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use 5 five times and any mathematical order to find the result 1

you can use 5 five times and use any mathematical order but the result should be 1

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Pop Quiz - Math - Math Tricks Explained -

Dec 29, 2009 · Amazing feats of mathematics. It’s magic. Skip to ... you have 5. 4. Use any A.T.M. card or credit card with 16 ... The grand total is 11 times 5 x + 8 y.
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By using the numbers 8,5 1 and 1 exactly once and with the ...

Using 1, 2, 3, 4 & 5 in any order and the ... how many numbers of five digits can be formed by using 1,2,3,4,5. ... find a mathematical expression with a result ...
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Mathematics 1 - Exeter Math - Phillips Exeter Academy

Mathematics 1 Mathematics Department ... IN ALL CLASSES THAT MEET FIVE TIMES A WEEK, INCLUDING MATHEMATICS, ... Mathematics 1 1 Phillips Exeter Academy ...
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Math Forum - Ask Dr. Math Archives: Number Sentences

Number Sentences (Order of Operations ... How can you get 73 by using 4 fours and any mathematical ... How can a result of 75 be reached by pushing these keys ...
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Numbers - Exponents - In Depth -

© 2000-2005 Math .com. All rights ... and means "five times five times five." The base can be any sort of ... rules to use with exponents. a 1 = a Any number raised ...
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CAHSEE on Target - Kern County

CAHSEE on Target Mathematics Curriculum ... math equation, 5 comes before 3. The order is reversed . ... five. x 5 Five times a number is equal to
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5.7 Translating English Sentences into Mathematical Equations ...

When six is subtracted from five times a number, the result ... arithmetic using the proper order of ... Translating English Sentences into Mathematical Equations ...
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Mathematical Puzzles: What is () + () + () = 30 using 1,3,5,7 ...

Mathematical Puzzles: What is + + = 30 using ... Or using all the same base: [math]13_5 + 11_5 + 1_5 = 30_5[/math] in base five, ... This question has come up many ...
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The Order of Operations: PEMDAS | Purplemath

Explains the order of operations ... But we can't have this kind of flexibility in mathematics; ... The order of operations was settled upon in order to prevent ...
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Math Forum: Ask Dr. Math FAQ: Order of Operations

What is the correct order of operations? Why use ... they give you the same result no matter what order ... Or search the Dr. Math archives for "order of operations" ...
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Suggested Questions And Answer :

use 5 five times and any mathematical order to find the result 1

answer is  [{(5-5)*5}+5]/5=1
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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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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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Probability Question

