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Hi,
In this thread, I'll be posting some non-routine algebra questions.
Non-Routine Maths (alias N-R Maths) is a type of math that has little to no practical significance, but helps immensely in increasing one's mathematical mental horizon.
I'll try to post at least one question a week at the very least but don't count on me.
If you find something worthwhile which fits in this category, you're most welcome to post, but just remember to number you're questions, so it's easier to navigate.
If possible also use the bold and italics tag:
[b][i] Bold, Italic Text [/i][/b]
Result:
Bold, Italic Text
Answers (with solutions) are most welcome!!
Question 1
Let the value of a certain number be equal to 0.1234567891011121314...997998999,
The digits are obtained via writing the integers from 1 to 999 in order.
The 2016th digit to the right of the decimal point is m.
What is the value of m?
Learning is fun - Exceptio probat regulam!
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Hi CurlyBracket;
Nice one!
I hope I got it right!
Edit 5/5/2022: Improved my solution (see post #4)
Last edited by phrontister (2022-05-04 04:12:40)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Three cheers for Phrontister! Hip Hip Hooray!
The answer is certainly perfect, but how did you get this expression?
m=RIGHT(x+y+z-(ax+by+cx-2016)/3)
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...how did you get this expression?
m=RIGHT(x+y+z-(ax+by+cx-2016)/3)
Sorry...was a bit hasty there.
This is better:
m=RIGHT((2016-ax-by)/3+x+y)
Is that what you got?
I've fixed my previous post.
Last edited by phrontister (2022-05-04 04:31:06)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Well, here's how I went about it.
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Hi CurlyBracket;
Yes, that's my reasoning also (see post #2), and my m=RIGHT((2016-ax-by)/3+x+y) expression is an attempt to state it algebraically.
I probably cheated by employing spreadsheet's RIGHT function to extract the last digit of 708, but used that to keep the expression small (though non-standard).
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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I'm sort of unfamiliar with most spreadsheet functions, so this was something new for me.
Yes, I can see the parallels between your solution and mine.
Here's the next question :
Question 2
A and B are two friends.
By subtracting B's age from the square of A, we get 158.
On the other hand, by subtracting A's age from the square of B gives us 108.
What is A's age?
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hi CurlyBracket
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob,
The answer is correct.
Could you give me a hint on how to solve it?
Learning is fun - Exceptio probat regulam!
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I called the ages A and B and made two equations. You can make, say, B the subject of one and substitute it into the other and solve the quadratic. I didn't, so I don't know about a second answer. My guess would be that the second solution is inadmissible in the context of the question (eg negative or out of range).
What I actually did, because it looked easier and was, was to factorise (A-B) out of the equation which left me with two factors = 50. It's reasonable to assume that there's an integer solution and 50 doesn't have that many factors, so I was able to spot one straight away that works.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob & CurlyBracket;
I solved it this way:
A = B^2 - 108
B = A^2 - 158
Solution 1:
Substituting B into the first equation resulted in A^4 - 316A^2 - A + 24856 = 0 ... which I couldn't solve longhand, only online (eg, WolframAlpha) or with CAS (eg, Mathematica, Excel).
Answer: A = 13
Solution 2:
Trial & error: Evaluating the first 2 equations with A = 13 yields B = 11, evaluating with B = 11 yields A = 13, and figures in either direction from those yield increasingly large errors.
Answer: A = 13
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Hi Bob and phrontister,
Okay, so trial and error is the way to go. I was hoping there would be another method to do it, since this question was picked out of a strictly "no calculator allowed" exam.
But I suppose simultaneous equations would be very difficult in a case where the squares are involved.
Learning is fun - Exceptio probat regulam!
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OK, I'll outline my complete method. I don't really consider it to be trial and error; rather it enables me to home in on possible solutions.
Let the numbers be A and B, with A>B wnlog
A^2 - B = 158
B^2 - A = 108
Subtracting
A^2 - B^2 - B + A = 50
(A-B)(A+B+1) = 50
Assuming we are seeking +ve integer solutions, with the factors of 50 being {50,25,10,5,2,1}
If A+B+1 = 50 then A = 25, B = 24. This is not a solution.
If A+B+1 = 25 then A = 13, B = 11. This is a solution.
If A+B+1 = 10 then A = 7, B = 2. This is not a solution,
It is unnecessary to check lower values of A+B+1 as A-B would come out bigger and lead to impossible soltions.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob,
Thank you, that makes sense.
Here's the next question:
Question 3
If n! has 4 zeroes at the end.
(n+1)! has 6 zeroes at the end.
Find the value of n.
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Why would a factorial gain zeros at all?
Whenever there is a factor of 5 and 2 in the calculation another zero is added.
eg. It isn't until 5! that we get the first zero .... 120.
factors of 2 are plentiful so I looked at when another factor of 5 is acquired
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi,
I was trying using this method:
I think it is almost the same method, just a little more direct.
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That is pretty much my way too.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Ahoy there! I'm back!
Question 4
If x²+ 2x + 5 is a factor of x^4 + px² + q.
p, q =?
Learning is fun - Exceptio probat regulam!
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Hi,
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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hi CurlyBracket
Here's a way to do this.
The x^3 term is zero and will give you a, and then the x term will give you b.
From there you can work out p and q.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I've got it now! Thanks for the help, Bob and Ganesh!
Here's another question. It's a bit different from the others. Flavours are the essence of life!
Question 5
Three teams of woodcutters have decided to organize a competition. The winner is the team who fells the maximum number of trees in the given time.
The first and third team together felled twice the number of trees felled by the second team.
The second and third team together felled three fold of the number of trees felled by the first team.
Who won? Or in the event of there being joint winners, who would then be named as such?
Last edited by CurlyBracket (2022-08-29 21:12:16)
Learning is fun - Exceptio probat regulam!
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Hi CurlyBracket;
Question 5: The Woodcutting Competition:
I think that there was no tie, and Third won, the felling ratio being Third:Second:First = 5:4:3.
EDIT: Included my solution method...
Last edited by phrontister (2022-08-07 21:35:53)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Hi phrontister,
That's the correct answer!
I solved it like this. It's quite similar.
Learning is fun - Exceptio probat regulam!
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Hi CurlyBracket;
Yes, I like that method.
Just a comment re the puzzle wording:
I know what you meant with it, but, in a strict interpretation of the wording, answering the second question could have been a bit awkward if there was a tie...and there are 3 scenarios with at least one winner in which a tie occurs.
Last edited by phrontister (2022-08-28 22:52:02)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Then should I edit it to just say "Who won?"
Learning is fun - Exceptio probat regulam!
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