Find the 8 digit number

This is a easy and good puzzle.
Find the 8 digit number which satisfies all of the following conditions
1-> the first 2 and last 2 digits are same.
2-> the total of 8 digit is 37.
3-> it is a square number.
4-> it is a cube number.

If anyone wants to make the code, here is the hint:
try making a series of square & cube numbers, then save the common numbers in series. Then you know the remaining task.


Ans:
16777216.


The square root is 4096 and cube root is 256.

i followed this approach as it decreases the chances , i checked for cubes of numbers which end up in 8 digits so the range of numbers from 216 - 464 on cubing gives the answer which is of 8 digits following the same approach in squaring 3163 - 9999 gives the numbers in 8 digits.

So i wrote a loop to check the cube roots against the square roots and i didnt use the condition the first two and last two are same ..
With using that i think the solution will be even simpler ..

Three people check into a hotel. They pay £30 to the manager and go to their room. The manager suddenly remembers that the room rate is £25 and gives £5 to the bellboy to return to the people. On the way to the room the bellboy reasons that £5 would be difficult to share among three people so he pockets £2 and gives £1 to each person. Now each person paid £10 and got back £1. So they paid £9 each, totaling £27. The bellboy has £2, totaling £29. Where is the missing £1?


Ans:

We have to be careful what we are adding together. Originally, they paid £30, they each received back £1, thus they now have only paid £27. Of this £27, £25 went to the manager for the room and £2 went to the bellboy.

Try it for three weightings

You have 12 coins, one of which is fake. The fake coin is indistinguishable from the rest except that it is either heavier or lighter, but you don't know which. Can you determine which is the fake coin and whether it is lighter or heavier using a balance scale and only 3 weightings?



Ans:

One solution is to label the coins with the letters FAKE MIND CLOT and weigh the coins in the following three combination:

MA DO -- LIKE
ME TO -- FIND
FAKE -- COIN

Logic will now allow you to find the fake coin based on the three results. For instance, if the results were left down, balanced, left down, we could work out which coin is fake in the following way:

From the middle weighing we know that the coins METOFIND are all normal. So one of the coins ACKL is fake. Therefore looking at these coins one at a time in the other two weighings, we can see that:

A - appears on the left twice and could be fake.
C - appears only once, therefore can't be fake (otherwise the first weighing would be balanced).
K - appears on opposite sides, so it can't make the left side go down both times.
L - appears only once, therefore can't be fake (otherwise the third weighing would be balanced).

Therefore the only possibility is A, which must be heavier. Any other combination of ups and downs will allow you to use the same logic to find the fake coin.

Matthew Shelborn frequently has to travel for his company, which gives him a chance to meet many people from all parts of the USA. In April, Matthew flew to five different US cities on business and he flew a different airline each time. During each trip he chatted with the person next to him, and no two people he talked to were in the same profession.

From the information, can you determine the date Matthew made each flight (each was on a Monday exactly one week apart starting on April 2nd), the airline he flew, his destination, and the profession of the person who sat next to him on each flight?

1. Three consecutive flights were, in order from first to last, the flight Matthew took with WTA Airways, the flight where he sat next to the teacher, and the flight he took to Atlanta.

2. Matthew sat beside the sports coach on a flight some time earlier in the month than the one he took to Seattle.

3. The week he flew Air Express was some time earlier in the month than the trip to Boston, which was some time earlier in the month than the trip where he sat next to the ballet dancer.

4. It wasn't on the trip to San Diego where Matthew sat next to the doctor.

5. Matthew didn't fly Skyways on his trip to Seattle, and he didn't fly WTA Airways on the trip where he sat next to the sports coach.

6. Atlanta was not Matthew's destination on the trip where he made the acquaintance of the ballet dancer.

7. The Fly America flight was exactly two weeks before the flight where Matthew passed the time chatting with the attorney.