There are 70 combinations of quartets of these 8 numbers, or, in other words, 70 different ways of grouping 4 outcomes from S. They all have the same probability of being selected as a group because each member of S has the same probability of selection. These are combinations of the possible outcomes of simply picking 4 numbers from 1 to 8 so that once drawn, a number cannot be returned to the pool. After a group of 4 has been drawn, then all the numbers are returned to the pool for the next group selection in this experiment. If we add together the numbers in each quartet we get a particular sum, associating each quartet with a sum. For example, 1+2+3+4=10 and 5+6+7+8=26. So the sum is in the range 10 to 26, and the sums are not usually unique. For example: 1+2+3+8=1+3+4+6=2+3+4+5=14. So, given a sum we can identify how many quartets have that sum. Or, we can find out how many have less or more than this sum. That gives us a probability based on the number of occurrences divided by the total number of possibilities. If we can find a sum that is shared by 49 quartets then the probability of selecting that sum will be 49/70=7/10=0.7. Alternatively, we can count the number of quartets that are below or above that sum. This will be one of the requirements of the experiment. But it doesn't have to be a sum. We can use other operators than plus. In fact, between 4 numbers we can have three operators, for example, 1*2+3*4=14, or 1*2*3*4=24. So two possible events for X will simply be the result of applying different operators. If we take the sample space for the operators as { + - * } then there will be 6 permutations of these, if all operators are to be applied in order, to ordered elements of each quartet. The elements can be ordered in ascending or descending order and the results will usually be different. The normal priority rules would apply (multiplication takes priority over addition and subtraction). All that's needed is a particular result (< X, ≤ X, = X, ≥ X, > X), so that the number of occurrences is 49 out of the possible 70 outcomes. Then P(X)=49/70=0.7. The question doesn't ask for specific X, just possibilities for X. If I can, I'll identify specific events for X...when I have time!  
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It's clear that the percentages exceed 100%, so I conclude that the dry matter content includes the crude ingredients, which are supplemented by another ingredient that is not specified. I also conclude that the dry matter content indicates that water is present, because none of the percentages for dry matter are 100%. First, remove water content and recalculate percentages: to do this we divide each percentage of the components by the percentage of dry matter. The result is expressed as a percentage. This enables us to mix the feed on a dry basis.  The adjusted percentages are: CORN: CP: 9.4909%, CF: 2.3812%, CFB: 1.1339% RICE: CP: 13.4427%, CF: 13.7776%, CFB: 8.3150% SBM: CP: 48.0682%, CF: 6.3636%, CFB: 5.1136% WHEAT: CP: 18.5698%, CF: 5.1903%, CFB: 8.5352% COPRA: CP: 20.7792%, CF: 13.2035%, CFB: 12.4459% We also have to adjust the percentages on the required feed: CP: 18.1818%, CF: 4.5455%, CFB: 13.6364% The last letter of each ingredient will be used to signify the fraction of that ingredient needed to make up the required feed (N, E, M, T, A). So we can write: CP REQUIREMENT: 9.4909N+13.4427E+48.0682M+18.5698T+20.7792A=18.1818 CF REQUIREMENT: 2.3812N+13.7776E+6.3636M+5.1903T+13.2035A=4.5455 CFB REQUIREMENT: 1.1339N+8.3150E+5.1136M+8.5352T+12.4459A=13.6364 N+E+M+T+A=1 (The ingredient fractions must add up to 1.) [To show these equations are valid, let's move away from percentages and consider actual amounts in the dry mix. For example, n grams of corn contains 9.4909n/100 grams (0.094909n grams) of crude protein. Similarly for the other ingredients: e grams of rice contains 13.4427e/100 grams (0.134427e grams) of crude protein. When we add together the crude protein contributions for all the ingredients we get the required amount, 18.1818x/100 (0.1818x grams), where x grams is the weight of the dry mix=n+e+m+t+a. So we can multiply through by 100 to leave the percentage numbers; if we divide through by x we get n/x, e/x, etc., and we can replace these by N, E, etc., where these are fractions of the total feed: N=n/x, E=e/x, ... X=x/x=1.] There are five variables but only four equations. The amount of copra is A=1-(N+E+M+T), so the equations can be reduced to omit A: CP REQUIREMENT: 11.2883N+7.3366E-27.2890M+2.2094T=2.5974 CF REQUIREMENT: 10.8222N-0.5741E+6.8398M+8.0132T=8.6580 CFB REQUIREMENT: 11.3120N+4.1308E+7.3323M+3.9107T=-1.1905 This last equation is suspicious because all the values on the left are positive but the value on the right is negative. Therefore at least one assumption in the logic is false. To resolve this difficulty we have to change the requirements to include inequalities. For example, the mix must contain at least a certain amount of an essential ingredient, or no more than a certain amount, as well as exactly a certain amount. We have no guide in the question to make any assumptions. Nevertheless, we'll continue with the solution and see where it leads us. The plan now is to treat T as an independent variable or constant, and to find each of N, E and M in terms of T. T is therefore a parameter from which the other fractions can be derived. We can eliminate M from CP and CF: 6.8398(11.2883N+7.3365E-27.2890M+2.2094T) + 27.2890(10.8223N-0.5714E+6.8398M+8.0132T)=6.8398*2.5974+27.2890*8.6580 77.2104N+50.1808E+15.1122T+295.3277N-15.6668E+218.6706T=17.7658+236.2681 372.5381N+34.5140E+233.7828T=254.0338 And we can similarly eliminate M from CF and CFB: 7.3323(10.8223N-0.5714E+6.8399M+8.0132T) - 6.8398(11.3120N+4.1308E+7.3323M+3.9107T)=7.3323*8.6580+6.8398*1.1905 79.3514N-4.2095E+58.7544T-77.3719N -28.2542E-26.7486T=63.4827+8.1427 1.9795N-32.4637E+32.0059T=71.6253. We now have two equations involving N, E and T, and we can eliminate E between them and so find N in terms of T: 32.4637(372.5381N+34.5140E+233.7828T) + 34.5140(1.9795N-32.4637E+32.0059T)=32.4637*254.0338+34.5140*71.6253, 12162.2997N+8694.1124T=10718.9626, 12162.2997N=10718.9626-8694.1124T, N=0.8813-0.7148T. If we substitute this value of N in 1.9795N-32.4637E+32.0059T=71.6253, we can find E in terms of T: 1.9795(0.8813-0.7148T)-32.4637E+32.0059T=71.6253; 1.7445-1.4149T-32.4637E+32.0059T=71.6253 -32.4637E+30.5910T=69.8808; E=(30.5910T-69.8808)/32.4637, E=0.9423T-2.1526. Substituting for E and N we can find M then A: using the CP requirement equation: 11.2883N+7.3366E-27.2890M+2.2094T=2.5974 11.2883(0.8813-0.7148T)+7.3366(0.9423T-2.1526)-27.2890M+2.2094T=2.5974, 9.9484-8.0689T+6.9133T-15.7928-27.2890M+2.2094T=2.5974, 1.0538T-27.2890M=8.4418, M=(1.0538T-8.4418)/27.2890=0.0386T-0.3093. We now have N, E and M in terms of T. We can find A from A=1-(N+E+M+T): A=1-(0.8813-0.7148T+0.9423T-2.1526+0.0386T-0.3093+T)=2.5806-1.2661T. All the ingredients have now been found in terms of T (wheat). These values are based on equality in the ingredient equations.  Summary N=0.8813-0.7148T, E=0.9423T-2.1526, M=0.0386T-0.3093, A=2.5806-1.2661T. These are all supposed to be positive fractions<1, and clearly they are not! This implies that the exact amounts required in the mix cannot be achieved. I've checked the arithmetic and logic fully and I can find no errors, so it does not appear to be possible to mix the desired quantities of the essential ingredients. It may be possible to get the right quantity of one particular ingredient at the expense of the others, or with an excess of at least one of the other ingredients. An excess could be as undesirable as a deficiency.
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How to Find Square Root

