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Brain Teasers : Mathemetics

RajdeepCE

@rajdeepce-7UdrG8 • Oct 25, 2024

Oct 25, 2024

1.6K

This is a easy and good puzzle.
Find the 8 digit number which satiesfies 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 remainimg task.

0 74

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 24, 2009

The answer is 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 numebers in 8 digits.

So i wrote a loop to check the cube roots aganist 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 ..
😀

Nice puzzle Rajdeep 😀

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RajdeepCE

@rajdeepce-7UdrG8 • May 26, 2009

Nice explanation,ES!!!!
Congrats!!!
Looks like I have to find more difficult puzzles for you.

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RajdeepCE

@rajdeepce-7UdrG8 • May 26, 2009

Here is the new puzzle for you all :
complete the following series,

*->1,2,7,18,41,74,___,___,___,...

fill all 3 blanks with proper explanation.

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 26, 2009

i am trying from 2 hours and didnt get atleast one number 😔

Good Puzzle will try 😀

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RajdeepCE

@rajdeepce-7UdrG8 • May 27, 2009

I am happy that atleast one CEan is interested in this puzzle.
Thanks ES & keep trying. I believe that you will get the solution first.
Let us have sometime & see that who has the guts to solve this.

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RajdeepCE

@rajdeepce-7UdrG8 • May 27, 2009

Hey CEans, try this simple one series :
-> 1,1,2,1,3,2,5,3,8,___,___,___,...

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 27, 2009

Are the next numbers 5 , 13 , 8 ,21

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Harshad Italiya

@harshad-ukH5ww • May 27, 2009

1,1,2,1,3,2,5,3,8,5,13,8

1+1=2
2-1=1
and so on

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RajdeepCE

@rajdeepce-7UdrG8 • May 27, 2009

Both are right!!!
Congrats Godfather & ES!!!
But what about the first one, it is quite hard, isnt it?

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Anil Jain

@CrazyBoy • May 27, 2009

RajdeepCEHere is the new puzzle for you all :
complete the following series,

*->1,2,7,18,41,74,___,___,___,...

fill all 3 blanks with proper explanation.
Rajdeep are you sure this series is correct...

What I can work out is it, is following pattern 3*2^n-n-3, where n starts from 0. (Though its not working for first element)
However 74 is not fitting in this solution; 6 member of the series should be should be 88. If yes then answer is 183, 374, 757.

Correct me if I am wrong.

-CB

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RajdeepCE

@rajdeepce-7UdrG8 • May 27, 2009

Thanks CB, for you reply and pointing out the error. I recalculated the whole series after your reply and I found one error in 6th number. Actually the series is right, but I misspelled the number. The original number is 84. And one more thing that series can be started from 0 but it doesnt affect the series. Sorry for that error.
@CrazyBoy, again thanks. And regarding about your answer, it isnt correct.
Here is the series,
#-> 0,1,2,7,18,41,84,161,__,__,__,...
Try finding this series.

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RajdeepCE

@rajdeepce-7UdrG8 • May 27, 2009

Here is another series. This series is identical with above. If you get through this series than you can easily solve the above series.
#-> 0,1,2,5,10,19,34,59,__,__,__,...

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 27, 2009

😁 i am fast than GF 😀

but the former one is tuff for me am not getting the logic behind 😔

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RajdeepCE

@rajdeepce-7UdrG8 • May 27, 2009

Here is hint: These series are based on Fibonacci Series.

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Harshad Italiya

@harshad-ukH5ww • May 28, 2009

English-Scared😁 i am fast than GF 😀

but the former one is tuff for me am not getting the logic behind 😔
Actually i spent my some time to formatting post 😉

Yep you are sandy 😁

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silverscorpion

@silverscorpion-iJKtdQ • May 28, 2009

Well, thanks very much for the hint rajdeep. I was breaking my head to get this right..