Dates: April 2nd, April 9th, April 16th, April 23rd, April 30th
Airline: Air Express, Fly America, MidUSA Air, Skyways, WTA Airways
Destination: Atlanta, Boston, Chicago, San Diego, Seattle
Seat Mate: Attorney, Ballet Dancer, Doctor, Sports Coach, Teacher

Ans:

Date Airline Destination Seat Mate
April 2nd WTA Airways Chicago Doctor
April 9th Fly America San Diego Teacher
April 16th Air Express Atlanta Sports Coach
April 23rd Skyways Boston Attorney
April 30th MidUSA Air Seattle Ballet Dancer
The trip to Atlanta (1), the trip where Matthew sat next to the ballet dancer (3), and the trip where Matthew sat next to the attorney (7) were among the last three trips [April 16th, 23rd, and 30th (intro)], and the trip to Atlanta wasn't the one where he met the ballet dancer (6) or the attorney [he flew WTA Airways exactly two weeks before the one to Atlanta, and he flew Fly America exactly two weeks before he sat by the attorney (7), thus, these three are separate flights, in some order. He didn't sit by the teacher on the trip to Atlanta (1), so the flight were he sat by the teacher was on one of the first two flights [April 2nd and April 9th (intro)], but it wasn't first [after WTA Airways](1), therefore, April 2nd flight was the WTA Airways flight, the one where he sat by the teacher was the April 9th flight, and the one to Atlanta was the April 16th flight, and he sat by the ballet dancer and attorney on the April 23rd and

April 30th flight, in some order. He didn't sit beside the sports coach on the WTA Airways flight (5) so that wasn't April 2nd, he sat beside the sports coach on the Atlanta flight [April 16th] and it was the April 2nd flight where he sat next to the doctor. April 2nd [first flight, WTA Airways, doctor] wasn't to Seattle (2), Boston (3), or San Diego (4), that flight was to Chicago. The April 9th flight wasn't to Seattle [after sports coach which was April 16th] or to Boston [after Air Express, which would have to be April 2nd, but WTA Airways was April 2nd (1)](3), thus, on April 9th he flew to San Diego. Boston wasn't the last destination (3), so it was his destination on April 23rd and Seattle was his destination on April 30th. Boston [April 23rd] was before the one where he sat by the ballet dancer (3), so he sat by the ballet dancer on April 30th flight to Seattle, and it was on the April 23rd flight to Boston that he sat by the attorney. Since he sat by the attorney on the April 23rd flight, then he flew Fly America on the 9th [exactly two weeks before](7). He flew Air Express before the flight when he flew to Boston [April 23rd](3) so he flew Air Express on the 16th. He didn't fly Skyways to Seattle [April 30th](5), thus, he flew Skyways to Boston on the 23rd and his last flight that month was on MidUSA Air (intro).

Try to trace the no of eggs

Once upon a time, and old lady went to sell her vast quantity of eggs at the local market.

When asked how many she had, she replied:

Son, I can't count past 100 but I know that.

If you divide the number of eggs by 2 there will be one egg left.
If you divide the number of eggs by 3 there will be one egg left.
If you divide the number of eggs by 4 there will be one egg left.
If you divide the number of eggs by 5 there will be one egg left.
If you divide the number of eggs by 6 there will be one egg left.
If you divide the number of eggs by 7 there will be one egg left.
If you divide the number of eggs by 8 there will be one egg left.
If you divide the number of eggs by 9 there will be one egg left.
If you divide the number of eggs by 10 there will be one egg left.

Finally. If you divide the Number of eggs by 11 there will be NO EGGS left!

How many eggs did the old lady have?




25,201 eggs. This puzzle has a few different methods for finding the solution, one of which is:

Find a number X into which all of the numbers from 2 to 10 divide evenly. You can do this by simply using 2*3*4*5*6*7*8*9*10, but you can find a smaller number by finding the prime factors, a subset of which can be used to form any number from 2 to 10. 2*2*2*3*3*5*7 will do. This comes out to be 2520, and is the lowest number into which all the numbers 2-10 divide evenly.

We can add 1 to this number to satisfy the first 9 constraints of the puzzle (the remainder of 2521/2, 2521/3 ... 2521/10 is one), but this does not satisfy the last constraint, divisibility by 11.

