The worlds first riddle!

there's light at the end of the tunnel
deep in thought
to be, or not to be

Mark and Shari each have marble collections. The number in Mark's collection in a square number (1,4,9,16, etc). Mark says to Shari, "If you give me all your marbles I'll still have a square number." Shari replies, "If you gave me the number in my collection you would still be left with an even square.

What is the least number of marbles Mark has

->25?<-

While I agree with TheInsanity's solution (and welcome to A2K, btw ),
the question, I feel, is somewhat oxymoronic, as shari lost all her marbles some time ago.

thank you

No offense, but I disagree with TheInsanity's solution. It solves the problem if there weren't the requirement that Mark be left with an even square.

Mark has 100 marbles. Shari has 96.

21978 * 4 = 87912

I think its even as in it's an even square number where it can be divided into two equal factors like all squares

TheInsanity, or as we might say; INTSAHNIETY. May I echo the welcome from Imur; who also agreed with the answer.

Marbles: 25

I made it: Mark has 25 and Shari has 24.

Mark disagreed, and thought; Mark has 100 marbles. Shari has 96.

I was at fault, the wording could have been far clearer.


Imur wrote, "shari lost all her marbles some time ago."

Oh, Now you have done it! In the interest of self-preservation, I wish to totally disassociate myself from this remark. RIP Imur.

BTW Mark, what have you done with Shari?



Tie break:

Imur wishes to sell a puppy for $11. TI wants to buy it but only has foreign currency. The exchange rate for the foreign currency is;

11 round coins = $15,
11 square coins = $16,
11 triangular coins = $17

How many of each coinage should the customer pay



ICI II


poCARPnd

Imur wishes to sell a puppy for $11. TI wants to buy it but only has foreign currency. The exchange rate for the foreign currency is;

11 round coins = $15,
11 square coins = $16,
11 triangular coins = $17

How many of each coinage should the customer pay

????????

ICI II= I See Eye to Eye


poCARPnd = Big fish in a small pond

The (voice) In sanity said:

???????? (Close, but no cigar)


ICI II= I See Eye to Eye


poCARPnd = Big fish in a small pond



AGE
A G E
AGE



LATE LATE
Go GO
TOWN TOWN

AGE
A G E
AGE

Split age between ages?

LATE LATE
Go GO
TOWN TOWN

Too Late to go to town

1 is an odd square

COINS
7 round, 1 square

middle age spread

'ABCDE * 4 = EDCBA. Solve for A,B,C,D, and E where each is a unique integer from 0 to 9'

Mark:
21978 * 4 = 87912


It is obvious that A can be no more than 2. If A were 3 then 3BCDE * 4 would be at least 120,000 which is more than five digits. Also A must be an even number because EDCBA is an even number since it is the product of at least one even number (4). We can eliminate A=0 because E would have to be 5 (5*4=0) but BCDE*4 could not hope to reach 50,000. So A must be 2.

Next consider E. E*4 must end in the digit 2. The only numbers that works for are 3 and 8. However with A=2 EDCBA must be at least 80,000. So 8 is the only number that satisfies both conditions.

Next consider B. We already know that 2BCD8*4 is at least 80000 and less than 90000. B can not be more than 2 because then 2BCD8 * 4 would be more than 80000. 2 is already taken so B must be 0 or 1. Lets consider the case that B=0. Then D8 * 4 must end in the digit 02. However there is no D that satisfies this condition. So B must be 1.
Next consider D. D8*4 must end in the digits 12. The only possibility is D=7 (78*4=312).

Now solve for C:
21C78 * 4 = 87C12.
84312+400C = 87012 + 100C
2700 = 300C
C=2700/300=9.

So ABCDE=21978.

COINS
7 round, 1 square

Let x be the number of circular coins, y be the number of square coins, and z be the number of triangular coins. Thus:
x*(15/11) + y*(16/11) + z*(17/11) = 11. Multiplying by 11:
15x + 16y + 17z = 121.

From here we have to use trial and error or educated guessing. Divided 121 by 15 to get 8 plus a remainder of 1. It was then obvious that x=7, y=1, and z=0 would solve the equation.

AGE
A G E
AGE

middle age spread


TheInsanity:
ICI II= I See Eye to Eye


poCARPnd = Big fish in a small pond

LATE LATE
Go GO
TOWN TOWN
Too Late to go to town



Mark has $1,000,000 he won in the Math challenge. He wishes to divide amongst his friends by giving each person named an amount of money in dollars, which is a power of 7 ($70=$1, $71=$7, $72=$49, $73=$343, ...). He does not want to give more than six people the same amount.