The solution to both the series are,

1) 0,1,2,7,18,41,84,161, 294,519,894

0+1+1; 1+2+4; 2+7+9; 7+18+16; 18+41+25; 41+84+36; 84+161+49; 161+294+64; 294+519+81

2) 0,1,2,5,10,19,34,59, 100,167,276

0+1+1; 1+2+2; 2+5+3; 5+10+4; 10+19+5; 19+34+6; 34+59+7; 59+100+8; 100+167+9;

Hope the answers are correct.. 😀😀

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silverscorpion

@silverscorpion-iJKtdQ • May 28, 2009

These are very interesting. Thanks for such puzzles. Give more difficult and interesting puzzles..
Cheers!!

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 28, 2009

RajdeepCEHere is another series. This series is identical with above. If you get through this series than you can easily solve the above series.
#-> 0,1,2,5,10,19,34,59,__,__,__,...

IS the answer 100 , 167 ,276 ??

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 28, 2009

oops looks like Scorpion already posted the answer 😀

Anyways thanks scorpion the first one didnt struck yet.got the second one.
😀

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RajdeepCE

@rajdeepce-7UdrG8 • May 28, 2009

Congratulation to SS, for solving both the series!!
@ES, cheers!!!! Atleast you struck one.
Here is the common equation,
for 1st series :- X(n+1)=X(n+1)+X(n)+n^2
and for 2nd series :- X(n+1)=X(n+1)+X(n)+n
take X(0)=0, X(1)=1.
I want to post some more series by replacing n by 2n,3n,or n^3, but I just thought that its enough. Thanks SS for your interest, I will post some more difficult series.

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RajdeepCE

@rajdeepce-7UdrG8 • May 28, 2009

Here is one another series for you all CEans, but here is one condition. You have to specify the common equation.
Let us start with modrate difficulty level, I will increase the difficulty latter, cause finding the common eqtation is little tough.

1-> 1,2,2,4,5,9,13,22,__,__,__,...
2-> 0,1,1,3,4,8,12,21__,__,__,...
3-> 1,0,2,2,5,7,13,21,__,__,__,...
4-> 2,1,4,3,6,8,14,21,__,__,__,...
Try this but dont forgot to reply with common equation.

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pradeep_agrawal

@pradeep-agrawal-rhdX5z • May 28, 2009

Below is my solution:

1-> 1,2,2,4,5,9,13,22,__,__,__,...
Considering x(0) = 1, x(1) = 2, the equation should be:
x(n+1) = x(n) + x(n-1) - (n % 2)
Hence,
x(8)
= x(7 + 1)
= x(7) + x(7-1) - (7 % 2)
= x(7) + x(6) - (7 % 2)
= 22 + 13 - 1
= 34

2-> 0,1,1,3,4,8,12,21__,__,__,...
Considering x(0) = 0, x(1) = 1, the equation should be:
x(n+1) = x(n) + x(n-1) + [(n+1) % 2)]
Hence,
x(8)
= x(7 + 1)
= x(7) + x(7-1) + [(7+1) % 2]
= x(7) + x(6) + [8 % 2]
= 21 + 12 + 0
= 33

3-> 1,0,2,2,5,7,13,21,__,__,__,...
Considering x(0) = 1, x(1) = 0, the equation should be:
x(n+1) = x(n) + x(n-1) + (n % 2)
Hence,
x(8)
= x(7 + 1)
= x(7) + x(7-1) + (7 % 2)
= x(7) + x(6) + (7 % 2)
= 21 + 13 + 1
= 35

4-> 2,1,4,3,6,8,14,21,__,__,__,...
I feel the series should be
2,1,3,3,6,8,14,21,__,__,__,...
If that is the case then considering x(0) = 2, x(1) = 1, the equation should be:
x(n+1) = x(n) + x(n-1) - [(n+1) % 2]
Hence,
x(8)
= x(7 + 1)
= x(7) + x(7-1) - [(7+1) % 2]
= x(7) + x(6) + (8 % 2)
= 21 + 14 + 0
= 35
If the series is not as i specified, then i need to give it another try.

-Pradeep

RajdeepCE

@rajdeepce-7UdrG8 • May 28, 2009

You are absolutely right!!!!
You get the another solution for series. Here is my solution:
take X(0)=1, X(1)=1,
1-> X(n+1)=X(n)+X(n-1)+Sin(n*pi/2)
2-> X(n+1)=X(n)+X(n-1)-Cos(n*pi/2)
3-> X(n+1)=X(n)+X(n-1)-Sin(n*pi/2)
4-> X(n+1)=X(n)+X(n-1)+Cos(n*pi/2).
Be prepared for the hard challenges.

