What mathematical symbol can be placed between 5 and 9, to get a number greater than 5 and smaller than 9?

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Symbol - solutiondecimal point – 5.9

There are infinite formulas that will fit any finite series. Try to guess the following number in each sequence (using the most simple mathematical operations, because as I mentioned, there is more than one solution for each number sequence).

* 8723, 3872, 2387, ?

* 1, 4, 9, 18, 35, ?

* 23, 45, 89, 177, ?

* 7, 5, 8, 4, 9, 3, ?

* 11, 19, 14, 22, 17, 25, ?

* 3, 8, 15, 24, 35, ?

* 2, 4, 5, 10, 12, 24, 27, ?

* 1, 3, 4, 7, 11, 18, ?

* 99, 92, 86, 81, 77, ?

* 0, 4, 2, 6, 4, 8, ?

* 1, 2, 2, 4, 8, 11, 33, ?

* 1, 2, 6, 24, 120, ?

* 1, 2, 3, 6, 11, 20, 37, ?

* 5, 7, 12, 19, 31, 50, ?

* 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, ?

* 126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, ?

* 4, 7, 15, 29, 59, 117, ?

* 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, ?

* 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, ?

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Number Series - solution• 8723, 3872, 2387, ?

7238 (moving of numerals)• 1, 4, 9, 18, 35, ?

68 (x*2+2, +1, +0, -1, -2)• 23, 45, 89, 177, ?

353 (x*2-1)• 7, 5, 8, 4, 9, 3, ?

10, 2 (two series – every second number: 7, 8, 9, 10 and 5, 4, 3, 2)• 11, 19, 14, 22, 17, 25, ?

20, 28 (two series – every second number: 11, 14, 17, 20 and 19, 22, 25, 28)• 3, 8, 15, 24, 35, ?

48 (x+5, +7, +9, +11, +13)• 2, 4, 5, 10, 12, 24, 27, ?

54, 58 (x*2, +1, *2, +2, *2, +3, *2, +4)• 1, 3, 4, 7, 11, 18, ?

29 (a+b=c, b+c=d, c+d=e …)• 99, 92, 86, 81, 77, ?

74 (x-7, -6, -5, -4, -3)• 0, 4, 2, 6, 4, 8, ?

6 (x+4, -2, +4, -2, +4, -2)• 1, 2, 2, 4, 8, 11, 33, ?

• 1, 2, 6, 24, 120, ?

720 (x*2, *3, *4, *5, *6)• 1, 2, 3, 6, 11, 20, 37, ?

• 5, 7, 12, 19, 31, 50, ?

81 (a+b=c, b+c=d, c+d=e …)• 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, ?

322, 161 (x*3+1, /2, *3+1, /2 …)• 126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, ?

485, 1456 (x/2, *3+1, /2, *3+1 …)• 4, 7, 15, 29, 59, 117, ?

235 (x*2-1, *2+1, *2-1 …)• 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, ?

2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5• 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, ?

4, 4

Find the number if:

1. The cipher is made of 6 different numerals.

2. Even and odd digits alternate, including zero (in this case as an even number).

3. The difference between two adjacent numerals is always greater than one (in absolute value).

4. The first two numerals (as one number) as well as the two middle numerals (as one number) are a multiple of the last two numerals (as one number).

What is the cipher? There is more than 1 solution.

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Cipher - solutionThe possible 2 last numerals are as follows: 03, 05, 07, 09, 14, 16, 18, 25, 27, 29 and 30. At least two multiples less than 100 (this condition is already accomplished), which consist of even and odd numeral (respecting all other conditions) are for 03, 07, 09 and 18 as follows:

03 – 27, 63, 69, 81

07 – 49, 63

09 – 27, 63, 81

18 – 36, 72, 90

There are 5 numbers that can be made of these pairs of numerals to create the cipher: 692703, 816903, 496307, 816309 and 903618. (If we assume, that also in the number 903618 is accomplished the requirement to alternate even and odd numbers, despite the opposite order.)

]]>Find the cipher if:

1. The cipher is made of 6 different numerals.

2. Even and odd figures alternate, including zero (in this case as an even number).

3. The difference of adjacent numerals is always bigger than one.

4. The first two numerals (as one number) as well as the two middle numerals (as one number) are a multiple of the last two numerals (as one number).

What is the cipher? (more than 1 solution)

Can you arrange 9 numerals - 1, 2, 3, 4, 5, 6, 7, 8 and 9 - (using each numeral just once) above and below a division line, to create a fraction equaling to 1/3 (one third)?