98=49*2, so sqrt(98)=sqrt(49)*sqrt(2)=7sqrt(2)=7*1.4142=9.8994 approx. There's another way using the binomial theorem. 98=100-2=100(1-0.02). sqrt(100)=10 so sqrt(98)=10(1-0.02)^(1/2) because square root is the same as power 1/2. (1+x)^n expands to 1+nx+(n(n-1)/1*2)x^2+(n(n-1)(n-2)/1*2*3)x^3+... Putting n=1/2 and  x=-0.02, we get sqrt(98)=10(1-0.02)^(1/2)=10[1-(1/2)0.02+((1/2)(-1/2)/2)0.0004+...]. This gives us: 10(1-0.01-0.00005+...)=10*0.98995=9.8995. A third method is to use an iterative process, which means you keep repeating the same action over and over again. Look at this: x=10-(2/(10+x)). If we solve for x we get x=sqrt(98); but we're going to find x in an iterative way. Start with x=0 and work out the right hand side: 10-2/10=9.8. This gives us a new value for x, 9.8, which we feed back into the right hand side: 10-(2/(10+9.8))=10-2/19.8=9.8989..., giving us another value for x, 9.8989... which we feed back into the right hand side: 10-(2/(10+9.8989...))=9.89949..., giving us yet another value for x and so on. Very quickly we build up accuracy with each x. You can do this on a calculator, a basic one that doesn't even have square roots, using the memory to hold values for you. Here's a very simple program, where STO means store in memory (if your calculator doesn't have STO use MC (memory clear) followed by M+ (add to memory)); MR means read memory (the steps show what calculator keys to press in order; / may be ÷ on your calculator): 0= +10=STO 10-2/MR= GO TO STEP 2 OR STOP (display should show the answer for sqrt(98)) Note: In STEP 3 the division must be carried out before subtracting from 10, otherwise you get the wrong answer. If your calculator doesn't do this you need to replace STEP 3 with: 0-2=/MR=+10= You should only have to go round the loop a few times before you get a really accurate result. To find the square root of 2 directly the iteration equation is x=1+1/(1+x) and the program is: 0= +1=STO 1/MR+1= GO TO STEP 2 OR STOP STEP 3 should work on all calculators.
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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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math triangle sums equals 20 on all sides using the numbers 1 - 9 only once

