CrazyBoy's Crazy series...

Anil Jain

@CrazyBoy • Oct 22, 2024

Oct 22, 2024

1.4K

If

23 = 15

54 = 36

83 = 33

Then

43 = ?

-CB

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RajdeepCE

@rajdeepce-7UdrG8 • Jul 1, 2009

The next number is 14.
The explanation behind this is that,
23 = (2+3)*5 = 15
54 = (5+4)*4 = 36
83 =(8+3)*3 = 33
so 43 = (4+3)*2 = 14

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Anil Jain

@CrazyBoy • Jul 1, 2009

RajdeepCE
The next number is 14.
The explanation behind this is that,
23 = (2+3)*5 = 15
so 43 = (4+3)*2 = 14

Its wrong...
(2+3)*5=25 😁

Also answer is wrong..

-CB
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RajdeepCE

@rajdeepce-7UdrG8 • Jul 1, 2009

Oops. I have done huge mathematical mistake in hurry.
Here is another try,
23 = (2+3)*3 = 15,
54 = (5+4)*4 = 36,
83 = (8+3)*3 = 33,
43 = (4+3)*4 = 28.
Let me know that this time I am right or not?

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safwan

@safwan-NH7W5Y • Jul 1, 2009

😕😕:?:
understood the serees of rajdeep but not of CB.
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shalini_goel14

@shalini-goel14-ASmC2J • Jul 1, 2009

crazyboy
If

23 = 15

54 = 36

83 = 33

Then

43 = ?

-CB
One trial from my side also
(2+3) * 3[last digit]=15
(5+4) * 4 =36
(8+3) * 3= 33

so 43 will be (4+3) * 3 =21

Am I right?
0
Anil Jain

@CrazyBoy • Jul 1, 2009

shalini_goel14
One trial from my side also
(2+3) * 3[last digit]=15
(5+4) * 4 =36
(8+3) * 3= 33

so 43 will be (4+3) * 3 =21

Am I right?
You got it right Shalini... , Have you taken Idea from Rajdeep's solution... Be true???

@ Rajdeep - I hope you noticed your error again.

@ Safan - Do you need explanation Now???

-CB
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Anil Jain

@CrazyBoy • Jul 1, 2009

How about this...

3 12 55 240 ??? ???

-CB
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ms_cs

@ms-cs-Ab8svl • Jul 1, 2009

whether it is 1200 4800 ?
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Anil Jain

@CrazyBoy • Jul 1, 2009

whether it is 1200 4800 ?
Wrong... .
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swapnakumar

@swapnakumar-lJkNkk • Jul 1, 2009

is it 987????
0
RajdeepCE

@rajdeepce-7UdrG8 • Jul 1, 2009

3,12,55,240,999,4060,...
Let me know, is it right or not?
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Anil Jain

@CrazyBoy • Jul 1, 2009

RajdeepCE
3,12,55,240,999,4060,...
Let me know, is it right or not?
Thats right...

Please explain this as welll....

-CB
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RajdeepCE

@rajdeepce-7UdrG8 • Jul 1, 2009

Ok, here is explanation,
3
(3*4)+0=12,
(12*4)+7=55,
(55*4)+20=240,
(240*4)+39=999,
(999*4)+64=4060,
The addition series 0,7,20,39,64,... is derived as follows,
0,
0+7=7,
7+7+6=20,
20+7+6+6=39,
39+7+6+6+6=64...and so on.
Hope you all understand the series now.
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Anil Jain

@CrazyBoy • Jul 2, 2009

😀 Good try...

Easier and another method to solve would be..

4^1-1^2 = 3
4^2-2^2 = 12
4^3-3^2 = 55

And so on.....

4^5-5^2 =
4^6-6^2 =

Answers 😀

-CB
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RajdeepCE

@rajdeepce-7UdrG8 • Jul 2, 2009

One doubt here CB,
4^5-5^5 and 4^6-6^6 both are negative as per me. What you say?
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Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • Jul 2, 2009

Hey the 4th element would be 4POW4 - 4pow4 =0 , not 240.

next elements will be negative .
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CEMember

@cemember-PlqlWH • Jul 2, 2009

RajdeepCE
One doubt here CB,
4^5-5^5 and 4^6-6^6 both are negative as per me. What you say?
There is nothing wrong in the explanation, just CB Sir has typed it wrong , it shoud be like following

4[sup]1[/sup]-1*1 = 3
4[sup]2[/sup]-2*2 = 12
4[sup]3[/sup]-3*3 = 55
4[sup]4[/sup]-4*4 = 240
4[sup]5[/sup]-5*5= 999
4[sup]6[/sup]-6*6 = 4060

And so on.....
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Anil Jain

@CrazyBoy • Jul 2, 2009

Updated my last post, I was supposed to write:

1^2, 2^2, 3^2, 4^2, 5^2, 6^2

-CB
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