Fortunately, we can multiply X (=2520) by any integer and add 1 and we will still satisfy constraints 1-9. So what Y do we multiply X by so that (X*Y) + 1 is divisible by 11. 2520/11 has a remainder of 1. Thus two 2520s divided by eleven would have a remainder of 1+1 = 2, and so forth...so ten 2520s divided by 11 would have a remainder of 10. This number plus one would divide eleven evenly, as well as also satisfy the first 9 constraints - thus 25201 is the answer.

minimum possible steps: Switching & entering

There are 2 separated rooms, one having 8 bulbs and the other having 8 switches.
You have to find out which switch is of which bulb, in minimum possible steps.
Switching & entering into the room once is defined as ONE STEP



Ans
Since we have 8 bulbs and 8 switches, we only need to find out 7 bulb/switch pairs and then we should know the 8th pair.

With the remaining 7 switches, I choose to switch on two at a time (switching everything else off before each step), like this:

1&2
2&3

And then we should be able to pair switches 1, 2 and 3 to their respectful bulbs with bulb no. 2 being the one lit in both steps and bulbs no. 1 and 3 lit in their respectful steps, and then continue with the same method:

4&5
5&6

Having figured out switches 4,5 and 6 also, we only need to flick switch number 7 to see which one that links to. And as stated previously we should also know by then that bulb no. 8 is the one that was never turned on.

Anybody that can do it in fewer steps?

I'm at work right know, which I should really be getting back to but looking forward to working on a formula for this problem later I'm thinking something - 1, then modulus 3 and then the rest dived by three?


WAIT! - Strike everything above

Having written this post I wondered if turning on three bulbs at a time would work better.... and yeah.... I'm guessing we could do it in 4 steps... :P

Turning on:
Step 1: 1 & 2 & 3
Step 2: 3 & 4 & 5 Giving us the bulb linked to no. 3 (lit first two steps)
Step 3: 5 & 6 & 7 Giving us bulb no. 5 (lit in step 2&3) AND bulb no. 4 (only lit in step 2)
Step 4: 7 & 8 & 1 Same as above, giving no. 6 (only lit in step 3), no. 7 (lit in steps 3&4), no. 8 (the only one first lit in step 4) and bulb 1 (lit in steps 1 and 4)

I could have just turned on switch 7&1 with switch 8 then being linked to the one bulb never lit. Better yet, the same algorithm could be used to link 9 switch/bulb pairs, then with the 9th bulb never being lit.

I then decided to write out the steps in respect to number of odds and even numbers, as so:
1 - odd number 1 = O1
2 - even number 1 = E1
3 - odd number 2 = O2
4 - even number 2 = E2...

With this notation, the steps for the above problem (with 8 or 9 bulbs that is) would be:
O1 & E1 & O2
O2 & E2 & O3
O3 & E3 & O4
O4 & E4 & O1

As some might notice, the number of steps required to pair X switches/bulbs is the same as the number of even numbers counting up to (and including) X. And because an even number divided by two gives us the number of even numbers counting up to that number, all we need to do is to subtract one from the number if X is an odd number, and I would guess the best mathematical notation for that would by x - modulus(x/2), right?

So the the number of steps required to pair X switches to X bulbs is (x - modulus(x/2)) / 2

So, for example, to figure out 13 switch/bulb pairs, we would need:
(13 - modulus(13/2)) / 2
(13 - 1) / 2
12 / 2 = 6 steps

23 Brilliant Doubts - Unanswered

23 Brilliant Doubts - Unanswered


1.If all the nations in the world are in debt(am not joking. even US has got debts), where did all the money go? (weird)

2.When dog food is new and improved tasting, who tests it? (to be give a thought)

3.What is the speed of darkness? (absurd)

4.If the "black box" flight recorder is never damaged during a plane crash, why isn't the whole airplane made out of that stuff? (very good thinking)

5.Who copyrighted the copyright symbol? (who knows)

6.Can you cry under water? (let me try)

7.Why do people say, "you've been working like a dog" when dogs just sit around all day? (i think they meant something else)

8.Why are the numbers on a calculator and a phone reversed? (God knows)

9.Do fish ever get thirsty? (let me ask and tell)

10.Can you get cornered in a round room? (by ones eyes)

12.Why do birds not fall out of trees when they sleep? (tonight i will stay and watch)

13.What came first, the fruit or the color orange? (seed)

14.If corn oil is made from corn, and vegetable oil is made from vegetables, then what is baby oil made from? (No comments)

15.What should one call a male ladybird? (No comments)

