Math+Challenge


 * This page is for me to post questions. Post your answers below each question, with your working shown. Please indicate your name beside the answer. Please do not change your friends' answers. Thank you.**

8 friends want to share the cost of a present. If 2 more friends were to join to split the cost, each person will pay $5 less. What is the initial amount paid by each person? (NMOS 2008)

8*5=40 40=2 40/2=20 20+5 =__**25**__ Adric

8*5 = 40 2u = 40 1u = 20 20 + 5 = __**25**__ Hari

x is the cost of the present. Then: x/8=x/10 +5 So: x/8=x+50/10 Multiply by 8: x=8x+400/10 Then by 10: 10x=8x+400 Take away 8x: 2x=400 Divide by 2: x=200 Now we know that the cost of the present is 200 dollars. Divide by 8: 200/8=25 Then divide by 10: 200/10=20 and the conditions are met. So each child got 25 dollars. Jeremy

8*5=40 40/2=20 20 is the sum of money in which each person has to pay after the two people joined. 20+5=25 Each child initialy had to pay __**25**__ dollars. Jun Da (ok, i know i ain't exactly in your class)

The sum of a few whole numbers is 16. Find the largest possible value of the product of these few whole numbers. (NMOS 2008)

64 Adric

What do you mean, //a few??// Jeremy (ß!)

256 2*2*2*2*2*2*2*2 Rayner

Change of answer 324 3*3*3*3*2*2 Rayner

ALERT, ALERT! Please indicate your name. Jeremy

change of answer 288 (invisible answer) Adric

Where is ur proof of getting 288??? Anyway, I have a bigger answer And Jeremy, we did indicate our name Rayner

oops, i meant 328, 4x4=16, 16x3=48,48x3=164,164x2=328 (invisible answer) BEAT THAT!!! Adric

3*3*3*3*2*2 Hari

LOL I already beat that 4*4=16 16*3=48 48*3=144 144*2=288 calculation error Hee Hee!!! Rayner

er.... why are u asking for miss hee? ok, i admit defeat. btw, i posted another question on PUPILS CONTRIBUTION. it was quite long b4 my father figured it out. Adric

The following is a 100-digit number: 298352983529835...29835 Find the sum of the first 78 digits.

424 (I calculated wrongly the other day) Zheng Xian

424 Sheldon

2+9+8+3+5=27 (5 numbers) 75/5=15 15x27=405 next 3 numbers =298=19 19+405=424 Adric

The unit 29835 is repeating. [78/5]=15 2+9+8+3+5=27 27*15=405 2+9+8=19 405+19=__424__ Jeremy

From 12.30pm to 7.45pm on a given day, how many times will the hour hand and the minute hand of a clock form a right angle?

13 times. (Used my mother's alarm clock) Isaac

Actually there is a mathematical way. A right angle is formed when the hour and minute hand form either 90 degrees or 270 degrees. Every minute, the minute hand moves 360/60 = 6 degrees and the hour hand moves 360/60/12 = 0.5 degree. At 12:30, the angle formed is 195 degrees. Every minute, the minute hand lessens the angle by 5.5 degrees. By 1 o'clock, the angle is already only 30 degrees, so they must have formed a right angle once and just once. Between one to two, the minute hand overtakes the hour hand and by 1.30, the angle is 135 degrees, so another right angle has been formed. By two, it has already exceeded 270 degrees, so a third right angle has already been formed. Another right angle is formed by 2:30 and, of course, a fifth at three sharp, the next one shortly after 3:30 (3.30 is 75 degrees). The next is not formed by four, but it is formed between 4:10 and 4.15. The eigth is formed before five, the ninth before 5.45 (that's 97.5 degrees), the tenth just after 6.15 (also 97.5 degrees), the next before 6:50 (96 degrees), the twelth around 7:20 (100 degrees). The thirteenth is after 7:45, so the answer is twelve. Sheldon

Matilda and Matthew share a bag of candies using the following process: Matthew takes a candy; Matilda takes two candies; then Matthew takes three, Matilda takes four, and so on, each taking one candy more than the other had taken. Matilda is the last one to take the appropriate number of candies, and the bag is then empty. She then has 10 candies more than Matthew. How many candies were in the bag?

210 candies. Jeremy

210 candies Hari

Samuel wants to distribute 174 identical marbles into a number of boxes. Given that each box can have 12 to 22 marbles and that each box contains different number of marbles, find the number of different ways to do the task.

1 way. It is 12+14+15+16+17+18+19+20+21+22. Lateef

Is it possible to divide a square into triangles so that all the triangles are acute angled? Show how you will divide the square into triangles.

