|
-
Feb 26th, 2006, 01:17 PM
#1
Thread Starter
Junior Member
-
Feb 27th, 2006, 05:22 AM
#2
Lively Member
Re: Manxy's easy challenge no.4
Answers:
A)118.
B)96.
C)2500.
D)30.
E)60 B.C.
-
Feb 27th, 2006, 05:24 AM
#3
Lively Member
Re: Manxy's easy challenge no.4
Do I get an iPod for correct answers?
-
Feb 27th, 2006, 08:29 AM
#4
Re: Manxy's easy challenge no.4
 Originally Posted by fahad k
Answers:
A)118.
B)96.
C)2500.
D)30.
E)60 B.C.
well, i dont have paper and pen here, but my question to you is how will you divide 30 sandwiches between 4 ppl evenly?
my guess would be 120, for question D, and for 1 also (i may be wrong, but i counted in my mind, so 120).
Harsh Gupta
-
Feb 27th, 2006, 09:04 AM
#5
Lively Member
Re: Manxy's easy challenge no.4
 Originally Posted by Harsh Gupta
well, i dont have paper and pen here, but my question to you is how will you divide 30 sandwiches between 4 ppl evenly?
my guess would be 120, for question D, and for 1 also (i may be wrong, but i counted in my mind, so 120).
Harsh Gupta
Nope.I guess the answer is 60 sandwiches.
Whereas I aint sure about A).
Ps:So I wont get the iPod afterall
-
Feb 27th, 2006, 03:56 PM
#6
Lively Member
Re: Manxy's easy challenge no.4
A: is 120
Here's how (VS.NET 2005; C#)
Create a new form, and add a button, and a textbox. For simplicity, I didn't rename anything. Here's the code for the button's press:
Code:
private string count()
{
string numbers = "";
for (int x = 300; x < 400; x++)
{
numbers += x.ToString();
}
char[] arrayNumbers = numbers.ToCharArray();
int totalNumber = 0;
for (int x = 0; x < arrayNumbers.Length; x++)
{
if (arrayNumbers[x] == Convert.ToChar("3"))
totalNumber++;
}
return totalNumber.ToString();
}
It can be tweaked very easily to be a more versatile method. I wrote this solely for this application.
-
Feb 27th, 2006, 03:58 PM
#7
Re: Manxy's easy challenge no.4
Wouldn't A be a trick question? I don't want to ruin it if it is
-
Feb 27th, 2006, 04:37 PM
#8
Lively Member
Re: Manxy's easy challenge no.4
 Originally Posted by szlamany
Wouldn't A be a trick question? I don't want to ruin it if it is 
Hmm.. Reading it again, I may suspect a "there may be the digit 3 written 120 times, but the number 3 is less than 300, thus never written".
-
Feb 28th, 2006, 06:37 AM
#9
-
Mar 1st, 2006, 03:27 AM
#10
Lively Member
Re: Manxy's easy challenge no.4
So these may be the answers. Isn't those?
a) 120
b) 2616
c) 2500
d) 60
e) 60 B.C.
-
Mar 7th, 2006, 04:12 PM
#11
Thread Starter
Junior Member
Re: Manxy's easy challenge no.4
Phrajib is the closest at the moment
Answers =
a) 120
b) 2616
c) 2500
d) 60
e) ?? (no-ones got this right yet)
I'll keep this open for another few days. In the meantime, I'm working on another 5 to occupy those who are challenged and those who are mentally challenged 
Manxy
-
Mar 7th, 2006, 05:12 PM
#12
Re: Manxy's easy challenge no.4
Sigh.
e) = 20 B.C.
The length of time from Hephaes-Ru's birth to Galat-Ra's death is Hephaes-Ru's age + Galat-Ra's age + time between HR's death and GR's birth. This total is given to be 120yrs. We know that their combined ages are 100yrs, hence the time between death and birth is 20 years. And since HR died in 40 B.C. GR must have been born in 20 B.C.
zaza
-
Mar 12th, 2006, 01:27 AM
#13
Lively Member
Re: Manxy's easy challenge no.4
Let,
Hephaes-Ru was born in X B.C.
Galat-Ra was born in Y B.C.
Now,
Hephaes-Ru was died in 40 B.C.
So,
Galat-Ra was died in (X+120) B.C.
Again,
(40-X) + (X+120-Y) = 100
-> 40-X+X+120-Y-100 = 0
-> 60 - Y = 0
So, Y=60
Galat-Ra was born in 60 B.C.
Is it wrong ? Please describe if u think otherwise.
-Rajib
-
Mar 12th, 2006, 01:38 AM
#14
Lively Member
Re: Manxy's easy challenge no.4
Sorry, I was wrong. It will be as follows
Let,
Hephaes-Ru was born in X A.C.
Galat-Ra was born in Y A.C.
Now,
Hephaes-Ru was died in 40 B.C. = -40 A.C.
So,
Galat-Ra was died in (X+120) A.C.
Again,
(-40-X) + (X+120-Y) = 100
-> -40-X+X+120-Y-100 = 0
-> -20 - Y = 0
So, Y= -20
Galat-Ra was born in 20 B.C.
So, zaza is wright. Sorry for that previous wrong answer.
-Rajib
-
Mar 12th, 2006, 04:05 PM
#15
Thread Starter
Junior Member
Re: Manxy's easy challenge no.4
congratulations on the correct answer being 20 B.C.
Love the example of your equation Rajib and yours also zaza
-
Mar 31st, 2006, 05:50 AM
#16
Frenzied Member
Re: Manxy's easy challenge no.4
I feel compelled to point out that in the Gregorian calendar (which I believe you're using) there is no year zero.
Therefore you are going to need to alter the equations to counter for the fact that AD0/BC0 doesn't actually exist. ie -40+80=41
"As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality." - Albert Einstein
It's turtles! And it's all the way down
Posting Permissions
- You may not post new threads
- You may not post replies
- You may not post attachments
- You may not edit your posts
-
Forum Rules
|
Click Here to Expand Forum to Full Width
|