Author |
Topic: Random mathemathical problems... (Read 1125 times) |
|
Mugwump101
Junior Member
 
Gender: 
Posts: 61
|
 |
Random mathemathical problems...
« on: Oct 22nd, 2005, 6:46pm » |
 Quote  Modify
|
I need help with both problems, and I can't seem to understand how to reach a point to solve for them.
1) a2-b2=50
a+b= 5
a - b= ?
2) Find the smallest number that will leave a remainder of 1 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5.
Thank you so much in advance!
|
|
 IP Logged |
"When I examine myself and my methods of thought, I come to the conclusion that the gift of fantasy has meant more to me than my talent for absorbing positive knowledge. "~ Albert Einstein
|
|
|
Neelesh
Junior Member
Gender: 
Posts: 147
|
|
Re: Random mathemathical problems...
« Reply #1 on: Oct 23rd, 2005, 1:02am » |
Quote Modify
|
on Oct 22nd, 2005, 6:46pm, Mugwump101 wrote:
1) a2-b2=50
a+b= 5
a - b= ?
|
|
Since you donot know (a-b), assume it to be x and then multiply the two values (a+b) and (a-b). You might get something that you want....
Quote:
2) Find the smallest number that will leave a remainder of 1 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5.
|
|
Simplest way is to try out each number from 1 to say, 30, 40 or till you get the answer.
Alternatively, assume the number to be x. Using only second and third condition,
x leaves remainder 3 when divided by 4
x leaves remainder 4 when divided by 5
i.e., x leaves remainder 1 less than the divisor.
So x can be found out by using LCM of 4 and 5. Then you can try to see whether the first condition is also satisfied or not.
Still better, use chinese remainder theorem.
|
|
IP Logged |
|
|
|
BNC
Uberpuzzler
Gender:
Posts: 1732
|
|
Re: Random mathemathical problems...
« Reply #2 on: Oct 23rd, 2005, 2:03am » |
Quote Modify
|
For 2, you can also do this:
1. Condition 3: Your number (call it X) is X=5N+4 (for any whole N).
2. Condition 2: See how X divides by 4: The remainder of 5N+4 / 4 is N. You want that N to leave remainder of 3 when divided by 4, hence N=4K+3.
The 2 condition gives: X=5N+4 = 20K + 19
3. This X, when divided by 3 should give remainder of 1. Xmod3 is 2K+1.
What is the smallest K that gives 2K+1 = 1?
|
|
IP Logged |
How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
|
|
|
Mugwump101
Junior Member
Gender:
Posts: 61
|
|
Re: Random mathemathical problems...
« Reply #3 on: Oct 23rd, 2005, 6:12am » |
Quote Modify
|
For the first one , my friend said it was 10. I'm not sure how he got it though.
For the second one, my friend did something like this.... is this valid?
3 6 9 12 15 18 21 24
Remainder of 1 -> 4 7 10 13 16 19 22 25
4 8 12 16 20 24
remainder of 3 -> 7 11 15 19 23 27
5 10 15 20 25 30
Remainder of 4 -> 9 14 19 24 29 34
And she said the only number that appears in all four statements is 19, so it has to be 19. Although, I'm not exactly sure why and how she used this method.
I'm so confused right now.
|
|
IP Logged |
"When I examine myself and my methods of thought, I come to the conclusion that the gift of fantasy has meant more to me than my talent for absorbing positive knowledge. "~ Albert Einstein
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Random mathemathical problems...
« Reply #4 on: Oct 23rd, 2005, 7:09am » |
Quote Modify
|
on Oct 23rd, 2005, 6:12am, Mugwump101 wrote:| For the first one , my friend said it was 10. I'm not sure how he got it though. |
|
let's say (a-b)=x (as suggested by Neelesh)
then (a+b)*(a-b)=5*x
but also (a+b)*(a-b)=a2- b2
and we already knew a2- b2=50
Quote:
For the second one, my friend did something like this.... is this valid?
3 6 9 12 15 18 21 24
Remainder of 1 -> 4 7 10 13 16 19 22 25
4 8 12 16 20 24
remainder of 3 -> 7 11 15 19 23 27
5 10 15 20 25 30
Remainder of 4 -> 9 14 19 24 29 34
And she said the only number that appears in all four statements is 19, so it has to be 19. Although, I'm not exactly sure why and how she used this method. |
|
It's correct.
You want a number that has all three properties (of having a certain remainder given a certain divider). So for each property you can list the first X numbers that have it. And then the first number to occur in all three lists must be the first to have all three of them.
Of course in other cases you might have to list a thousand number before you find an answer.
|
| « Last Edit: Oct 23rd, 2005, 7:48am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Random mathemathical problems...
« Reply #5 on: Oct 23rd, 2005, 12:21pm » |
Quote Modify
|
The first problem was easy for your friend, and also for neelash and towr, because all three are familiar with the useful identity:
(a + b)(a - b) = a2 - b2.
(Multiply it out yourself to see why this is true.)
Now, what you are given is a2 - b2 = 50, and by the identity, this means that (a + b)*(a - b) = 50. Since you also know that (a + b) = 5, we have 5*(a - b) = 50. Divide by the 5, and you get (a - b) = 10.
A second method (generally usable when solving two equations in two unknowns) is to solve one equation for one of the variables in terms of the other, substitute into the second equation, and solve it for the value of the other variable. In this case:
a + b = 5, so b = 5 - a.
a2 - b2 = 50
a2 - (5 - a)2 = 50
a2 - (25 - 10a + a2) = 50
10a - 25 = 50
10a = 75
a = 7.5
b = 5 - a = 5 - 7.5 = -2.5
So (a - b) = 7.5 - (-2.5) = 10.
As you can see, this is a lot easier if you recognize the identity. But the direct solving method works too.
A little more explanation on what your friend did to solve the second: In the first set of numbers, he or she (you used both pronouns in your post - I will use "she" from now on, assuming that you dropped the s accidently) is trying to find a list of all numbers (up to some stopping point she is hoping is high enough) satisfying the first condition: A remainder of 1 when divided by 3.
She starts by listing the multiples of 3:
3 6 9 12 15 18 21 24.
For a number to leave a remainder of 1 when divided by 3, it has to be 1 higher than a multiple of 3, so she next adds 1 to each of the multiples listed:
4 7 10 13 16 19 22 25
The number we are looking for has to be one of these (or higher).
Now, the number has to leave a remainder of 3 when divided by 4, so she does the same thing with these numbers: start by listing all the multiples of 4:
4 8 12 16 20 24
To get numbers leaving a remainder of 3, she adds 3 to each of these numbers:
7 11 15 19 23 27
Again, the number we are looking for has to be one of these (or higher).
Last, the third condition is a remainder of 4 when divided by 5. So, first list the multiples of 5:
5 10 15 20 25 30
Then add 4 to get numbers leaving a remainder of 4:
9 14 19 24 29 34.
Now she compares the 3 final lists:
4 7 10 13 16 19 22 25 (remainder of 1 when divided by 3)
7 11 15 19 23 27 (remainder of 3 when divided by 4)
9 14 19 24 29 34 (remainder of 4 when divided by 5)
and she notes that the smallest number which is in all three lists is 19.
Had she continued the lists into higher numbers, she would have found more common numbers (79 is the next one, followed by 139, 199, 259, ... - notice a pattern? Can you think of a reason they all differ by 60?). But 19 is still the smallest, which is what you were asked for.
|
| « Last Edit: Oct 23rd, 2005, 12:27pm by Icarus » |
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print