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RajdeepCE

@rajdeepce-7UdrG8 • May 28, 2009

Try your mind on this:

#->> 5,17,37,65,101,__,__,__,....

It will be better if you submit the solution with common equation cause it is easy to explain others.

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pradeep_agrawal

@pradeep-agrawal-rhdX5z • May 28, 2009

Below is my solution:

5,17,37,65,101,__,__,__,....
Considering x(0) = 5, the equation should be:
x(n+1) = x(n) + 12 + 8*n
Hence,
x(5)
= x(4 + 1)
= x(4) + 12 + 8*4
= 101 + 12 + 32
= 145

-Pradeep

RajdeepCE

@rajdeepce-7UdrG8 • May 28, 2009

Great Pradeep!!!! You are blazzing fast for solving the series. Now I have to increase the difficulty to intermediatle level.

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 28, 2009

Pradeep you are lightining fast excellent .

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RajdeepCE

@rajdeepce-7UdrG8 • May 28, 2009

Here is the another series, you dont need to give common equation. So just give the answer,

#->>> 0,1,3,2,6,7,5,__,__,__,__,.....

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pradeep_agrawal

@pradeep-agrawal-rhdX5z • May 28, 2009

That's a tough one. I guess it should be 4 but not sure.

Reason:
The series is: 0,1,3,2,6,7,5,__,__,__
If we take difference of each consecutive number the series of differences is
1,2,-1,4,1,2
So i feel the pattern it follows is
1,2,-1,4,1,2,-1,4...
And so the next value in the given series should be (5-1) = 4.

If you specify few next element of the series then may be i can comment more on that.

-Pradeep

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 28, 2009

Dude if you consider it 5-7 is -2 not 2 .
So i guess the series may not be correct wat say?

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pradeep_agrawal

@pradeep-agrawal-rhdX5z • May 28, 2009

English-ScaredDude if you consider it 5-7 is -2 not 2 .
So i guess the series may not be correct wat say?
Thanks for bringing that up. I did a typo.

Actually i mean that the pattern should be
1,2,-1,4,1,-2,-1,-4...
or something similar.

For the current series i will still go with 4 as guess for next number. But to determine the correct pattern i will need more data as input.

-Pradeep

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 28, 2009

Yep this sounds right , may be we need the next few numbers to determine the sequence 😀

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CEMember

@cemember-PlqlWH • May 28, 2009

RajdeepCEHere is the another series, you dont need to give common equation. So just give the answer,

#->>> 0,1,3,2,6,7,5,__,__,__,__,.....
If I assume that this given series does not have any wrong numbers then after putting my 2 stupid explanations into it, series would be somewhat like following:

0,1,3,2,6,7,5,12,13,14,15,16,11...

or

0,1,3,2,6,7,5,12,13,14,9,...

Answer Please?

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RajdeepCE

@rajdeepce-7UdrG8 • May 29, 2009

Series is perfectly correct, there is no errors in the series. I am giving some more elements, hope this will help you:
#->>> 0,1,3,2,6,7,5,4,12,13,__,__,__,__,....
Now you will have to specify the next four elements of the series. I will give you hint in few days if I dont get the answer.

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RajdeepCE

@rajdeepce-7UdrG8 • May 29, 2009

Here is few another series,
#->>> 1,2,4,9,17,38,__,__,__,....
#->>> 1,2,4,5,7,9,10,__,__,__,...

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RajdeepCE

@rajdeepce-7UdrG8 • May 30, 2009

Here is Weekend Bonanza!!!
Put each number 1 to 9 in the X in order to make the equation correct.

XX/XXX + XX/XX=7

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silverscorpion

@silverscorpion-iJKtdQ • May 30, 2009

Can the X's take any values from 1 to 9 or is there some order or condition to be followed???

