Official Quant thread for CAT 2013

MBA CAT 2024
21586
@Calvin4ever said:
nice but is this general or shud it be applied in particular cases alone ??
it can be applied to find the tens digit only provided you know the cyclicity of numbers.
To find the hundred's digit you must either find the remainder when divided by 1000 or use binomial expansion :P
21587
@Calvin4ever said:
Inspite of there being no step marks, pls do tell me the method u used. mine is taking a lot of time !
Mine also long....looked at pattern of first few powers ke last 2...
89, 21, 69, 41...
...isse samjha ki aage ka hoga 49, 61, 29, 81, 09, 01. Matlab 10 steps mein 01 aake repeat hoga. So 90 steps mein bhi 01 aayega and hence 91st mein 89 aayega.

Could also have speeded a bit by realising that 5 steps mein 49 aaya to 49 ka square ends in 01.

regards
scrabbler
21588
@amresh_maverick said:
1> Ten's digit of 24^257 - 13^131 2> 97^97^97 when divided by 11 leaves rem3> X = 1! +2!+3!.....+100! , unit digit of x^xxxx4> rem when 19^92 is divided by 92PS: OA at 2:20
bhai 2:19 hogaye.. OA post kar do
21589
@amresh_maverick said:
1> Ten's digit of 24^257 - 13^131 2> 97^97^97 when divided by 11 leaves rem3> X = 1! +2!+3!.....+100! , unit digit of x^xxxx4> rem when 19^92 is divided by 92PS: OA at 2:20
OA
1> 87
2> 4
3> 3
4> 49
21590
@amresh_maverick said:
OA 1> 872> 43> 34> 49
1> ka sol??
21591
@Dexian said:
1> ka sol??
as usual but a bit lengthy , find last two digit for both

24^257 - 13^131

For 24^257 -------> 24

For 13^131 --------->37

So last two digits for 24^257 - 13^131 ------> 24 - 37 = 87

Hence , 8 is the ten's digit


21592

1> Sum of first 40 terms of 2 +10 + 30 + 68 + 130 .......

2> Rem when 23574^(1^2 +2^2 +3^2 +....+77^2)) is divided by 5

Ps: OA by 2:40

21593
@amresh_maverick said:
1> Ten's digit of 24^257 - 13^131 2> 97^97^97 when divided by 11 leaves rem3> X = 1! +2!+3!.....+100! , unit digit of x^xxxx4> rem when 19^92 is divided by 92PS: OA at 2:15

1> 8 ?
2> 4 ?
3> 3
4> 49

21594
@amresh_maverick said:
as usual but a bit lengthy , find last two digit for both 24^257 - 13^131 For 24^257 -------> 24For 13^131 --------->37So last two digits for 24^257 - 13^131 ------> 24 - 37 = 87Hence , 8 is the ten's digit
bro are you sure 24^257 ka tens digit is 2.. i am getting it as 4
21595
@amresh_maverick said:
1> Sum of first 40 terms of 2 +10 + 30 + 68 + 130 .......2> Rem when 23574^(1^2 +2^2 +3^2 +....+77^2)) is divided by 5Ps: OA by 2:40
2>4
21596
@iLoveTorres said:
bro are you sure 24^257 ka tens digit is 2.. i am getting it as 4
I am solving by another method to see If 24 is correct or not

For last two digits of 24^257 we can divide it by 100

Rem 24^257/100 --> 24 * 24^256 /100 ----> 6 * 24^256/25 --- > 6 * (-1)^256 ----> 6

Rem = 6*4=24
21597
@amresh_maverick said:
1> Sum of first 40 terms of 2 +10 + 30 + 68 + 130 .......2> Rem when 23574^(1^2 +2^2 +3^2 +....+77^2)) is divided by 5Ps: OA by 2:40
1 - 673220
2 - 4

regards
scrabbler
21598
@amresh_maverick said:
1> Sum of first 40 terms of 2 +10 + 30 + 68 + 130 .......2> Rem when 23574^(1^2 +2^2 +3^2 +....+77^2)) is divided by 5Ps: OA by 2:40
1) Common term -> n(n^2 + 1)
now find the sum...

2) 4
sure nahi hun...notepad pe solve kara hai sabb...
21599
@amresh_maverick said:
1> Sum of first 40 terms of 2 +10 + 30 + 68 + 130 .......2> Rem when 23574^(1^2 +2^2 +3^2 +....+77^2)) is divided by 5Ps: OA by 2:40
@scrabbler said:
1 - 673220 2 - 4regardsscrabbler
OA
21600
@Logrhythm said:
1) Common term -> n(n^2 + 1) now find the sum...2) 4 sure nahi hun...notepad pe solve kara hai sabb...
2nd to oral hai....y u use notepad...

regards
scrabbler
21601
@amresh_maverick said:
1> Sum of first 40 terms of 2 +10 + 30 + 68 + 130 .......2> Rem when 23574^(1^2 +2^2 +3^2 +....+77^2)) is divided by 5Ps: OA by 2:40
1> felt tough finding the series itself
2> 4
21602
@Calvin4ever said:
1> felt tough finding the series itself 2> 4
same here
21603
@scrabbler said:
2nd to oral hai....y u use notepad...regardsscrabbler
(-1)^odd :p

strike nahi kara....meine sum nikala...4 se uska remainder liya...fir answer nikala...bad manager i am...wasting time...
21604
@iLoveTorres said:
bro are you sure 24^257 ka tens digit is 2.. i am getting it as 4
24 * 24 = 576. Now next power will be 76*24 ~ -24 *24 ~-76 ~ 24.

So last 2 digits will repeat as 24, 76, 24, 76, 24, 76...so for odd power 257, it will be 24.

regards
scrabbler
21605
@scrabbler said:
2nd to oral hai....y u use notepad...regardsscrabbler
bro 1st wale ka third order difference is same.. how to frame a AP if the third order difference is same?