How can he divide the money



…………………………………………………………………Quite



CCCCCsailingCCCCC

$1,000,000
1000000 base 10 = 11333311 base 7

quite right

CCCCCsailingCCCCC
Sailing in the seas

Mark:

…………………………………………………………………Quite
quite right


TheInsanity

CCCCCsailingCCCCC
Sailing in the seas



TLOIOKBMACEK


DDUOWPNS


A plane flies from Athens to Brussels at maximum speed. Normally, flying at maximum speed would enable it to reach Brussels in four-fifths of the time that it takes to fly there at cruising speed. On this occasion, however, the velocity of a favorable wind enables it to get there in only half the time it would normally take at maximum speed.

On the return journey, it leaves Brussels at 1 p.m... Ignoring time zones, and encountering the same velocity and direction of wind, at what time will the plane arrive back at Athens

TLOIOKBMACEK

Look back in time

DDUOWPNS

Down in the dumps

A plane flies from Athens to Brussels at maximum speed. Normally, flying at maximum speed would enable it to reach Brussels in four-fifths of the time that it takes to fly there at cruising speed. On this occasion, however, the velocity of a favorable wind enables it to get there in only half the time it would normally take at maximum speed.

On the return journey, it leaves Brussels at 1 p.m... Ignoring time zones, and encountering the same velocity and direction of wind, at what time will the plane arrive back at Athens

5 p.m.

ATHENS TO BRUSSELS
The return trip will take four times as long as the first leg. Since I don't know how long the first leg took, I don't know how long the return trip takes.

Maybe I'm missing something.

i looked it up and the average flight is about 4 hours long

"$1,000,000"

Mark:
1000000 base 10 = 11333311 base 7

In my humble opinion, that was almost the best answer ever in the history of the world. However, does it have one ?'3' too many? (not for base 7, but the answer)

He should give:

• 1 person $1
• 1 person $49
• 3 people $343
• 3 people $2401
• 3 people $16807
• 1 person $117649
• 1 person $823543.

Lets call a the number of people he leaves $1 to, b the number of people he leaves $7 to, c the number of people he leaves $49 to, etc. The solution equation is thus:

1,000,000 = a + 7b + 49c + 343d + 2401e + 16807f + 117469g + 822283h.

Next divide 1,000,000 by 7. If 1,000,000 is divisible by 7 we could divide the money without giving anyone only $1. However 1,000,000/7 = 142857 plus a remainder of 1. By giving one person $1 (a=1) we now have:

999,999 = 7b + 49c + 343d + 2401e + 16807f + 117469g + 822283h.
Next divide by 7:
142,857 = b + 7c + 49d + 343e + 2401f + 16807g + 117469h.
Dividing 142,857 by 7 we get 20,408 plus a remainder of 1. So let b=1, divide by 7, and we get:
20,408 = c + 7d + 49e + 343f + 2401g + 16807h .
Dividing 20,408 by 7 we get 2,915 plus a remainder of 3. So let c=3, divide by 7, and we get:
2915 = d + 7e + 49f + 343g + 2401h.
Dividing 2915 by 7 we get 416 plus a remainder of 3. So let d=3, divide by 7, and we get:
416 = e + 7f + 49g + 343h.
Dividing 416 by 7 we get 59 plus a remainder of 3. So let e=3, divide by 7, and we get:
59 = f + 7g + 49h.

Dividing 59 by 7 we get 8 plus a remainder of 3. So let f=3, divide by 7, and we get:
8 = g + 7h.

Obviously g=1 and h=1.



"A plane flies from Athens to Brussels at maximum speed. Normally, flying at maximum speed would enable it to reach Brussels in four-fifths of the time that it takes to fly there at cruising speed. On this occasion, however, the velocity of a favorable wind enables it to get there in only half the time it would normally take at maximum speed.

On the return journey, it leaves Brussels at 1 p.m... Ignoring time zones, and encountering the same velocity and direction of wind, at what time will the plane arrive back at Athens"

I posted this riddle the same time I received it, hoping to answer first. I sought clarification on a point and was sent the apparent solution:

"Never! Most of the information given is irrelevant. If the tails wind doubled its speed going, its velocity must be equalled that of the plane. On the way back, such a wind would effectively reduce the speed of the plane to zero."

Not one I agree with, but I'm not a pilot. What say you guy's?



You are standing on a rock in the middle of a circular lake of radius 1. There is a tiger on the shore of the lake that can run four times as fast you can swim, however the tiger can not swim. The tiger is hungry and always attempts to keep the distance between the two of you at a minimum.

How can you safely swim to shore


Done
Very

Grtoatituadle

Vsearichn