16.If a person suffered from amnesia and then was cured would they remember that they forgot? (can somebody help )

17.Can you blow a balloon up under water? (yes u can)

18.Why is it called a "building" when it is already built? (strange isn't it)

19.If you were traveling at the speed of sound and you turned on your radio would you be ! able to hear it? (got to think scientifically)

20.If you're traveling at the speed of light and you turn your headlights on, what happens? (i don have a change to try)

21.Why is it called a TV set when theres only one? (very nice)

22.If a person owns a piece of land do they own it all the way down to the core of the earth? (this is nice)

23.Why do most cars have speedometers that go up to at least 130 when you legally can't go that fast on any road? (stupid, break the law)

How many positive factors of 36,000,000 are not perfect squares?

Question: How many of the positive factors of the number 36,000,000 are not perfect squares.

Any answer without an explanation is spam.





Ans.

My answer is 149.Is it right or wrong??

Here goes my explanation:

36000000=6 x 6 x 10 x 10 x 10 x 10 x 10 x 10
=2 x 3 x 2 x 3 x 2 x 5 x 2 x 5 x 2 x 5 x 2 x 5 x 2 x 5 x 2 x 5
=2⁸ x 3² x 5⁶
Factors possible are:
1
2, 2², 2³, 2⁴, 2⁵, 2⁶, 2⁷, 2⁸
3, 3²
5, 5², 5³, 5⁴, 5⁵, 5⁶
For each (2 to 2⁸ and 3 to 3² ) there will be total 16 (8 x 2) combinations of factors having subfactors 2 and 3 only.

2 x 3, 2² x 3, 2³ x 3, 2⁴ x 3, 2⁵ x 3, 2⁶ x 3, 2⁷ x 3, 2⁸ x 3

2 x 3², 2² x 3², 2³ x 3², 2⁴ x 3², 2⁵ x 3², 2⁶ x 3², 2⁷ x 3², 2⁸ x 3²


Similarly for each (2 to 2⁸ and 5 to 5⁶ ) there will be total 48(8 x 6) combinations of factors having subfactors 2 and 5 only.

2 x 5, 2² x 5, 2³ x5, 2⁴ x 5, 2⁵ x 5, 2⁶ x 5, 2⁷ x 5, 2⁸ x 5

2 x 5², 2² x 5², 2³ x 5², 2⁴ x 5², 2⁵ x 5², 2⁶ x 5², 2⁷ x 5², 2⁸ x 5²

2 x 5³, 2² x 5³, 2³ x 5³, 2⁴ x 5³, 2⁵ x 5³, 2⁶ x 5³, 2⁷ x 5³, 2⁸ x 5³

2 x 5⁴, 2² x 5⁴, 2³ x 5⁴, 2⁴ x 5⁴, 2⁵ x 5⁴, 2⁶ x 5⁴, 2⁷ x 5⁴, 2⁸ x 5⁴

2 x 5⁵, 2² x 5⁵, 2³ x 5⁵, 2⁴ x 5⁵, 2⁵ x 5⁵, 2⁶ x 5⁵, 2⁷ x 5⁵, 2⁸ x 5⁵

2 x 5⁶, 2² x 5⁶, 2³ x 5⁶, 2⁴ x 5⁶, 2⁵ x 5⁶, 2⁶ x 5⁶, 2⁷ x 5⁶, 2⁸ x 5⁶

For each (3 to 3² and 5 to 5⁶ ) there will be total 12 (6 x 2) combinations of factors having subfactors 3 and 5 only.

3 x 5, 3² x 5
3 x 5², 3² x 5²
3 x 5³, 3² x 5³
3 x 5⁴, 3² x 5⁴
3 x 5⁵, 3² x 5⁵
3 x 5⁶, 3² x 5⁶

Again for each (2 to 2⁸ and 3 to 3² and 5 to 5⁶ ) there will be total 96 (8 x 2 x 6) combinations of factors having subfactors 2,3 and 5 only.