How many triangles can be in the divided square? Isaac

Any number, as long as they are all acute angled. Mdm Teo

Yes, I think. Jun Da

It is impossible to divide a square into triangles so that all angles are acute. This is because of the fact that if you bisect the four corners, the centre will have //four// right angles!This process could go on forever if you tried to split the right angles into half again, and then you would have even more right angles! Jeremy

It is possible. Try harder. Mdm Teo

I know. Look at the picture. Rayner

But the picture clearly conveys the fact that some angles are obtuse. So, Rayner, you are wrong. Try again. Jeremy

I agree with Jeremy. I tried a few times and when a triangle has an obtuse angle, I can't seem to divide it into an acute angle because the new triangle that I create also has an obtuse angle. How to do??? Isaac :-(

I found out the answer!!! (YAY ME) You have to divide it into 60-triangles. Look at the picture below. Lateef P.S. From Lateef: I could thank http://www.ics.uci.edu/~eppstein/junkyard/acute-square/60-square.gif.

Well, you may only need eight. Jeremy

Hey! Why is nobody responding? Jeremy

There are 3 boxes. One box contains a white ball, one a black ball and one a doll. A sentence is written on the cover lid of each of the boxes. The sentence written on the box containing the white ball is always true. The sentence written on the box containing the black ball is always false. The sentence written on the box containing the doll can be true or false.

//Box 1: The sentence written on Box 2 is true.// Box 2: Box 1 contains a black ball. Box 3: The doll is in Box 1.

Which box contains the doll?

box 1 and 2 as the snetence in box 3 may be true or false jian rong Box 1 contains the ball. Zheng Xian (Yun Yee: I think you meant Box 1 contains the __doll__) Yes I made a mistake. Zheng Xian The answer is Box 2 Harish Box 1 says that Box 2 is true, that means Box 1 has the black ball which means that Box 1 is false, which means Box 1 either contain a white ball or a doll. But if Box 1 was lying Box 2 would be lying and thus Box 1 contains the doll. Illio

Either Box 1 or 2 There is no evidence for this. Rayner

I thought it should be Box 1!! (Jarret)

It IS Box 1. If Box 1 is telling the truth, then Box 2 is telling the truth too. But Box 2 says that Box 1 contains the black ball, so they are both lying. Hence, Box 3 is telling the truth and so the doll is in Box 1. Actually I like these type of questions.^^ Yun Yee

Yup, I got Box 1 too! Isaac mi 2! adric

I also got Box 1! Sheldon

I got box 1 like all of you! Jun Da

You are all right! It is Box 1! Shivanand

Listen, if Box 1 is right, Box 1 should contain a black ball. But, if Box 1 contains the black ball, the sentence would be false. So, Box 1 does not contain the black ball. According to your concepts, the sentence is as follows : Box 1 contains the doll, Box 2 contains the black ball, and Box 3 contains the white ball. But, if Box 2 contains the black ball, when he said Box 1 contains the black ball, Box 2 would be telling the truth if Box 1's sentence is false. If Box 3 contains the white ball ( the last item ), then Box 1 contains the doll. This is very confusing..... My answer is : Box 1 : Black Ball, Box 2 : White Ball, Box 3 : Doll.

So, Box 3 contains the doll. Neville P.S. Mdm Teo, what's the answer ? (Mdm Teo: The doll is in Box 1. This is a past Math Olympiad question.)

But neville, you say that box one contains thee black ball so it is telling a lie, so it will mean that the white ball is telling a lie, but then if it contains the white ball, it cannot lie so my answer is like everyone's answer. Box 1 contains the doll. WY

It should be in Box 1. Mudzakkir

Box 1 Zhirui

Absolutely. The doll is in Box 1. Tricia (I think you mean __DOLL)__ Shivanand

Include me in Hari

Why the doll is in box 1: Box 1 says that the sentence written on 2 is true. But we do not know whether it is true or not. Let us assume that 1 is true. Then 1, by 2, would contain the black ball. But 3 gives a contradiction. So, the sentence on 2 is false, and so contains the black ball. From this we also know that 1 is false and so 3 is true. So: Box 1 contains the doll, Box 2 contains the black ball, Box 3 contains the white ball. Jeremy (is there a Boolean algebra solution?)

I've found a easy way. Let's take that Box 1 contains the white ball Means that the statement on Box 2 is true. But if it is true then Box 1 contains the black ball, which not true. So, Box 1 does not contain the white ball Now, let's take it that Box 1 contains the black ball. Means that the statement on Box 2 is not true (OR false) If the statement is false, then Box 1 does not contain the black ball. So, Box 1 does not contain the black ball. The last object left is the doll. So, Box 1 contains the doll. MY ANSWER!!! Box 1: Doll, Box 2: Black ball, Box 3: White ball Rayner

Huh? Jeremy (ß!)

