Guide :

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

by using the ! or the square route or exponents

## Research, Knowledge and Information :

### Ruth Carver - Four 4's Puzzle - Math Forum

From the Math Forum Ruth Carver's . Math Tips & Tricks

### 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 ...

### 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 ...

### 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 ...

### 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 ...

### 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 ...

### 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 ...

## 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

### 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.

### 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.

### how di you use four fours to get 5

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