Note that if we take the numbers 1 to 19 in 9 pairs, the sums of the paired numbers come to 20: (1,19), (2,18), (3,17), (4,16), (5,15), (6,14), (7,13), (8,12), (9,11). Using the numbers 1 to 9 only, we can break down 4 of these pairs by splitting the larger number of each pair: (3,8,9), (4,7,9), (5,6,9), (5,7,8). These are the only unique groups of three, but they do not include 1 and 2. But if we include 1 and 2 we know the remaining 2 numbers to make up a group of 3 must add up to 19, which can't be achieved using a pair of numbers 1 to 9. Therefore, in some cases we need 4 numbers to make up the sum. We can only use  the numbers 1 to 9 and we need to use them all and we only have 3 sides of the triangle, so if we have 4 numbers on one side, one of the other sides will have only 2 numbers, because we're only left with 5 when we've taken out 4 for one side. What to do? We have to use the vertices so that the numbers at each vertex are used twice. Let's assume that all three vertices contain such numbers, then we will have 4 numbers a side. Instead of looking at groups of three numbers, we have to look for groups of 4: (1,2,8,9), (1,3,7,9), (1,4,6,9), (2,3,6,9), (1,4,7,8), (2,3,7,8), (1,5,6,8), (2,4,5,9), (2,4,6,8), (3,4,5,8), (3,4,6,7). Out of these we need to find where different groups contain a common number. These common numbers will form the vertices. Let's try putting the numbers 1, 2 and 3 at the vertices. This gives us a choice of 5 groups for each of the numbers 1, 2 and 3 (we can arrange the order so that the numbers occupy the first element in each set). Here are the sets: (1,2,8,9), (1,3,7,9), (1,4,6,9), (1,4,7,8), (1,5,6,8) (2,1,8,9), (2,3,6,9), (2,3,7,8), (2,4,5,9), (2,4,6,8) (3,1,7,9), (3,2,6,9), (3,2,7,8), (3,4,5,8), (3,4,6,7) Now we have to rearrange the numbers so that there are two vertices within each set, so that means we pick groups containing 1 and 2, 2 and 3, and 3 and 1. Yes, there are some, but unfortunately some contain duplicates of the two numbers that are not at the vertices. Look at the following, for example: (1,8,9,2), (2,6,9,3), (3,7,9,1) 9 is duplicated and we have no 4 or 5. In fact, we can"t have 1 as a vertex with 2 and 3 as the other vertices, because there's always a duplicate to spoil the arrangement. The same thing happens with 2, 3 and 4, and 3, 4 and 5. Symmetry is a common occurrence in mathematics and there's nothing more pleasing to a mathematician than an "elegant" solution. So I'm going to assume that symmetry applies to this problem and go for vertices in the middle of the range 1 to 9. The middle is occupied by numbers 4, 5 and 6. Let's see what happens if we use them as vertices: (4,2,9,5), (5,1,8,6), (6,3,7,4) This time it works! There are no duplications in the central pair of numbers so all the numbers from 1 to 9 are there. The other combination of 4 and 5 in the same group, (4,3,8,5), leads to a duplicated 8 in (5,1,8,6). So one answer is:  The vertices are A, B and C=4, 5 and 6 respectively, with 2 and 9 along AB, 1 and 8 along BC, and 3 and 7 along AC.  
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Whats the calculations for the Pythagorean spiral project?