Write down any number between 100 and 999. Multiply it by 11. Multiply your answer by 91. What do you notice about your answer? Try other numbers. What do you observe?
 * __A Number Trick__**

The first three numbers and the last three numbers ot the answer are the same, and each repeated number is the same as the number you thought of. Example: 572 x 11 x 91 = 572 572 = Same number Mandy

The reason for this pattern is that 11 x 91 = 1001. When you multiply a 3-digit number (say //abc//) by 1001, the first 1 will give the same 3-digit number, the two zeros will equal 0, and the last 1 will again give the same 3-digit number, except that the ones digit will be in the thousands place, the tens digit will be in the ten thousands place, and the hundreds digit will be in the hundred thousands place. Since there are no overlaps, the resulting 6-digit number will be of the form //abcabc//. Sheldon

Not fair, Mandy deleted my edit here!!! Zheng Xian

I did not!!!!! It is because we posted our answers at the same time and yours was not saved!!!!!!!!!! Mandy

Did I warn you on the Science wiki, Mandy?¿!¡ Jeremy (ß!)

This works for 10101 (3x7x13x37) with a 2 digit number too. Wayne

The ages of 3 girls, when multiplied together, equal 1680. The sum of their ages equals their mother's age, which is a prime number less than 42. How old are the 3 girls and their mother?

Children: 14, 15, 8 Mother: 37 Mandy

Factoring 1680 I have: 2*2*2*2*3*5*7 Factors are: 2,3,4,5,6,7,8,10,12,14,15,16,20,21,24,28,30,35,40 (included only those <=40) or 2,3,2*2,5,2*3,7,2*2*2,2*5, 2*2*3,2*7,3*5,2*2*2*2,2*2*5, 3*7,2*2*2*3,2*2*7,2*3*5,5*7,2*2*2*8 Prime numbers (<42): 2,3,5,7,11,13,17,19,23,29,31,37,41 Testing: 16,15,7 16+15+7=38 14,24,5 14+24+5=43 8,14,15 8+14+15=**37** BINGO! Children are of ages 8, 14, 15 Mother is 37 Jeremy

There are 2006 oranges. The 1st group of people consumed 1/2 of the oranges. The 2nd group of people consumed 1/3 of the remaining oranges. The 3rd group..... and the 2005th group of people consumed 1/2006 of the remaining oranges. Find the number of oranges left.

There is 1 orange left. Sheldon (Mdm Teo: Sheldon, would you explain your working please?)

The first group of people consumed 1/2 of the oranges, so the number of oranges left is 1/2 x 2006. The second group of people consumed 1/3 of the remaining oranges, so the number of oranges left is 2/3 x 1/2 x 2006. The third group of people consumed 1/4 of the remaining oranges, so the number of oranges left is 3/4 x 2/3 x 1/2 x 2006 and so on. Therefore the number of oranges left in the end = 2005/2006 x 2004/2005 x 2003/2004 x 2002/2003.........x 4/5 x 3/4 x 2/3 x 1/2 x 2006 = 1(because all the numbers cancel out except 1/2006 x 2006 which is 1). Sheldon

The number 24 has the property that it is one short of a square number, and its double is also one short of a square number.
 * __Square relations__**

24 + 1 = 25 = 5 x 5 (24 x 2) + 1 = 49 = 7 x 7

What is the next number with the same property?

24 + 12 = 36 = 6 x 6 (24 x 5) + 1 = 121 =11 x 11 From Huang Zhong

(Dear Huang Zhong, please read the question carefully. Your answer is not correct as you are using the same number again. Mdm Teo)

840 + 1 = 841 = 29 x 29 (840 x 2) + 1 = 1681 = 41 x 41 Sheldon

Express as many numbers as you can from 1 to 100 using exactly four 4s and any mathematical symbols you know. For example: 15 = 44/4 + 4 16 = (4 x 4) + 4 - 4 (Please add on to the list below)
 * __Four 4s__**

0 = 4 - 4 + 4 - 4 1 = 4 / 4 + 4 - 4 8 = 4 + (4 - 4) + 4 16 = 4 X 4 X (4 / 4) 17 = 4 X 4 + 4 / 4 24 = 4 + 4 + (4 X 4) From Alex

60= 4 x 4 x 4 - 4 44= 44 + 4 - 4 11= 44 / 4 - 4 (?) 68= 4 x 4 x 4 + 4 From Geng Leong

