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A doubt
Remainder when 22^33 /45 ??...help plzz.




Originally Posted by mohasan02 First, sorry for the late posting. Seems you got the answer right !!
Jus one doubt  In the equations for CP & SP, i guess (50+x) is to be replaced by (50+N) b`cuz, I got the same equations
S.P. = [1(20+1.5x/100)][50+N].100
C.P. = [50+N][40+40.(x/50)]
Though on equating, [50+N] will cancel out !!.. But yeah the equations are correct frm d eqn
new cp old cp =increase in productn cost
(50+n)40  2000=x/50 *2000 u get
n= x so,it doesnt mattr whethr u write x or u write n.. i think dats y he has writn 50+x instead of 50+n ..

The Following User Says Thank You to quixotic_me For This Useful Post:

Originally Posted by amrit.pati Remainder when 22^33 /45 ??...help plzz. i think itz 43.. is it correct?


Q 11.
Kindly follow question number from hereon . I have considered numbering all the questions after Q.5 !! Here is a question from my side 
The function "f" is defined on the set of integers and
f(n) = (n4) ; if n>2008
= f{f(n+6)} ; if n<2008
Find f(95)
a. 2003
b. 2007
c. 2005
d. 2006
e. 2004


Originally Posted by mohasan02 Kindly follow question number from hereon . I have considered numbering all the questions after Q.5 !! Here is a question from my side 
The function "f" is defined on the set of integers and
f(n) = (n4) ; if n>2008
= f{f(n+6)} ; if n<2008
Find f(95)
a. 2003
b. 2007
c. 2005
d. 2006
e. 2004 op c.2005
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by amrit.pati Remainder when 22^33 /45 ??...help plzz. Denominator = 5*9
When 22^33 is div by 5 remainder is 2
When 22^33 is div by 9 remainder is 1
So actual rem is the smallest no which gives a rem of 2 when div by 5 & rem of 1 when div by 9.
Hence actual rem is 37
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by shaunak_87 op c.2005 plz post ur solution !


Originally Posted by shaunak_87 op c.2005 .... kindly post your solution. evn i`ll post my method.... Infact, I am not very satisfied with my method. IN any case, it would be great to share different approaches for a problem


My method for Q.11
Originally Posted by mohasan02 Kindly follow question number from hereon . I have considered numbering all the questions after Q.5 !! Here is a question from my side 
The function "f" is defined on the set of integers and
f(n) = (n4) ; if n>2008
= f{f(n+6)} ; if n<2008
Find f(95)
a. 2003
b. 2007
c. 2005
d. 2006
e. 2004 From the given expression, we infer that
f(2009) = 2005
f(2007) = f{f(2013)} = f {2009} = 2005 ....... (i)
f(2005) = f{f(2011)} = f{2007} = 2005 ........ obtained from (i)
f(2003) = f{f(2009)} = f{2005} = 2005
f(2001) = f{f(2007)} = f{2005} = 2005
.
.
.
.
.
f(odd terms) = 2005
therefore f(95) = 2005 

Q12
plz solve ds one ...
Find the highest natural number N less than 400,such that N can be written as sum of consecutive natural numbers in 11 ways ,but cannot be written as sum of 11 consecutive numbers.


Q13
and ds one too..
A number N when divided by divisor D gives a remainder of 52.The number 5N when divided by D gives a remainder of 4.How many values of D are possible?


Hedonist
Originally Posted by quixotic_me and ds one too..
A number N when divided by divisor D gives a remainder of 52.The number 5N when divided by D gives a remainder of 4.How many values of D are possible? a nice Q.... getting 9 values as the answer... so well if its right wld post my method.....


Hedonist
Originally Posted by quixotic_me plz solve ds one ...
Find the highest natural number N less than 400,such that N can be written as sum of consecutive natural numbers in 11 ways ,but cannot be written as sum of 11 consecutive numbers. getting answer as 315 .... nice set of Qs...


Originally Posted by quixotic_me and ds one too..
A number N when divided by divisor D gives a remainder of 52.The number 5N when divided by D gives a remainder of 4.How many values of D are possible? Ans should be 3
D={64,128,256}
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by quixotic_me plz solve ds one ...
Find the highest natural number N less than 400,such that N can be written as sum of consecutive natural numbers in 11 ways ,but cannot be written as sum of 11 consecutive numbers. Getting 315 = 3^2*5*7 as the No.
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by grimreaper a nice Q.... getting 9 values as the answer... so well if its right wld post my method..... @Grim:
D should be >52 and the values of D will be D=64,128,256 i guess.
Pardon me if i'm wrong...
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Virtuoso
Originally Posted by grimreaper Q4) 3*2^p + 5^q + 7^r + 11^s =2008 what is the value of p+q+r+s, where all are greater than 0
a)9
b)10
c)24
d)15
e)unique solution does not exist Originally Posted by shaunak_87 op e)unique solution does not exist
p q r s
0 4 2 3
9 0 3 2 hi guys,
Isn't mentioned that all values are to be greater than zero.
Is there any other method other than trial and error for this question.


Hedonist
Originally Posted by shaunak_87 @Grim:
D should be >52 and the values of D will be D=64,128,256 i guess.
Pardon me if i'm wrong... oh ya,... sigh such mistakes :....
D has to be more than 52....
so 3 such values..... 64 128 256


Hedonist
Originally Posted by Rajasekaran hi guys,
Isn't mentioned that all values are to be greater than zero.
Is there any other method other than trial and error for this question. oh sorry about that.. inlcude 0 too ... greater than equal to zero...
and yes there is a proper method to do this prob.....
try it... wld post the solution then... just that noone tried it at all...apart from shaunak... so didnt post the solution as of yet......
@all  give it a try

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