Starting with an isosceles right-angled triangle of equal sides of, say, 10cm, the length of the hypotenuse is sqrt(10^2+10^2)=sqrt(2*10^2)=10sqrt(2), using Pythagoras' theorem c^2=a^2+b^2. Now building on the hypotenuse as one side of a new right-angled triangle, we make the other side 10cm. So all we're doing is using c as a new b and finding the next value for c. This time the hypotenuse is sqrt(2*10^2+10^2)=sqrt(3*10^2)=10sqrt(3). Then the process is repeated, and the hypotenuse of the next triangle is 10sqrt(4)=20cm. Each segment of the spiral is 10cm long, but the radius increases as the square root of the natural numbers: 1, 2, 3, 4, etc. so the nth radius is 10sqrt(n). The length of the spiral is 10n. The spiral is usually shown mathematically as just the unit side, so the perimeter or length of the spiral is just n and radius sqrt(n). It's not a true spiral, because the edges are straight, whereas a true mathematical spiral is a continuous curve. The angle between consecutive radii is tan^-1(1/sqrt(n)) so it depends on where you are on the spiral. The largest angle is the first one at 45 degrees. The next is tan-1(0.7071)=35.3 degrees approximately, then 30 degrees, then 26.6, 24.1, 22.2, 20.7, 19.5, 18.4, 17.5,... When we reach sqrt(16)=4, the sum of the angles comes to about 351.15 degrees, and the next segment of the spiral takes us a little past a complete revolution at 364.78 degrees. The project normally stops at this point, and the final result looks like the shell of a nautilus mollusc (which is rounded).
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What's a function of function?

If you just mean the purpose of a function then see immediately below. If you mean f of g where f and g are two functions, go to the end. The purpose of a function is like giving someone a to-do list. A function (normally shown as f(x)= meaning "function of x", or y=) will usually contain one variable, usually represented by x. Some functions may have more than one variable. If x isn't the variable, it will usually be a different letter like t or a or any letter. Let's say it is x. This variable will appear in a formula with numbers. The formula is the function and it contains instructions in symbols telling you what to do with the variable. For example, multiply the variable by 2 then add 3 and divide the result by 4. This would be written f(x)=(2x+3)/4 or y=(2x+3)/4. The equals sign means that the function is defined as the expression on the right. Just like someone might say to you, think of a number (that's x), but don't tell me what it is, then double it and add 3 and divide the result by 4. Functions can be plotted on a graph. The idea of this is that on the horizontal axis (x axis) you have markers for 0, 1, 2, etc. On the vertical axis (y or f(x)) you also have markers. The graph is usually a continuous line, and every point on the line is made by putting a different value for x and marking the point (on a rectangular grid f(x) units vertically and x units horizontally) for the result of working out the value of the function for different values of x. The points make up the graph. You can use the graph to find the value of x for any value of the function, and the value of the function for any value of x, provided the graph extends far enough. This is just a brief example of the purpose of a function. A practical example is the conversion of temperature in degrees Fahrenheit (F) to degrees Celsius: C=5(F-32)/9. This function tells you what do with the value of the temperature. You could plot this as a straight line graph and you can read off degrees C for any temperature in degrees F, and vice versa. Another example is d=30t where d=distance, speed=30mph, t=time. The function is 30t, and from it you can work out the distance a car moves in a particular time t when its speed is 30mph. Function of a function: this simply means you work out the value of applying one function and then feed this answer as the variable into the other function. An example could be that one function is used to find out how many miles a car travelling at an average speed of 30mph in a given time. The second function could be to find out how much fuel is used to travel that distance. So g(t)=30t is the first function. The second function is h(d)=d/24 where fuel consumption is 24 miles per gallon. h of g is h(g(t))=30t/24=5t/4.
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