1 = 44/44 2 = 4/4 + 4/4 3 = (4 + 4 + 4) / 4 4 = 4 + (4 - 4) x 4 5 = (4 x 4 + 4) / 4 6 = (4 + 4) / 4 + 4 7 = 4 + 4 - 4/4 8 = 4 + 4 + 4 - 4 9 = 4 + 4 + 4/4 10 = (44 - 4) / 4 11 = 44 / ((square root of 4) + (square root of 4)) 12 = 4 x 4 - (square root of 4) x (square root of 4) 13 = (44 / 4) + (square root of 4) 14 = 4 x 4 - 4 + (square root of 4) 15 = 4 x 4 - 4/4 16 = 4 + 4 + 4 + 4 17 = 4 x 4 + 4/4 18 = 4 x 4 + 4 - (square root of 4) 19 = 4! - 4 - 4/4 20 = (4 + 4) x (square root of 4) + 4 21 = (44 - (square root of 4)) / (square root of 4) 22 = 4 x 4 + 4 + (square root of 4) 23 = 4! + 4/4 - (square root of 4) 24 = 4 x 4+ 4 + 4 25 = ((square root of 4) + (square root of 4))! + 4/4 26 = 44/(square root of 4) + 4 27 = 4! + (square root of 4) + 4/4 28 = 4! + 4 + 4 - 4 29 = 4! + 4 + 4/4 30 = 4! + 4 + 4 - (square root of 4) 31 = 32 = (4 x 4 x (square root of 4)) - 4 33 = 34 = 4 + 4 + 4! + (square root of 4) 35 = 4! + 44 / 4 36 = 4! + 4 + 4 + 4 37 = 4! + (4! + (square root of 4)) / (square root of 4) 38 = 44 - 4 - (square root of 4) 39 = 4! + (4 + 4/4)!! 40 = 44 - (square root of 4) - (square root of 4) 41 = 44 - (4!/4!!) 42 = 44 - 4 + (square root of 4) 43 = 44 - 4/4 44 = 44 + 4 - 4 45 = 44 + 4/4 46 = 44 + 4 - (square root of 4) 47 = 4! + 4! - 4 / 4 48 = 44 + (square root of 4) + (square root of 4) 49 = (4! x (square root of 4)) + 4/4 50 = 44 + 4 + (square root of 4) 51 = 4! x (square root of 4) + (4!/4!!) 52 = 44 + 4 + 4 53 = ((4!!)!! + 4!)/ 4!! + (square root of 4) 54 = 4! x (square root of 4) + (square root of 4) + 4 55 = ((4!!)!! + 4!)/ 4!! + 4 56 = 4! x (square root of 4) + 4 + 4 57 = (4!/4!!)^4 - 4! 58 = 4!! x 4!! - 4 - (square root of 4) 59 = ((4!!)!! + 4!)/ 4!! + 4!! 60 = 44 + (4 x 4) 61 = 4!! x 4!! - (4!/4!!) 62 = 4 x 4 x 4 - (square root of 4) 63 = 4!! x 4!! - 4/4 64 = 4! x (square root of 4) + 4 x 4 65 = 4!! x 4!! + 4/4 66 = 4 x 4 x 4 + (square root of 4) 67 = 4!! x 4!! + (4!/4!!) 68 = 4 x 4 x 4 + 4 69 = ((4!)^(square root of 4) - 4!)/ 4!! 70 = 44 + 4! + (square root of 4) 71 = ((4!)^(square root of 4) - 4!!)/ 4!! 72 = (4 x 4 + (square root of 4)) x 4 73 = ((4!/4!!)^4 - 4!! 74 = 4!! x 4!! + 4!! + (square root of 4) 75 = ((4!)^(square root of 4) + 4!)/ 4!! 76 = 4! x (4!/4!!) + 4 77 = (4!/4!!)^4 - 4 78 = 4!! x (4!! + (square root of 4)) - (square root of 4) 79 = (4!/4!!)^4 - (square root of 4) 80 = (4 x 4 + 4) x 4 81 = (4 - 4/4)^4 82 = 4!! x (4!! + (square root of 4)) + (square root of 4) 83 = (4!/4!!)^4 + (square root of 4) 84 = 44 x (square root of 4) - 4 85 = (4!/4!!)^4 + 4 86 = 44 x (square root of 4) - (square root of 4) 87 = ((4 + (square root of 4))! - 4!) / 4!! 88 = 44+44 89 = (4!/4!!)^4 + 4!! 90 = 4! x 4 - 4 - (square root of 4) 91 = ((4 + (square root of 4))! + 4!!) / 4!! 92 = 4! x 4 - (square root of 4) - (square root of 4) 93 = 4! x 4 - (4!/4!!) 94 = 4! x 4 - 4 + (square root of 4) 95 = 4! x 4 - 4/4 96 = 4! x 4 + 4 - 4 97 = 4! x 4 + 4/4 98 = 4! x 4 + 4 - (square root of 4) 99 = 4! x 4 + (4!/4!!) 100 = (4 + 4 + (square root of 4))^(square root of 4) Rayner (Wei Yang actually wrote on my section and most are actually wrong. (._. |||) )

