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Old Feb 26th, 2006, 12:17 PM   #1
manxy1
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Manxy's easy challenge no.4

A/ If you wrote all of the numbers from 300 to 400 on a piece of paper, how many times would you have written the number 3?

B/ There are six donkeys travelling to Greece, and each donkey is carrying six baskets. In each basket are six cats, and each cat has had two baby kittens. How many legs are there in total between all the donkeys and cats/kittens?

C/ Charlie has just won a load of money playing Poker! He puts 1/5 of his winnings in the Bank, and 1/2 of his winnings he invests into shares.
After those investments Charlie realises he still has quite a lot left. He then goes to London Central and spends 1/10 of his winnings on chocolate eggs! At the end of the day he goes home and puts his remaining 500 Pence in his mattress.
How much did Charlie win?

D/ Mrs. Doyle the Housekeeper has made a big, big pile of sandwiches for the party of friends visiting her house tonight. When she counts them, she realises that she can equally split the sandwiches between two friends and have none left over. She can also split them between three friends, four friends, five friends and six friends equally and every time there will be no sandwiches left over.
What is the smallest amount of sandwiches that Mrs. Doyle could have made?

E/ Two kings of Sakhmet from history (fiction) were King Galat-Ra and King Hephaes-Ru. Galat-Ra died 120 years after Hephaes-Ru was born. Their combined ages when they died was 100 years. Hephaes-Ru died in the year 40 B.C.
In what year was King Galat-Ra born?

Hope that keeps you busy for a while
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Old Feb 27th, 2006, 04:22 AM   #2
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Re: Manxy's easy challenge no.4

Answers:

A)118.

B)96.

C)2500.

D)30.

E)60 B.C.
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Old Feb 27th, 2006, 04:24 AM   #3
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Re: Manxy's easy challenge no.4

Do I get an iPod for correct answers?
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Old Feb 27th, 2006, 07:29 AM   #4
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Re: Manxy's easy challenge no.4

Quote:
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
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Old Feb 27th, 2006, 08:04 AM   #5
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Re: Manxy's easy challenge no.4

Quote:
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
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Old Feb 27th, 2006, 02:56 PM   #6
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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.
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Old Feb 27th, 2006, 02:58 PM   #7
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Re: Manxy's easy challenge no.4

Wouldn't A be a trick question? I don't want to ruin it if it is
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Old Feb 27th, 2006, 03:37 PM   #8
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Re: Manxy's easy challenge no.4

Quote:
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".
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Old Feb 28th, 2006, 05:37 AM   #9
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Re: Manxy's easy challenge no.4

If you want a better proof to #1, {apart from the trick interpretation} - it appears 100 times in the 100s column, 10 times in the 10s column (33x) and 10 times in the 1s column (3x3) = 120.
Talking of trick questions, i say D is ZERO sandwiches
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Old Mar 1st, 2006, 02:27 AM   #10
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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.
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Old Mar 7th, 2006, 03:12 PM   #11
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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
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Old Mar 7th, 2006, 04:12 PM   #12
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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.

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Old Mar 12th, 2006, 12:27 AM   #13
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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
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Old Mar 12th, 2006, 12:38 AM   #14
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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
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Old Mar 12th, 2006, 03:05 PM   #15
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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
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Old Mar 31st, 2006, 04:50 AM   #16
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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

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