Brainy Puzzles

Puzzle Solvers: Interesting brainy puzzles shared by fellow puzzlers

304 Members

Join this group to post and comment.

# Brain Teasers : Mathemetics

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.

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.

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 ..

π

Nice puzzle Rajdeep π

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 ..

π

Nice puzzle Rajdeep π

Nice explanation,ES!!!!

Congrats!!!

Looks like I have to find more difficult puzzles for you.

Congrats!!!

Looks like I have to find more difficult puzzles for you.

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.

complete the following series,

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

fill all 3 blanks with proper explanation.

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

Good Puzzle will try π

Good Puzzle will try π

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.

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.

Hey CEans, try this simple one series :

-> 1,1,2,1,3,2,5,3,8,___,___,___,...

-> 1,1,2,1,3,2,5,3,8,___,___,___,...

Are the next numbers 5 , 13 , 8 ,21

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

1+1=2

2-1=1

and so on

**5,13,8**1+1=2

2-1=1

and so on

Both are right!!!

Congrats Godfather & ES!!!

But what about the first one, it is quite hard, isnt it?

Congrats Godfather & ES!!!

But what about the first one, it is quite hard, isnt it?

Rajdeep are you sure this series is correct...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.

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

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.

@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.

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,__,__,__,...

#-> 0,1,2,5,10,19,34,59,__,__,__,...

π i am fast than GF π

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

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

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

Actually i spent my some time to formatting post πEnglish-Scaredπ i am fast than GF π

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

Yep you are sandy π

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,

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,

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.. ππ

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.. ππ

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

Cheers!!

Cheers!!

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 ??

oops looks like Scorpion already posted the answer π

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

π

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

π

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.

@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.

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.

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.

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

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.

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.

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.

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

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

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

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

Pradeep you are lightining fast excellent .

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

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

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

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

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

Dude if you consider it 5-7 is -2 not 2 .

So i guess the series may not be correct wat say?

So i guess the series may not be correct wat say?

Thanks for bringing that up. I did a typo.English-ScaredDude if you consider it 5-7 is -2 not 2 .

So i guess the series may not be correct wat say?

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

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

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

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

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?

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.

#->>> 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.

Here is few another series,

#->>> 1,2,4,9,17,38,__,__,__,....

#->>> 1,2,4,5,7,9,10,__,__,__,...

#->>> 1,2,4,9,17,38,__,__,__,....

#->>> 1,2,4,5,7,9,10,__,__,__,...

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

Put each number 1 to 9 in the X in order to make the equation correct.

XX/XXX + XX/XX=7

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

86/13 + 95/247 = 7

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

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

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.

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.

I feel the series follow pattern where the difference from previous number can be represented as: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.

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

@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.

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

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

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...

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...

Still two series left...RajdeepCEHere is few another series,

#->>> 1,2,4,9,17,38,__,__,__,....

#->>> 1,2,4,5,7,9,10,__,__,__,...

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!!!!

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!!!!

Is the use of square root allowed??

If so, then here's the answer..

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

If so, then here's the answer..

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

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

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

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

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.

What about my answer RajDeep ? π

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

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

Are you sure we are supposed to exactly one 1,3,4,6 or can we use multiple values like two 1 and rest only once ?

No, we have to use all four numbers exactly one time.

Solving this..

Answer of Series 2 is

Am I right?

**1,2,4,5,7,9,10,12,14,16,17,19,21,23,25,...****One Odd,Two Even,Three Odd,Four even,Five odd... and So On....**Am I right?

Yes GF!!! you struck it right!!!

Congrats.!!!

To all CEans, only two teasers left to solve. Hurry up!

Congrats.!!!

To all CEans, only two teasers left to solve. Hurry up!

6/(1-(3/4))

is 24 ππ

is 24 ππ

Congrats ES!!

Still one series left...

I will post new puzzle after it is solved.

Still one series left...

I will post new puzzle after it is solved.

Modulus (%) is Arithmetic operatorRajdeepCEHere 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!!!!

(3%1)+(4*6)=24

@Sheeko, see the answer given by English-Scared, it will clear your doubt.

Still one series is unsolved...

Still one series is unsolved...

*->>> 1,2,4,9,17,38,__,__,__,...

(1*2)+0=2

(2*2)-0=4

(4*2)+1=9

(9*2)-1=17

(17*2)+4=38

(38*2)-4=72

(72*2)+9=153

(153*2)-9=297

1 ,2 ,4 ,9 ,17 ,38 ,72 ,153 ,297

(2*2)-0=4

(4*2)+1=9

(9*2)-1=17

(17*2)+4=38

(38*2)-4=72

(72*2)+9=153

(153*2)-9=297

1 ,2 ,4 ,9 ,17 ,38 ,72 ,153 ,297

Great Sheeko, well done but you discovered the whole new series. So I have to give some elements to resolve the ambiguity error. Here it is,

#->>> 1,2,4,9,17,38,72,161,305,___,___,___,....All the best this time...