1 = (4 / 4) x (4 / 4) 1 = ((4 / 4) / 4) x 4 1 = (4 - 4) + (4 / 4) 1 = ((square of 4 + 4) / 4) - 4 2 = (4 - square root of 4) + 4 - 4 2 = ((4 + 4) - 4) - square root of 4 2 = ((square of 4 - 4) -4) / 4 2 = ((4 - 4) + 4) - square root of 4 3 = (4 + 4 + 4) / 4 4 = (((square of 4 - 4) - 4) - 4) 4 = ((square root of 4 + 4) - 4) + square root of 4 5 = ((square of 4) + 4) / 4 6 = (4 + 4 + 4) / square root of 4 6 = ((square of 4)+ 4 + 4) / 4 7 = (4 + 4) - 4/4 8 = ((4 + 4) x 4) / 4 8 = ((square of 4 + 4) - 4) / 4 9 = 44/4 - square root of 4 10 = (4 + 4 + 4) - square root of 4 11 = (square of 4) - 4 - 4/4 12 = (square of 4) + 4 - 4 - 4 13 = (square of 4) - 4 + 4/4 14 = ((4 - 4) + square of 4) - square root of 4 14 = (square root of 4) + 4 + 4 + 4 15 = (4/4 + square of 4) - square root of 4 16 = 4 + 4 + 4 + 4 16 = (4/4) x ( square of 4) 17 = (square of 4 +) 4/4 18 = (square of 4) / 4 x 4 + (square root of 4) 19 = (square of 4) + (4 - (4/4) <span style="color: #0b0eda; font-family: Impact,Charcoal,sans-serif;">By : Hari

1 = (4 / 4) x (4 / 4) 2 = (4 x 4 / 4) / (square root of 4) 2 = [4 - (4 / 4)]! - 4 2 = (4 / 4) + (4 / 4) 3 = (4 + 4 + 4) / 4 4 = (square root of 4 x 4 x 4) - 4 5 = (4 x 4 + 4) / 4 6 = (4 + 4 + 4) / (square root of 4) 7 = 4 + [(4 + (square root of 4)) / (square root of 4) 8 = 4 x 4 / 4 + 4 9 = 4 + 4 + (4 / 4) 10 = 4 x 4 - 4 - (square root of 4) 11 = 44 / ((square root of 4) + (square root of 4)) 12 = (square root of 4 x 4 x 4) + 4 13 = (44 / 4) + (square root of 4) 14 = 4 + 4 + 4 + (square root of 4) 15 = (4 x 4) - (4 / 4) 15 = 44 / 4 + 4 16 = 4 + 4 + 4 + 4 16 = (4 / 4) x (4 x 4) 16 = (square root of 4) x (square root of 4) x (square root of 4) x (square root of 4) 17 = 4 x 4 + (4 / 4) 18 = (square of 4) / 4 x 4 + (square root of 4) 19 = (square of 4) + [4 -(4/4) ( I ,Hari changed this with the permission of Isaac) (Did you ask me? Anyway, thanks! Isaac) 20 = (square of 4) + 4 - 4 + 4 21 = (4 x 4) + 4 + (square root of 4) 22 = (square of 4) + 4 + (square root of 4) 24 = 4 x 4 + 4 + 4 24 = (square of 4) + (square of 4) - 4 - 4 25 = [4 + (4 / 4)] ^ (square root of 4) 26 = [(square of 4) x (square root of 4)] - 4 - (square root of 4) 27 = (square of 4) + 44 / 4 28 = (square of 4) + 4 + 4 + 4 28 = [(4 x 4) x (square root of 4)] - 4 30 = [(4 x 4) x (square root of 4)] - (square root of 4) 31 = [(square of 4) x (square root of 4)] - (4 / 4) 32 = [(square of 4) x (square root of 4)] + 4 - 4 32 = 4! + 4! - (4 x 4) 33 = [(square of 4) x (square root of 4)] + (4 / 4) 34 = [(4 x 4) x (square root of 4) + (square root of 4) 36 = [4 + (square root of 4)] + [4 + (square root of 4)] 40 = 4! + 4! - 4 - 4 43 = 44 - (4 / 4) 44 = 4! + 4! - (square root of 4) - (square root of 4) 44 = 44 + 4 - 4 45 = 44 + (4 / 4) 45 = [4 + (square root of 4)] ! / (4 x 4) 52 = 44 + 4 + 4 60 = 4 x 4 x 4 - 4 64 = 4 ^ [4 - (4 / 4)] 64 = (square of 4) + (square of 4) + (square of 4) + (square of 4) 68 = 4 ^ 4 / 4 + 4 88 = 44 + 44 96 = (4! / 4) x 4 x 4 96 = 4! + 4! + 4! + 4! From Isaac

