The PaGaLGuY UnderDogs Team 2012

MBA CAT 2024
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@allan89 said:
I am not the one to suggest here ...still....attempt sectional tests in QA.. before going into a mock..
Warm uP/Net Practice before the match??? great suggestion....thanks!
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@[522923:allan89]: Havent joined any coaching dis season, any idea how to do dat??

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@Brooklyn said: @allan89: Havent joined any coaching dis season, any idea how to do dat?
try previous years mock papers.. cl or time...and try those.. else buy one , cl test gym or time sectional.. if u can practice even 20 in a span of 1 month, even that would be worth it..
664

answer it

500,1800,100,?,20,50,4
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@knitmaneesh said: answer it
500,1800,100,?,20,50,4
300?
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@[584040:FSOG] how it comes 300...reply soon

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@knitmaneesh said: @FSOG how it comes 300...reply soon
2 series are going on.. first third and fifth term are one fifth of previous and second fourth and sixth are one sixth of previous term...so T4=T2/6 =300
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@[447351:anupam001]thnx


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wat will be next term ....21,38,46,48,96,98,? rply with ans

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@knitmaneesh said: wat will be next term ....21,38,46,48,96,98,? rply with ans
1st 3rd 5th term form pattern of ((mult by 2) + 4)....4th and 6t terms are simply previous term plus 2...7th term should be...96*2 + 4=196...what is the answer ?
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@[447351:anupam001]though i dont hve ans ,me too solved by ur way...

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my problem is:
the first 44 integers r written in order to form the large number N = 123456............424344. what is the remainder when N is divided by 45?

(a) 4 (b) 9 (c) 14 (d) 18
plz whatever conclusion you come to plz let me know!!!!!!!
thanx
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@knitmaneesh said: my problem is: the first 44 integers r written in order to form the large number N = 123456............424344. what is the remainder when N is divided by 45?(a) 4 (b) 9 (c) 14 (d) 18 plz whatever conclusion you come to plz let me know!!!!!!! thanx
Not the ideal thread to discuss problems in plenty 😛

nevertheless solving ..

the given number leaves remainder 4 when divided by 5

the given number is perfectly divisible by 9 since 44*45/9 is divisible by 9

hence final remainder 5x + 4 = 9y

which is 9



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15, 66, 132, 363, 726,?
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@[169132:naga25french].....sir.....the digit sum will be 1+2+......+1+0+1+1+1+2+.....+4+2+4+3+4+4=266 which leaves a remainder of 5.

Now,
5q+4=9p+5
5q-9p=1
q=2,p=1.
ie.14.....
plz correct me if m wrong....
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@[610748:knitmaneesh]....Its the sum of the previous no & its palindromic form.....ie. 15+51.....66+66.....132+231....hence....726+627=1353

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@sid2222000 said: @naga25french.....sir.....the digit sum will be 1+2+......+1+0+1+1+1+2+.....+4+2+4+3+4+4=266 which leaves a remainder of 5.
Now,
5q+4=9p+5
5q-9p=1
q=2,p=1.
ie.14.....
plz correct me if m wrong....
Beauty of 9 is we can be flexible while calculating its divisibility

like say u need to find 123 / 9

1+2+3 ===> 6

12 + 3 = 15 ===> 6

1 + 23 = 24 ===> 6

In the above prob we have first 44 integers ,

sum = 44*45/2 = 22*45 ==> 9k --> perfectly divisible ..

so the number is perfectly divisible by 9

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@[169132:naga25french].....ahh....i missed a 4.....:P

samajh gaya sir....thnx.....i hate these counting sums....thnx fr enlightening me sir.....:)
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@knitmaneesh said: my problem is: the first 44 integers r written in order to form the large number N = 123456............424344. what is the remainder when N is divided by 45?(a) 4 (b) 9 (c) 14 (d) 18 plz whatever conclusion you come to plz let me know!!!!!!! thanx
I prefer playing with the options :mg:

The number when is divided by 5 will give a remainder 4. Hold option (a) (b) (c)
The number is divisible by 9 negate (a) & (c)
Hence (b)
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and the last bastion will fall this season.