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In the Name of God بسم الله

SOLVE THESE RIDDLES

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

Using the numerals 1,7,7,7 and 7 (a "1" and four "7"s) create the number 100. As well as the five numerals you can use the usual mathematical operations (+, -, x, ÷ and brackets (). For example: (7+1) × (7+7) = 112 would be a good attempt, but not right, because it is not 100.

I am thinking of a 6-digit number. The sum of the digits is 43. And only two of the following three statements about the number are true:

(1) it's a square number,

(2) it's a cube number, and

(3) the number is under 500000.

If one and a half hens lay one and a half eggs in one and a half days, how many eggs does one hen lay in one day?

Guess the next number in series: 1,2,6,42,1806,? What is the next number?

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  • Advanced Member
What mathematical symbol can be put between 5 and 9, to get a number bigger than 5 and smaller than 9?

Using the numerals 1,7,7,7 and 7 (a "1" and four "7"s) create the number 100. As well as the five numerals you can use the usual mathematical operations (+, -, x, ÷ and brackets (). For example: (7+1) × (7+7) = 112 would be a good attempt, but not right, because it is not 100.

I am thinking of a 6-digit number. The sum of the digits is 43. And only two of the following three statements about the number are true:

(1) it's a square number,

(2) it's a cube number, and

(3) the number is under 500000.

If one and a half hens lay one and a half eggs in one and a half days, how many eggs does one hen lay in one day?

Guess the next number in series: 1,2,6,42,1806,? What is the next number?

A decimal point. So, 5.9.

177-77 = 100

No idea.

One? Though I don't know why it says how many eggS, which shows it might be plural. Again, no idea.

3263442

I take procrastination to a whole new level.

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

Using the numerals 1,7,7,7 and 7 (a "1" and four "7"s) create the number 100. As well as the five numerals you can use the usual mathematical operations (+, -, x, ÷ and brackets (). For example: (7+1) × (7+7) = 112 would be a good attempt, but not right, because it is not 100.

I am thinking of a 6-digit number. The sum of the digits is 43. And only two of the following three statements about the number are true:

(1) it's a square number,

(2) it's a cube number, and

(3) the number is under 500000.

If one and a half hens lay one and a half eggs in one and a half days, how many eggs does one hen lay in one day?

Guess the next number in series: 1,2,6,42,1806,? What is the next number?

Nice post.

Questions 2, 4 and 5 are pretty straight forward.

Question 2 is a nice little one:

7+7=14

14 = 98/7.....7=49/7

49/7+(1/7)= 50/7

98/7 x 50/7 = 4900/49 = 100

Thus ((1/7) + 7) x (7 + 7) = 100

Question 4 is simple from a mathematical point of view but unsure whether it requires one to take in account of the biological point of view haha but since the question already presupposes 1/2 hens and 1/2 eggs then answer is as follows:

3/2 x 3/2M = 3/2eggs as number of eggs is directly proportional to number of hens and number of days.

M = (3/2) / (3/2)^2 = 2/3

Therefore 1 hens x 1 day x (M=2/3) = 2/3 eggs. Or strictly speaking no eggs as it requires more than 1 day to yield the first egg or we can say 1 egg which is two thirds of the way to being laid lol.

Question 5, very simple. Whenever there seems exponential increase in a sequence (or decrease for that matter) then powers obviously involved. Like with any sequence evaluate the difference and we see (2-1) = 1, (6-2) = 4 , (42-6) = 36 and (1806-42) = 1764. Quick inspection yields the relation 1 sqrd = 1, 2sqrd =4, 6sqrd = 36 and 42sqrd=1764.

Thus 1+1=2.....2+4=6.....6+36=42......42+1764=1806.....1806+(1806)sqrd=3,263,442 .....etc

I dont get the point of question 3 whatsoever (not just because i dont know the answer lol). Wouldnt it be better to state which of the statements are true or at least guarantee one of them to be true to save spending unworthy time going through all the various possibilities. With six digits adding up to 43 being the only given condition there simply too many numbers to go through lol to work it out. Tell us which two statements are true and i might have a stab at it.

As for question 1, would really like to know the answer to this one. Apart from the obvious 5 < x < 9 which involves 3 symbols there is nothing that can satisfy the question. We can always cheat by using the symbol of "." lol to create 5.9 but of course the decimal point isnt a mathematical symbol.

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Nice post.

Questions 2, 4 and 5 are pretty straight forward.