(1*2)+0=2

(2*2)-0=4

(4*2)+1=9

(9*2)-1=17

(17*2)+4=38

(38*2)-4=72

(72*2)+17=161

(161*2)-17=305

(161*2)+72=394

(394*2)-72=716

1 , 2 , 4 , 9 , 17 , 38 , 72 , 161 , 305 , 394 , 716

(2*2)-0=4

(4*2)+1=9

(9*2)-1=17

(17*2)+4=38

(38*2)-4=72

(72*2)+17=161

(161*2)-17=305

(161*2)+72=394

(394*2)-72=716

1 , 2 , 4 , 9 , 17 , 38 , 72 , 161 , 305 , 394 , 716

@Sheeko, sorry buddy your series is wrong.

The next element should be 682.

The next element should be 682.

(1*2)+0=2

(2*2)-0=4

(4*2)+1=9

(9*2)-1=17

(17*2)+4=38

(38*2)-4=72

(72*2)+17=161

(161*2)-17=305

(305*2)+72=682

(682*2)-72=1292

1 , 2 , 4 , 9 , 17 , 38 , 72 , 161 , 305 , 682 , 1292

please tell me that i'm right and this isn't a new seriesπ

(2*2)-0=4

(4*2)+1=9

(9*2)-1=17

(17*2)+4=38

(38*2)-4=72

(72*2)+17=161

(161*2)-17=305

(305*2)+72=682

(682*2)-72=1292

1 , 2 , 4 , 9 , 17 , 38 , 72 , 161 , 305 , 682 , 1292

please tell me that i'm right and this isn't a new seriesπ

At last you got it!!!

Congratulation!!

Here is the new series,

All the best.

Congratulation!!

Here is the new series,

#->>> 1,4,27,256,___,___,___,....This time you have to specify the common equation of series too, cause its damn easy to find the next number but for finding common equation, you have to mess with your mind.

#->>> 2,8,7,28,___,___,___,....

#->>> 2,8,27,85,___,___,___,....

All the best.

1) The first series is n[SUP]n[/SUP].

so, the next terms are 3125, 46656, etc..

2) The second series is,

2+(n-1)*5/2 if n is odd,and

(n-1)*4 if n is even.

So, the next terms are 12,48,17,68, etc..

3) I can't find the common equation for this. I managed to find the next terms in the series.. They came to be

208, 422..

Let me know if they are correct..

so, the next terms are 3125, 46656, etc..

2) The second series is,

2+(n-1)*5/2 if n is odd,and

(n-1)*4 if n is even.

So, the next terms are 12,48,17,68, etc..

3) I can't find the common equation for this. I managed to find the next terms in the series.. They came to be

208, 422..

Let me know if they are correct..

Only 1st series is right. Other two series were wrong.

2,8,7,28,12,48,17,68,...

this is 2 and 2*4 =8 and 2+5 =7 and 7*4 = 28 format am i correct??

this is 2 and 2*4 =8 and 2+5 =7 and 7*4 = 28 format am i correct??

Sorry ES, your logic is also wrong. Anyone else?

Where are the brilliant & crazy minds of CE????????

* bumping the thread *

Guys still two series are remained unsolved, why dont you try your brain on this. Here it is,

Guys still two series are remained unsolved, why dont you try your brain on this. Here it is,

#->>> 2,8,7,28,27,__,__,__,...C'mon CEans, give it a tryπ

#->>> 2,8,27,85,__,__,__,...

Any series grows how you want to make it grow...

My try for first one...

2, 8, 7, 28, 27, 108, 107, 428........

Solution: 2*4 =8 -1 =7*4 =28-1 =27*4 =108-1 =107*4 =428..... and so on...

-CB

My try for first one...

2, 8, 7, 28, 27, 108, 107, 428........

Solution: 2*4 =8 -1 =7*4 =28-1 =27*4 =108-1 =107*4 =428..... and so on...

-CB

I also thought of the same solution πcrazyboyMy try for first one...

2, 8, 7, 28, 27, 108, 107, 428........

I feel the second series should go as:

2, 8, 27, 85, 260, 786, 2365 ...

2 * 3 + 2 = 8

8 * 3 + 3 = 27

27 * 3 + 4 = 85

85 * 3 + 5 = 260

260 * 3 + 6 = 786

786 * 3 + 7 = 2365

-Pradeep

Yeah!! You both are right.

π

congrats crazyboy and pradeep!!!

π

congrats crazyboy and pradeep!!!