arrangements, we have uniqueness. Suppose we carry out weighings 2 and 3 on this other arrangement and get the results > and =, then ABCD is 8 3 11 5. (The arrangement for weighings 2 and 3 is 11 5 8 3.) Sometimes you won't need to look for other arrangements to establish uniqueness, because the arrangements you find from weighings 5 and 6 already contain the uniqueness you need. And it doesn't always have to be weighings 2 and 3, it could be any two of the weighings between 1 and 4 that are used for the weighing test, so long as it's always the same two weighings that are used for the six possibilities. Hint: = and > are much less common than <, so = and > in the weighings between 1 and 4, so look out for them when establishing uniqueness.
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how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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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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using the digits 1-9 only once and use each digit with only addition to make a 100?

The sum of the digits 1 to 9=45, which is divisible by 9. The 9's remainder (or digital root) has to be preserved during arithmetical operations. If, for example, we add 23 to 87 we get 110. The digital root for each operand is found by simply adding the digits of the number together: 2+3=5 and 8+7=15, we keep doing this until we have a single figure: 1+5=6, so the digital root of 87 is 6. The remainder after dividing by 9 equals the digit root. 23 divided by 9 is 2 rem 5; 87/9=9 rem 6. When numbers are added together the result has a digital root equal to the sum of the digital roots of the operands; so 5+6=11 and 1+1=2, so the result of 23+87=110 and 110 has a digital root of 2 (1+1+0=2).  Where is all this leading? Since we can only use the digits 1 to 9 once, we know that no matter what numbers we choose to make out of those digits the final result will have a digit root of 9 because 4+5=9. But the digital root of 100 is 1; therefore it is not possible to find numbers using the digits 1 to 9 once only such that the sum is 100, because the digital roots don't add up. The closest we can get is 99, which has a digital root of 9 because 9+9=18=1+8=9, matching the digital root of 45. One answer (not the only one) is: 59+12+3+4+6+7+8=99.
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how di you use four fours to get 5

4 +(4/4)^4 ..............
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