2 x 3 x 5, 2²x 3 x 5, 2³x 3 x 5, 2⁴ x 3 x 5, 2⁵ x 3 x 5, 2⁶ x 3 x 5, 2⁷ x 3 x 5, 2⁸ x 3 x 5

2 x 3² x 5 , 2²x 3² x 5,2³x 3² x 5, 2⁴ x 3² x 5, 2⁵ x 3² x 5, 2⁶ x 3² x 5, 2⁷ x 3² x 5, 2⁸ x 3² x 5


2 x 3x 5², 2²x 3 x 5², 2³x 3 x 5², 2⁴ x 3 x 5², 2⁵ x 3 x 5², 2⁶ x 3 x 5², 2⁷ x 3 x 5², 2⁸ x 3 x 5²

2 x 3² x 5² , 2²x 3² x 5², 2³x 3² x 5², 2⁴ x 3² x 5², 2⁵ x 3² x 5², 2⁶ x 3² x 5², 2⁷ x 3² x 5², 2⁸ x 3² x 5²


2 x 3x 5³, 2²x 3 x 5³, 2³x 3 x 5³, 2⁴ x 3 x 5³, 2⁵ x 3 x 5³, 2⁶ x 3 x 5³, 2⁷ x 3 x 5³, 2⁸ x 3 x 5³

2 x 3² x 5³ , 2²x 3² x 5³, 2³x 3² x 5³, 2⁴ x 3² x 5³, 2⁵ x 3² x 5³, 2⁶ x 3² x 5³, 2⁷ x 3² x 5³, 2⁸ x 3² x 5³


2 x 3x 5⁴, 2²x 3 x 5⁴, 2³x 3 x 5⁴, 2⁴ x 3 x 5⁴, 2⁵ x 3 x 5⁴, 2⁶ x 3 x 5⁴, 2⁷ x 3 x 5⁴, 2⁸ x 3 x 5⁴

2 x 3² x 5⁴, 2²x 3² x 5⁴, 2³x 3² x 5⁴, 2⁴ x 3² x 5⁴, 2⁵ x 3² x 5⁴, 2⁶ x 3² x 5⁴, 2⁷ x 3² x 5⁴, 2⁸ x 3² x 5⁴


2 x 3x 5⁵, 2²x 3 x 5⁵, 2³x 3 x 5⁵, 2⁴ x 3 x 5⁵, 2⁵ x 3 x 5⁵, 2⁶ x 3 x 5⁵, 2⁷ x 3 x 5⁵, 2⁸ x 3 x 5⁵

2 x 3² x 5⁵ , 2²x 3² x 5⁵, 2³x 3² x 5⁵, 2⁴ x 3² x 5⁵, 2⁵ x 3² x 5⁵, 2⁶ x 3² x 5⁵, 2⁷ x 3² x 5⁵, 2⁸ x 3² x 5⁵


2 x 3x 5⁶, 2²x 3 x 5⁶, 2³x 3 x 5⁶, 2⁴ x 3 x 5⁶, 2⁵ x 3 x 5⁶, 2⁶ x 3 x 5⁶, 2⁷ x 3 x 5⁶, 2⁸ x 3 x 5⁶

2 x 3² x 5⁶, 2²x 3² x 5⁶, 2³x 3² x 5⁶, 2⁴ x 3² x 5⁶, 2⁵ x 3² x 5⁶, 2⁶ x 3² x 5⁶, 2⁷ x 3² x 5⁶, 2⁸ x 3² x 5⁶

And finally 2⁸ x 3² x 5⁶ itself is also a factor
So counting all those factors whose subfactor's exponential power is an even no, I get total no of factors of 36000000 which are not perfect squares as 149.
All such factors are marked in bold.

The Ant problem

An ant is at a fixed end 'P' of a 1 meter-long rubber band.
Just as the ant starts walking along the rubber band at a constant 1 cm/sec,
the other end 'Q' is pulled away from 'P' at a steady speed of 100 cm/sec.

Assume a uniformly stretching, infinitely elastic rubber band is used in this puzzle.

Will the ant ever reach 'Q', and if yes, when?

Try this kids.

You have to use the mathematical operators in between the numbers to get the correct output....

1 1 1 =6
2 2 2 =6
3 3 3 =6
4 4 4 =6
5 5 5 =6
6 6 6 =6
7 7 7 =6
8 8 8 =6
9 9 9 =6

its very simple ,you all can easily solve it....