Question 2 is a nice little one:

7+7=14

14 = 98/7.....7=49/7

49/7+(1/7)= 50/7

98/7 x 50/7 = 4900/49 = 100

Thus ((1/7) + 7) x (7 + 7) = 100

Question 4 is simple from a mathematical point of view but unsure whether it requires one to take in account of the biological point of view haha but since the question already presupposes 1/2 hens and 1/2 eggs then answer is as follows:

3/2 x 3/2M = 3/2eggs as number of eggs is directly proportional to number of hens and number of days.

M = (3/2) / (3/2)^2 = 2/3

Therefore 1 hens x 1 day x (M=2/3) = 2/3 eggs. Or strictly speaking no eggs as it requires more than 1 day to yield the first egg or we can say 1 egg which is two thirds of the way to being laid lol.

Question 5, very simple. Whenever there seems exponential increase in a sequence (or decrease for that matter) then powers obviously involved. Like with any sequence evaluate the difference and we see (2-1) = 1, (6-2) = 4 , (42-6) = 36 and (1806-42) = 1764. Quick inspection yields the relation 1 sqrd = 1, 2sqrd =4, 6sqrd = 36 and 42sqrd=1764.

Thus 1+1=2.....2+4=6.....6+36=42......42+1764=1806.....1806+(1806)sqrd=3,263,442 .....etc

I dont get the point of question 3 whatsoever (not just because i dont know the answer lol). Wouldnt it be better to state which of the statements are true or at least guarantee one of them to be true to save spending unworthy time going through all the various possibilities. With six digits adding up to 43 being the only given condition there simply too many numbers to go through lol to work it out. Tell us which two statements are true and i might have a stab at it.

As for question 1, would really like to know the answer to this one. Apart from the obvious 5 < x < 9 which involves 3 symbols there is nothing that can satisfy the question. We can always cheat by using the symbol of "." lol to create 5.9 but of course the decimal point isnt a mathematical symbol.

I'm scared.

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Ok, true, it leaves the second statement of having a whole cube root as the false one. So how did you work it out?

found out that a number with the digit sum of 43 could not be a cube and then trial and error. Took the square root of 500,000 and worked down.

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Where are the answers by the OP?

Question 5, very simple. Whenever there seems exponential increase in a sequence (or decrease for that matter) then powers obviously involved. Like with any sequence evaluate the difference and we see (2-1) = 1, (6-2) = 4 , (42-6) = 36 and (1806-42) = 1764. Quick inspection yields the relation 1 sqrd = 1, 2sqrd =4, 6sqrd = 36 and 42sqrd=1764.

Thus 1+1=2.....2+4=6.....6+36=42......42+1764=1806.....1806+(1806)sqrd=3,263,442 .....etc

I did it a bit differently although yours could be more of an 'expert method':

1 - 2(*3) - 6(*7) - 42(*43) - 1806(*1807) - 3263442

The first number doesn't hold true for the calculation but the rest do.

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found out that a number with the digit sum of 43 could not be a cube and then trial and error. Took the square root of 500,000 and worked down.

Aha..well done. It is indeed true that any figure with digit sum of 43 cannot be a cube as 43 = 7 (mod 9) and thus no cube roots. Very good spot. Not a fan of a "trial and error" method at the best of times for obvious reasons but particularly so here. Luckily you started from 500,00 which immediately gives you the answer as 707 is the first square root working down.

Even with the conditions given i find it impossible to cut it down to any number below 55 or so possibilities that necessitate "trialling" through. Hardly a "riddle".

Where are the answers by the OP?

I did it a bit differently although yours could be more of an 'expert method':

1 - 2(*3) - 6(*7) - 42(*43) - 1806(*1807) - 3263442

The first number doesn't hold true for the calculation but the rest do.

The two methods are identical. I merely explained it to demonstrate how one can very easily spot whats happening in the sequence without having to "brainstorm".

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Aha..well done. It is indeed true that any figure with digit sum of 43 cannot be a cube as 43 = 7 (mod 9) and thus no cube roots. Very good spot. Not a fan of a "trial and error" method at the best of times for obvious reasons but particularly so here. Luckily you started from 500,00 which immediately gives you the answer as 707 is the first square root working down.

Thanks. It wasn't exactly luck that made me start from the square root of 500,000 but more because the number (43) was so large that there needed to be some 9's in it - and 9 couldn't be the first number so it would probably be 49.... something.

Please excuse my primitive solving technique - I have no skills other than logic based trial and error - I am not a mathematician past second year level - and given that it was a few years ago, I vaguely remember that. :blink:

Oh - but I got the number series one immediately :angel: :P - by looking at how fast the numbers increased I could figure out that it was something added or subtacted from a square.....so again just trial and error.