1 = (square root of 4)^4 / (4 x 4) 2 = (square root of 4) x (square root of 4) x (square root of 4) / 4 4 = (4 - 4) x 4 + 4 4 = (4 - 4) / 4 + 4 12 = 4 - (4 / 4) x 4 16 = 4^4 / (4 x 4) 16 = 44 - 4! + 4 16 = (4 + 4 - 4) x 4 20 = 4 + (4 / 4) x 4 24 = 44 - 4! + 4 32 = 4^4 / (4 + 4) 32 = (square root of 4) x (square root of 4) x (square root of 4) x 4 44 = 44 x 4 / 4 48 = (4 + 4 + 4) x 4 56 = (4! - 4) x 4 - 4! 64 = (square root of 4) x (square root of 4) x 4 x 4 76 = (4! - 4) x 4 - 4 84 = (4! - 4) x 4 + 4 88 = (4! + 4) x 4 - 4! 88 = (square root of 4)^4 x 4 + 4! From Sheldon

48=4^4-4x4 From Huang Zhong (Rayner:^ means to the power of and 4 to the power of 4 = 256 not 64)

0=4+4-4-4 1=4/4+4-4 2=(4x4)/(4+4) 3=(4+4+4)/4 4=4x(4-4)+4 5=(4x4+4)/4 By Yu Jun

1=4-4+4/4 2=4+4-4-sqrt4 3=4-sqrt4+4/4 4=4+4-sqrt4-sqrt4 5=sqrt4+sqrt4+4/4 6=4+4-4+sqrt4 7=4+4-4 8=4*4-4-4 9=4+4+4/4 10=4+4+4-sqrt4 11=44/sqrt4/sqrt4 12=4+4+sqrt4+sqrt4 13=44/4+sqrt4 14=4+4+4+sqrt4 15=4*4-4/4 16=4+4+4+4 17=4*4+4/4 18=4*4+4-sqrt4 19=4!-4-4/4 20=4*4+sqrt4+sqrt4 21=4!-sqrt4-4/4 22=4*4+4+sqrt4 23=4!-sqrt4+4/4 24=4*4+4+4 25=4!+sqrt4-4/4 26=4!+4-4+sqrt4 27=4!+4-4/4 28=4!+4+4-4 29=4!+4+4/4 30=4!+4+4-sqrt4 31=4!+(sqrt4/.4)+sqrt4 32=4*4+4*4 33=4!+(sqrt4/.4)+4 34=4!+4+4+sqrt4 35=4!+(44/4) 36=4!+4+4+4 37=F(4/.4)+F(sqrt4/.4) 38=4!+4!-4/.4 39=F(4/.4)+sqrt4/.4 40=sqrt4/.4*4*sqrt4 41=L(4+4)-4-sqrt4 42=4!+4!-4-sqrt4 43=L(4+4)-sqrt4-sqrt4 44=44/4*4 45=L(4+4)-4+sqrt4 46=4!+4!-4+sqrt4 47=4!+4!-4/4 48=4!+4!+4-4 49=4!+4!+4/4 50=4/.4*sqrt4/.4 where F(1)=0, F(2)=1, F(n)=F(n-1)+F(n-2) and L(1)=1, L(2)=3, L(n)=L(n-1)+L(n-2). Jeremy

1 is 4/4+4-4 2 is 4-4square root+4-4 3 is (4+4+4)/4 4 is 4 square root+4square root+4-4 5 is (4*4+4)/4 6 is 4 square root+4+4-4 7 is 4+4 square root+(4/4) 8 is 4*4-4-4 9 is 4+4+(4/4) 10 is 4+4square root+4square root+4square root 11 is 44/4square root/4square root 12 is 4+4+4square root+4square root 13 is 44/4+4square root 14 is 4+4+4+4square root 15 is 44/4+4 16 is 4*4+4-4 17 is 4*4+(4/4) 18 is (4square root+4square root)*4+4square root 19 is 4!-4-(4/4) 20 is 44/4square root-4square root 21 is 4!-4square root-(4/4) 22 is 4*4+4+4square root 23 is (44+4square root)/4square root 24 is 4*4+4+4 25 is 4!+4square root-(4/4) 26 is 24+4square root+4-4 27 is 4!+4-(4/4) 28 is 4!+4-4+4 29 is 4!+4+(4/4) 30 is 4!+4square root+4square root+4square root 31 is ((4+4square root)!+4!)/4! 32 is 4!+4+4square root+4square root 33 is 4!+4+4square root/.4 34 is 4!+4+4+4square root 35 is 4!+44/4 36 is (4+4square root)*(4+4square root) 37 is 4!+(4!+4square root)/4square root 38 is 44-4-4square root 39 is 4!+4!/(4*.4) 40 is 44-4square root-4square root 41 is (4!+4square root)*.4-4! 42 is 4!+4!-4+4square root 43 is 44-(4/4) 44 is 4!+4!-4square root-4square root 45 is 44+(4/4) 46 is 44+4-4square root 47 is 4!+4!-(4/4) 48 is 44+4square root+4square root 49 is 4!+4!+(4/4) 50 is 44+4+4square root 51 is (4!-4+.4)/.4 52 is 44+4+4 53 is 4!+4!+4square root/.4 54 is 4!+4!4square root+4 55 is (4!-4+4square root)/.4 56 is 4!+4!+4+4 57 is (4!-4square root)/.4+4square root 58 is (4!+4)*4square root+4square root 59 is (4!-4square root)/.4+4 60 is 4*4*4-4 61 is (4!+4square root)/.4-4 Jun Da

