Problems and Solutions

Wanna be math wiz · 2005-11-30 05:33:56

very simple and very handy. Thanks Ganesh

Jai Ganesh · 2005-11-30 16:09:14

Mathsyperson's solution to problem # k + 50 is correct. Wanna be math wiz, thanks for your post.

Problem # k + 51

What would be the coefficient of the middle term of (a+b)^10?

mathsyperson · 2005-12-01 05:18:53

Jai Ganesh · 2005-12-01 16:54:17

Correct!

Problem # k + 52

Thereare a number of rabbits in a room. A leopard enters the room and eats a certain number of rabbits. The remaining rabbits run towards the next room. As they move from the first room to the second, the number of rabbits doubles. The leopard then enters the second room and eats the same number of rabbits as it did in the first room. The remaining rabbits run towards the third room. As they land in the third room, the number of rabbits multiplies by 3. Then the leopard comes and eats the same number of rabbits as it did in the first and the second. The remaining rabbits run to the fourth room and as they go from the third to the fourth room, the numbers or rabbits multiplies by 4. When the leopard comes to the fourth room, and eats the same number of rabbits as it did in the first, second and the third, there are no rabbits left.

How many rabbits were there in the first room before the leopard went and how many did the leopard eat in each room?

mathsyperson · 2005-12-02 05:10:44

In the third room, does the leopard eat twice the amount that he ate in the second room or the same amount?

Well, I have the answer for either situation.

There are actually many answers. All that matters is that you need to get the ratio of initial rabbits to rabbits eaten right.

Jai Ganesh · 2005-12-02 16:40:08

Mathsyperson is correct! The question was the leopard eats the same number of rabbits in all the four rooms. I had been given five minutes to answer this question in a science exhibition at my niece's school. I was told by a twelve year old student that I would get a big prize if I answered it within five minutes! I did and the reward I got was Rs.10 (INR10 ~USD0.20)! I returned the reward for someone after me who may answer the puzzle!

Jai Ganesh · 2005-12-02 20:36:45

Problem # k + 53

Prove that the sum of a side of an n x n magic square (a magic square containing numbers 1 to n²) is (n³+n)/2.

krassi_holmz · 2005-12-04 07:15:24

The sum of all cells in the magic square is ∑=n²(n²+1)/2. We have n rows with same sum in each and the sum of the rows is equal to the sum of the cells, so the sum in each row is equal to the sum of the cells divided by the number of the rows, e.a. A=n²(n²+1)/2n=(n³+n)/2
sorry for the syntax, I don't know English very well.

krassi_holmz · 2005-12-04 08:16:42

About the problem k+43:
x^2+1/x^2=22
x^2+2+(1/x)^2=25
(x+1/x)^2=25
Let x > 0
x+1/x=A=5
A^3=125=(x^3+1/x^3)+3*5
x^3+1/x^3=125-15=110

A^5=3125=(x^5+1/x^5)+5*110+10*5
x^5+1/x^5=3125-5*110-10*5=2525

A^7=78125=(x^7+1/x^7)+7*2525+21*110+35*5
x^7+1/x^7=78125-(7*2525+21*110+35*5) =57965

A^9=1953125=(x^9+1/x^9)+9*57965+36*2525+84*110+126*5
x^9+1/x^9=1953125-(9*57965+36*2525+84*110+126*5)=1330670

Is this true? Check it. I'm sure there's an easier way.

Last edited by krassi_holmz (2005-12-04 08:23:05)

mathsyperson · 2005-12-04 09:20:26

You've got a typo in the first line, but apart from that it looks fine.

You can do it a bit quicker by going from ^3 to ^9, rather than ^3, ^5, ^7, ^9.

x³ + 1/x³ = 110

(x³ + 1/x³)³ = (x^9 + 1/x^9) + 3*110

∴ (x^9 + 1/x^9) = 110³ - 3*110 = 1330670.
Our answers match, so that's a good sign.

Jai Ganesh · 2005-12-04 16:44:28

Well done, krassi_holmz,
your solutions to both Problem # k + 53 and Problem # K + 43 are correct.
As Mathsyperson suggested, there existed a shorter route.

Jai Ganesh · 2005-12-04 16:47:10

Problem # k + 54

Why is any four digit number abcd written again adjacent to it, resulting in abcdabcd, always divisible by 73? (a,b,c,d ∈ 0,1,2,3,4,5,6,7,8, and 9).

krassi_holmz · 2005-12-04 17:16:53

#k+54
Because N[abcdabcd]=N[abcd]+10000.N[abcd]=10001.N[abcd]=73.137.N[abcd], where N[abcd] means number abcd.

Jai Ganesh · 2005-12-04 18:22:25

Absolutely right!
Very well done, krassi_holmz!

Jai Ganesh · 2005-12-04 18:44:48

Problem # k + 55

Give a list of 1,000 consecutive non-prime numbers.

krassi_holmz · 2005-12-04 19:25:43

2/1001!+2
3/1001!+3
...
1000/1001!+1000
1001/1001!+1001

Jai Ganesh · 2005-12-04 20:52:56

If you meant
1001! + 2, 1001! + 3, 1001!+4 .........1001! + 1001,
you are correct.
Well done, got three in a row right!
A set of slightly smallers numbers would be
999! + 2, 999! +3, 999! + 4,.............999! + 999, 999! + 1000, 999! + 1001.
Because, 999! + 999 would be divisible by 999,
999!+1000 would be divisible by 1000,
999! + 1001 would be divisble by 7, 11, 13.
999! + 1002 would be divisible by 2.

BTW, you got a nice avatar!

Jai Ganesh · 2005-12-05 00:29:17

Problem # k + 56

A tennis championship is played on a knock-out basis, i.e., a player is out of the tournament when he loses a match.

(a) How many players participate in the tournament if 255 matches are totally played?
(b) How many matches are played in the tournament if 63 players totally participate?

mathsyperson · 2005-12-05 04:01:57

Jai Ganesh · 2005-12-05 16:30:39

Mathsyperson is correct!

Problem # k + 57

The circumference of the front wheel of a cart is 30 feet long and that of the back wheel is 36 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel?

Ricky · 2005-12-05 17:22:59

900ft.

Jai Ganesh · 2005-12-05 19:31:15

You're Correct! Well done, Ricky!
You may use the hide tag for your replies, if you want to.

Jai Ganesh · 2005-12-05 20:45:52

Problem # k + 58.

What is the area of the biggest (i) equilateral triangle and (ii) square that can be cut out of a circular metal plate of radius 'a'?

ryos · 2005-12-06 04:46:01

Last edited by ryos (2005-12-07 15:46:42)

Jai Ganesh · 2005-12-06 20:04:05

ryos,
Please check your solution again!

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