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Fraction - solution5832/17496 = 1/3

Find the four-digit number designated by asterisks, given the following:

* All four digits of the unknown number are different.

* None of the digits is zero.

* Each “0” on the right of each four-digit number below indicates that the number has a matching digit in a non-matching position with the unknown number.

* Each “+” on the right of each four-digit number below indicates that the number has a matching digit in a matching position with the unknown number.

6152 +0

4182 00

5314 00

5789 +

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Master Mind - solution6741

]]>What is the number replaced by asterisks, if:

* symbol 0 represents a guessed number in the line,

* symbol + represents a guessed number in the line, that is on its right place,

* the result has no number 0,

* no numeral is more than once in the result.

6152 +0

4182 00

5314 00

5789 +

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Find a 9-digit number, which you will gradually round off starting with units, then tenth, hundred etc., until you get to the last numeral, which you do not round off. The rounding alternates (up, down, up ...). After rounding off 8 times, the final number is 500000000. The original number is commensurable by 6 and 7, all the numbers from 1 to 9 are used, and after rounding four times the sum of the not rounded numerals equals 24.

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9-Digit Number - solution473816952 – if rounding changes the next numeral character

What 5-digit number has the following property? If we put numeral 1 in front of the number, we get a number three times smaller, than if we put the numeral 1 behind this number.

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5-Digit Number - solutionUsing an easy equation: 3(x+100000) = 10x+1 we find out that the number is 42857.

An easier number puzzle is as follows. Two friends are chatting:

- Peter, how old are your children?

- Well Thomas, there are three of them and the product of their ages is 36.

- That is not enough ...

- The sum of their ages is exactly the number of beers we have drunk today.

- That is still not enough.

- OK, the last thing is that my oldest child wears a red hat.

How old were each of Peter's children?

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Children - solutionLet’s start with the known product – 36. Write on a sheet of paper the possible combinations giving the product of 36. Knowing that the sum is not enough to be sure, there are two possible combinations with the same sum (1-6-6 a 2-2-9). And as we learned further that the oldest son wears a hat, it is clear that the correct combination of ages is 2-2-9, where there is exactly one of them the oldest one.

Using four sevens (7) and a one (1) create the number 100. Except the five numerals you can use the usual mathematical operations (+, -, x, , root and brackets ().

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100 - solution100 = 177-77 = (7+7)x(7+(1:7))

I do not know other solutions.

Using the numerals 1, 9, 9 and 6, mathematical symbols +, -, x, :, root and brackets create the following numbers:

29, 32, 35, 38, 70, 73, 76, 77, 100 and 1000.

All the numerals must be used in the given order (each just once) and without turning upside down.

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1996 - solutionI used [brackets] as the symbol for root.

This is definitely one of the harder number puzzles on this site.

A teacher says: I'm thinking of two natural numbers greater than 1. Try to guess what they are.

The first student knows their product and the other one knows their sum.

First: I do not know the sum.

Second: I knew that. The sum is less than 14.

First: I knew that. However, now I know the numbers.

Second: And so do I.

What were the numbers?

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Savoury - solutionThe numbers were 2 and 9. And here comes the entire solution.

There shall be two natural numbers bigger than 1. First student knows their product and the other one knows their sum.

The sum is smaller than 14 (for natural numbers bigger than 1), so the following combinations are possible:

2 2 ... NO - the first student would have known the sum as well

2 3 ... NO - the first student would have known the sum as well

2 4 ... NO - the first student would have known the sum as well

2 5 ... NO - the first student would have known the sum as well

2 6

2 7 ... NO - the first student would have known the sum as well

2 8

2 9

2 10

2 11 ... NO - the first student would have known the sum as well

3 3 ... NO - the first student would have known the sum as well

3 4

3 5 ... NO - the first student would have known the sum as well

3 6

3 7 ... NO - the first student would have known the sum as well

3 8 ... NO - the product does not have all possible sums smaller than 14 (eg. 2 + 12)

3 9 ... NO - the first student would have known the sum as well

3 10 ... NO - the product does not have all possible sums smaller than 14

4 4

4 5

4 6 ... NO - the product does not have all possible sums smaller than 14

4 7 ... NO - the product does not have all possible sums smaller than 14

4 8 ... NO - the product does not have all possible sums smaller than 14

4 9 ... NO - the product does not have all possible sums smaller than 14

5 5 ... NO - the first student would have known the sum as well

5 6 ... NO - the product does not have all possible sums smaller than 14

5 7 ... NO - the first student would have known the sum as well

5 8 ... NO - the product does not have all possible sums smaller than 14

6 6 ... NO - the product does not have all possible sums smaller than 14

6 7 ... NO - the product does not have all possible sums smaller than 14

So there are the following combinations left:

2 6 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 (it is impossible to create a pair of numbers from sum 8, so that the product would have an alternative sum bigger than 14 ... eg. if 4 and 4, then there is no sum – created from their product 16 – bigger than 14 – eg. 2 + 8 = only 10)

2 8

2 9

2 10

3 4 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14

3 6 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14

4 4 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14

4 5 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14

The second student (knowing the sum) knew, that the first student (knowing the product) does not know the sum and he thought that the first student does not know that the sum is smaller than 14.

Only 3 combinations left:

2 8 ... product = 16, sum = 10

2 9 ... product = 18, sum = 11

2 10 ... product = 20, sum = 12

Let’s eliminate the sums, which can be created using a unique combination of numbers – if the sum is clear when knowing the product (this could have been done earlier, but it wouldn’t be so exciting) - because the second student knew, that his sum is not created with such a pair of numbers. And so the sum can not be 10 (because 7 and 3) – the second student knew, that the first student does not know the sum – but if the sum was 10, then the first student could have known the sum if the pair was 7 and 3.

The same reasoning is used for eliminating sum 12 (because 5 and 7).

So we have just one possibility – the only solution – 2 and 9.

And that’s it.

]]>This is definitely one of the hardest number puzzles on this site.

A teacher says: I'm thinking of two natural numbers bigger than 1. Try to guess what they are.

The first student knows their product and the other one knows their sum.

First: I do not know the sum.

Second: I knew that. The sum is less than 14.

First: I knew that. However, now I know the numbers.

Second: And so do I.

What were the numbers?

Rectify the following equality 101 - 102 = 1 by moving just one digit.

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Equation - solutionMove the numeral 2 half a line up to achieve 101 - 10

^{2}= 1.

* Find a 10-digit number, where the first figure defines the count of zeros in this number, the second figure the count of numeral 1 in this number etc. At the end the tenth numeral character expresses the count of the numeral 9 in this number.

* Find a 10-digit number, where the first numeral character expresses the count of numeral 1 in this number, ..., the tenth numeral the count of zeros in this number.

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10-Digit Number - solution* Sum of all numerals must be ten because each numeral stands for the count of other numerals and because this number shall have ten numerals. Beginning to choose reasonable numerals for the first figure you can come across the correct number: 6210001000.

* 2100010006.

A teacher thinks of two consecutive numbers between 1 and 10 (1 and 10 included). The first student knows one number and the second student knows the second number. The following exchange takes place:

First: I do not know your number.

Second: Neither do I know your number.

First: Now I know.

What are the 4 solutions of this easy number puzzle?

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Easy Savoury - solutionNone of the students can have numbers 1 or 10, since they would guess the other one’s number with no problems. I will describe solutions at one end of the interval of numbers 1-10 (the same can be done on the other end).

Information that the second student does not know must be important for the first student. So the first one must expect that the second one has 1 or 3 (if the first one has 2). And as the second student does not know, then he has certainly not 1. So the first pair is 2 and 3.

If the first one had 3, then he would expect the other one to have either 2 or 4. But if the second one had 2 (and the second one would have known that the first one does not have 1), then he would know the number of the first student. However, neither the second student knows the answer – so he has 4. The second pair of numbers is 3 and 4.

Solutions at the other end of interval are 9 and 8 or 8 and 7.

]]>A teacher thinks of two consecutive numbers between 1 and 10 (Edit: 1 and 10 included). The first student knows one number and the second student knows the second number. The conversation of the students is as follows:

First: I do not know your number.

Second: Neither do I know your number.

First: Now I know.

Will you find all 4 solutions?

The day before yesterday I was 25 and the next year I will be 28. This is true only one day in a year. What day is my birthday?

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Birthday - solutionHe was born on December 31st and spoke about it on January 1st.

The grid below is to be filled with six numbers (3 vertical and 3 horizontal) from the given list. You can use each number more than once. After completing the number puzzle sum up all digits in the grid. This is defined as the score. What is the maximum possible score?

the grid and available numbers

Example: Using the numbers 40067 04802 78215 twice

The score is 73 (of course other solutions with higher scores are possible).

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The Number Puzzle - solutionThis one is not verified – I used 2 numbers 39543 and 89398. And this is what the grid looks like:

So the total score is 147.