1 is (4-4)+(4/4) 2 is (4square root+4square root)/(4square/4) 3 is 4*4/4square root+4/(4/4) 4 is (4square root+4square root)/(4/4) 5 is (4square+4)/(4square/4) 6 is 7 is 8 is (4+4)-(4-4) 9 is 10 is 11 is 12 is 13 is 14 is 15 is 16 is (4+4)+(4+4) 17 is 18 is 19 is 20 is 21 is 22 is 23 is 24 is 25 is 26 is 27 is 28 is 29 is 30 is 31 is 32 is (4x4)+(4x4) 33 is 34 is 35 is 36 is 37 is 38 is 39 is 40 is 41 is 42 is 43 is 44 is 45 is 46 is 47 is 48 is 49 is 50 is 51 is 52 is 53 is 54 is 55 is 56 is 57 is 58 is 59 is 60 is 61 is 62 is 63 is 64 is 65 is 66 is 67 is 68 is 69 is 70 is 71 is 72 is 73 is 74 is 75 is 76 is 77 is 78 is 79 is 80 is 81 is 82 is 83 is 84 is 85 is 86 is 87 is 88 is 89 is 90 is 91 is 92 is 93 is 94 is 95 is 96 is 97 is 98 is 99 is 100 is Illio

I got it!!! 1= 44/44 2= 4/4 + 4/4 3= (4 + 4 + 4) / 4 4= 4 + (4 - 4) x 4 5= (4 x 4 + 4) / 4 6= (4 + 4) / 4 + 4 7= 4 + 4 - 4/4 8= 4 + 4 + 4 - 4 9= 4 + 4 + 4/4 10= (44 - 4) / 4 11= 44 / (SQRT4 + SQRT4) 12= 4 x 4 - SQRT4 x SQRT4 13= 4 x 4 - (4! / 4!!) 14= 4 x 4 - 4 + SQRT4 15= 4 x 4 - 4/4 16= 4+ 4 +4 + 4 17= 4 x 4 + 4/4 18= 4 x 4 + 4 - SQRT4 19= 4! - 4 - 4/4 20= (4 + 4) x SQRT4 + 4 21= (44 - SQRT4) / SQRT4 22= 4 x 4 + 4 + SQRT4 23= 4! + 4/4 - SQRT4 24= 4 x 4+ 4 + 4 25= (SQRT4 + SQRT4)! + 4/4 26= 4 x 4 + 4!! + SQRT4 27= 4! + SQRT4 + 4/4 28= 4! + 4 + 4 - 4 29= 4! + 4 + 4/4 30= 4! + 4 + 4 - SQRT4 31= 4! + 4!! - 4/4 32= 4! + 4!! + 4 - 4 33= 4! + 4!! + 4/4 34= 4! + 4!! + 4 - SQRT4 35= 4! + 4!! + (4!/4!!) 36= 4! + 4 + 4 + 4 37= 4! + (4! + SQRT4) / SQRT4 38= 4! + 4!! + 4 + SQRT4 39= 4! + (4 + 4/4)!! 40= 44 - SQRT4 - SQRT4 41= 44 - (4!/4!!) 42= 44 - 4 + SQRT4 43= 44 - 4/4 44= 44 + 4 - 4 45= 44 + 4/4 46= 44 + 4 - SQRT4 47= 44 + (4!/4!!) 48= 44 + SQRT4 + SQRT4 49= (4! x SQRT4) + 4/4 50= 44 + 4 + SQRT4 51= 4! x SQRT4 + (4!/4!!) 52= 44 + 4 + 4 53= ((4!!)!! + 4!)/ 4!! + SQRT4 54= 4! x SQRT4 + SQRT4 + 4 55= ((4!!)!! + 4!)/ 4!! + 4 56= 4! x SQRT4 + 4 + 4 57= (4!/4!!)^4 - 4! 58= 4!! x 4!! - 4 - SQRT4 59= ((4!!)!! + 4!)/ 4!! + 4!! 60= 4!! x 4!! - SQRT4 - SQRT4 61= 4!! x 4!! - (4!/4!!) 62= 4!! x 4!! - 4 + SQRT4 63= 4!! x 4!! - 4/4 64= 4!! x 4!! + 4 - 4 65= 4!! x 4!! + 4/4 66= 4!! x 4!! + 4 - SQRT4 67= 4!! x 4!! + (4!/4!!) 68= 4!! x 4!! + SQRT4 + SQRT4 69= ((4!)^SQRT4 - 4!)/ 4!! 70= 4!! x 4!! + 4 + SQRT4 71= ((4!)^SQRT4 - 4!!)/ 4!! 72= 4!! x 4!! + 4 + 4 73= ((4!/4!!)^4 - 4!! 74= 4!! x 4!! + 4!! + SQRT4 75= ((4!)^SQRT4 + 4!)/ 4!! 