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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • May 30, 2009

86/13 + 95/247 = 7

it almost took 45 mins to me to arrive at the answer awesome puzzle 😀

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RajdeepCE

@rajdeepce-7UdrG8 • May 30, 2009

Cheers ES!!!!
But dont forgot to solve the series.
P.S: 3 series are left to solve.
I dont want to give more mathematical puzzles until I get the solution of atleast two series.
Try to get rid of it, first one is tough & demands different logic but remaining can be solved with few logics.

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pradeep_agrawal

@pradeep-agrawal-rhdX5z • May 30, 2009

RajdeepCEI am giving some more elements, hope this will help you:
#->>> 0,1,3,2,6,7,5,4,12,13,__,__,__,__,....
Now you will have to specify the next four elements of the series.
I feel the series follow pattern where the difference from previous number can be represented as:
1,2,-1,4,1,-2,-1,8,1 ,2,-1,-4,1,-2,-1,16...,

The series of differences follow the pattern:
- The difference will always be (+/-)2^n, (n >= 0)

- n is the highest number for which the position of difference is completely divisible by 2^n, e.g., if we consider position 8, 8 is completely by 2^3 = 8. So we put 8 there. Similarly if we consider position 12, 12 is completely divisible by 2^2 = 4 so we put 4 there.

- The +/- is determined from the previous sign of same difference value and the first appearance of any difference value is always positive, e.g., at position 12 we have -4 because the previous appearance of 4 at position 4 was positive. At position 16 we have 16 (positive) because it appears for first time.

So the next four number in the series should be 15,14,10,11.

-Pradeep

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RajdeepCE

@rajdeepce-7UdrG8 • May 30, 2009

@Pradeep, your answer is right but I cant get through the logic you used in series. I have different logic for series. I have little doubt about the integrity of the logic you used. Can you please specify more elements of the series (preferebly 10-12) to check the integrity of the series? I am looking forward for this.

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pradeep_agrawal

@pradeep-agrawal-rhdX5z • May 30, 2009

As per my logic the series of differences will be:

1,2,-1,4,1,-2,-1,8,1,2,-1,-4,1,-2,-1,16,1,2,-1,4,1,-2,-1,-8...

And hence the series of actual numbers will be:
0,1,3,2,6,7,5,4,12,13,15,14,10,11, 9,8,24,25,27,26,30,31,29,28,20...

-Pradeep

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RajdeepCE

@rajdeepce-7UdrG8 • May 30, 2009

Congrats, Pradeep!!!! You are absolutely right.
This series is actually a gray code series. Here it is,
0000=0,
0001=1,
0011=3,
0010=2,
0110=6,
0111=7,and so on...

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RajdeepCE

@rajdeepce-7UdrG8 • May 30, 2009

RajdeepCEHere is few another series,
#->>> 1,2,4,9,17,38,__,__,__,....
#->>> 1,2,4,5,7,9,10,__,__,__,...
Still two series left...

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RajdeepCE

@rajdeepce-7UdrG8 • Jun 1, 2009

Here is another mathematics teaser :
Use only 1, 3, 4, & 6 to derive the number 24. You have to follow some rules,
->> use only standard arithmetic expression
->> use of trignometric and logrethmic functions is not allowed here
->> you can use infinite numbers of bracket
Go CEans!!!!

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silverscorpion

@silverscorpion-iJKtdQ • Jun 2, 2009

Is the use of square root allowed??

If so, then here's the answer..

6 * (sqrt(4)) * (3-1) = 6*2*2 = 24

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CEMember

@cemember-PlqlWH • Jun 2, 2009

If square root is not allowed but exponential is allowed then 😉

1[sup]3[/sup] * 4 * 6 = 24

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RajdeepCE

@rajdeepce-7UdrG8 • Jun 2, 2009

Sorry SS, your answer is right, but we have to strictly use the simple arithmetic operations. Anyway keep on scratching your head, you will definatley get the answer.

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CEMember

@cemember-PlqlWH • Jun 2, 2009

What about my answer RajDeep ? 😔

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RajdeepCE

@rajdeepce-7UdrG8 • Jun 2, 2009

@CEMember, only simple arithmetic operat¨¬on is allowed, you can only use addition(+), subtraction(-), multiplication(*), & division(/), and use lots of brackets() as many as you need.

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