Ans:
(1+1+1)! = 6
2+2+2 = 6
3x3-3 = 6
4+4-(sqrt 4) = 6 [positive square root]
(5/5)+5 = 6
6-6+6 = 6
7-(7/7) = 6
8-sqrt(8/(cube root(8))) = 6
(squrt 9) x (sqrt 9) - (sqrt 9) =6

Mirror puzzle

Imagine you are standing in front of a mirror, facing it. Raise your left hand. Raise your right hand. Look at your reflection. When you raise your left hand your reflection raises what appears to be his right hand. But when you tilt your head up, your reflection does too, and does not appear to tilt his/her head down. Why is it that the mirror appears to reverse left and right, but not up and down?






Ans.

Find the number?

The series is -

12,12,18,45,180,1170,?

What's the next number in above series?






Ans.

Yep, my answer matches with that of bayazidahmed, 12285.

Here's the solution:

Lets name the general term of the sequence T(n)

The sequence (say R(n)) where R(n)=T(n+1)/T(n)
is 1,1.5,2.5,4,6.5...

The sequence (say S(n)) where S(n)=R(n+1)-R(n)
is 0.5,1,1.5,2.5....


Thus S(n) can be thought of a Fibonacci style sequence where S(n)=S(n-1)+S(n-2)
Considering this relation, the next term in S(n) would be 1.5+2.5=4

Considering this, the next term in R(n) would be 6.5+4=10.5

Thus the next term in T(n)=1170*10.5 = 12285

Where lies the flaw in the logic?

You can imagine an arrow in flight, toward a target. For the arrow to reach the target, the arrow must first travel half of the overall distance from the starting point to the target. Next, the arrow must travel half of the remaining distance.

For example, if the starting distance was 10m, the arrow first travels 5m, then 2.5m.

If you extend this concept further, you can imagine the resulting distances getting smaller and smaller. Will the arrow ever reach the target?

Ponder this?

Imagine a prisoner in a prison. He is sentenced to death and has been told that he will be killed on one day of the following week. He has been assured that the day will be a surprise to him, so he will not be anticipating the hangman on a particular day, thus keeping his stress levels in check.

The prisoner starts to think to himself, if I am still alive on Thursday, then clearly I shall be hanged on Friday, this would mean that I then know the day of my death, therefore I cannot be hanged on Friday. Now then, if I am still alive on Wednesday, then clearly I shall be hanged on Thursday, since I have already ruled out Friday. The prisoner works back with this logic, finally concluding that he cannot after all be hanged, without already knowing which day it was.
Casually, resting on his laurels, sitting in his prison cell on Tuesday, the warden arrives to take him to be hanged, the prisoner was obviously surprised!

Ponder this?

Find the numbers

I'm thinking of two integer numbers, each of them is more than 1 and their sum is less than 100.

I tell my friend CEan - Ash the sum of these two numbers, and another friend, CEan - MaRo, product of these two numbers.

Then such a dialog took place:

Ash: I can't determine what are these numbers.
MaRo: Ah, i knew you wouldn't be able to do this.
Ash: Oh, then i know what they are!
MaRo: Oh, then i know them too!

Can you determine the numbers?













Ans:
Assumption : Both numbers should be more than 1

since Ash and Mac knew sum and other the product...but they never exchange information abt sum/product....

if numbers are 2,2

Ash knows 4....but basic priciple is both numbers are more than one so
1,3(it differs because...one number is equal to one)
2,2

and Mach know 4...
so it can be
1,4 (it differs because...on number is equal to one)
2,2
............................
if numbers are 2,3

Ash knows 5....but basic priciple is both numbers are more than one so
1,4(it differs because...one number is equal to one)
2,3

and Mach know 6...
so it can be
1,6 (it differs because...on number is equal to one)
2,3

1 < x < y and x+y < 100 [x,y are the numbers].

But, is the answer, right?

Think harder!

Try this lock puzzle

You want to get through a security door, where you have to enter 3 digits as password, where each digit can be any from 0 to 9. The checking mechanism is, however, defective, and so will let you in if any 2 of your digits match with the password's. e.g. if the password is 087 and you enter 057, then you will be let in. You don't have much time and hence want to make as few tries as possible. What is the minimum number of tries in which you can definitely enter, and what are they (you need not give explicitly, if you describe a pattern).