76= 4! x (4!/4!!) + 4 77= (4!/4!!)^4 - 4 78= 4!! x (4!! + SQRT4) - SQRT4 79= (4!/4!!)^4 - SQRT4 80= 4!! x (4 + 4 + SQRT4) 81= (4 - 4/4)^4 82= 4!! x (4!! + SQRT4) + SQRT4 83= (4!/4!!)^4 + SQRT4 84= 4!! x (4!! + SQRT4) + 4 85= (4!/4!!)^4 + 4 86= (4 + SQRT4)! / 4!! - 4 87= ((4 + SQRT4)! - 4!) / 4!! 88= (4 + SQRT4)! / 4!! - SQRT4 89= (4!/4!!)^4 + 4!! 90= (SQRT4 + SQRT4 + SQRT4)! / 4!! 91= ((4 + SQRT4)! + 4!!) / 4!! 92= 4! x 4 - SQRT4 - SQRT4 93= 4! x 4 - (4!/4!!) 94= 4! x 4 - 4 + SQRT4 95= 4! x 4 - 4/4 96= 4! x 4 + 4 - 4 97= 4! x 4 + 4/4 98= 4! x 4 + 4 - SQRT4 99= 4! x 4 + (4!/4!!) 100= (4 + 4 + SQRT4)^SQRT4 '!!': With an even number such as 4, you just multiply the even numbers, so 4!! = 4x2 =8 or 6!! = 6x4x2 = 48 or 8!! = 8x6x4x2=384. With an odd number such as 7 you just multiply the odd numbers so 7!! = 7x5x3x1 =105. Zheng Xian (Rayner: Look at my section. It looks the similar. I know how you "calculated" them.) (ZX: I know. Don't reveal it. You wrote them in white so people won't suspect.) (Rayner: Hey! You just asked me not to reveal but said that it is in white.) (ZX: As in how we got the answer.)

SEAM x T = MEATS
 * __Cryptarithms__**

What do the letters stand for?

4973 x 8 = 39784 Sheldon

Can you put two numbers made from one each of the ten digits (1234567890) to replace the 'y's to make a multiplication sum? Zero is not to be at the start or end of either number.

yyyyyyyyyy x 2 = yyyyyyyyyy

4563728190 x 2 = 9127456380 Rayner (Mdm Teo: Good effort, Rayner. Please read the last sentence of the question.)

1638024579 x 2 = 3276049158 Mandy

1023456789 x 2 = 2046913578 Sheldon

1203456789 x 2 =2401913578 Zhi Rui (Mandy: Zhi Rui, I think you meant to put a 6 at- 240__1__913578.)

Can you arrange the numerals 1 to 9 in a single fraction that equals exactly 1/3 (one-third)?
 * __Fraction puzzle__**

5823/17469 Mandy

(Henry Ernest Dudeney was a mathematician)__** Make a set of prime numbers which will add to the smallest possible total. You must use each of the 9 digits once and once only. For eg. 61 + 283 + 47 + 59 = 450. Can you make a smaller total than this?
 * __Dudeney puzzle

89+43+61+5+7+2=207 Blue Diamond and Mandy

5+29+47+61+83=225 Given by Cyan Blue in class

2+5+47+61+89+3=207 Wei Yang (Mdm Teo: Wei Yang, are 87 and 49 prime numbers?)

2+5+7+43+61+89=207 Zhi Rui

7+5+61+89+43+2=207 Rayner

2+3+5+41+67+89=207 Zheng Xian

5+61+23+47+89=225 Huang Zhong