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


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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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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 a
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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 valu
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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) represent
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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
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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 &nb
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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 ex
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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, an
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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=
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how di you use four fours to get 5

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