Don't know about the hens one... ??

Edited by Maryaam
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Thanks. It wasn't exactly luck that made me start from the square root of 500,000 but more because the number (43) was so large that there needed to be some 9's in it - and 9 couldn't be the first number so it would probably be 49.... something.

Please excuse my primitive solving technique - I have no skills other than logic based trial and error - I am not a mathematician past second year level - and given that it was a few years ago, I vaguely remember that. :blink:

Oh - but I got the number series one immediately :angel: :P - by looking at how fast the numbers increased I could figure out that it was something added or subtacted from a square.....so again just trial and error.

Don't know about the hens one... ??

A two fold apology is due.

Firstly, you spotted something I simply glossed over at the time which warrants a big well done and the basis for your solution is an impressive one. I did not intend to use "luck" as a mitigating factor whatosever. I was merely expressing dissatisfaction that such a so called "riddle" must involve "trialling" of some kind or another.

I also misunderstood your method at first for i thought you were working with the roots (i.e ascertaining the legitimate roots between 320 and 707 that satisfy the conditions stated through exploring modular activity) as oppose to starting with the sum itself. I would slightly question your reasoning that the most "probable" digit after "4" would be 9 particularly if starting with "4". I am sure you can quickly work out that there are many more six digit figures that sum to 43 that do not have second digit "9" as oppose to those that do. However, you are right to start from 500,000 and again well done.

I hope that whilst it has been a few years lol, that all the Maths you do remember continues to stand you in good stead.

Well done in getting the series riddle correct. They are not always so obvious to a mathematician so credit to anyone who manages to spot them with ease.

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A two fold apology is due.

Firstly, you spotted something I simply glossed over at the time which warrants a big well done and the basis for your solution is an impressive one. I did not intend to use "luck" as a mitigating factor whatosever. I was merely expressing dissatisfaction that such a so called "riddle" must involve "trialling" of some kind or another.

I also misunderstood your method at first for i thought you were working with the roots (i.e ascertaining the legitimate roots between 320 and 707 that satisfy the conditions stated through exploring modular activity) as oppose to starting with the sum itself. I would slightly question your reasoning that the most "probable" digit after "4" would be 9 particularly if starting with "4". I am sure you can quickly work out that there are many more six digit figures that sum to 43 that do not have second digit "9" as oppose to those that do. However, you are right to start from 500,000 and again well done.

I hope that whilst it has been a few years lol, that all the Maths you do remember continues to stand you in good stead.

Well done in getting the series riddle correct. They are not always so obvious to a mathematician so credit to anyone who manages to spot them with ease.

Thanks again :) Just one small comment - re: the 49... reasoning.... I thought the second number had a good likelihood to be a 9 as the first number could not be greater than a 4 and that left 5 digits to add to 39; it was square so the last number had to be even - so couldn't be a 9 (highest would be an 8) - but yet I figured there was probably two or three 9's, so I thought it probable (not absolute - but a good guess) that the second number was a 9..at least it was a good start point. Edited by Maryaam
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Thanks again :) Just one small comment - re: the 49... reasoning.... I thought the second number had a good likelihood to be a 9 as the first number could not be greater than a 4 and that left 5 digits to add to 39; it was square so the last number had to be even - so couldn't be a 9 (highest would be an 8) - but yet I figured there was probably two or three 9's, so I thought it probable (not absolute - but a good guess) that the second number was a 9..at least it was a good start point.

Last number can be a 9 and indeed it was.

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Last number can be a 9 and indeed it was.

Of course it can be :D and it was!! What was I thinking? I guess I was remembering the beginning of my thinking of why the first and last number could not be 9, but it was faulty. Told you I was a math illiterate!! I just stumble around until I land in the right place...very labour intensive.

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Of course it can be :D and it was!! What was I thinking? I guess I was remembering the beginning of my thinking of why the first and last number could not be 9, but it was faulty. Told you I was a math illiterate!! I just stumble around until I land in the right place...very labour intensive.

Lol, not to worry.

Your reasoning still stands, however i was merely hinting in the original post that 49 is by no means the most "probable" out of the six digit figures under 500,000 to sum to 43. It is more of a 2/5 chance.

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Of course it can be :D and it was!! What was I thinking? I guess I was remembering the beginning of my thinking of why the first and last number could not be 9, but it was faulty. Told you I was a math illiterate!! I just stumble around until I land in the right place...very labour intensive.

Don't be so self-deprecating. You solved the problem, which the overwhelming majority of people could not have done.

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