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Originally Posted by Rajasekaran shaunak,
Guess the question should be numbers less than 1000. Try it with this condition.
Yes, that should be ok......
diff of sqrs of 2 consecutive natural nos. is of form (k+1)^{2}  k^{2} = 2k+1 i.e. ODD
Se we are looking at n(n+1)/2 which are odd i.e. neither n nor (n+1) is a mult of 4.
The border case no is 44*45/2 just less than 100.
So from N=2 to N=44, there are total 43 nos. and exactly 11 multiples of 4 within them....and for every multiple of 4, there are 2 such nos such that N(N+1)/2 is even
Also ans should be 4311*2 = 21 CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Hedonist
Originally Posted by Rajasekaran Grim,
I tried the same approach as yours, but its incorrect
Will post solution once others give a try. as i said... am blind..... the Q asks for number lesser than N that dont divide it...... thts why.... we are finding number greater than N.... thats all.....
am blind these days

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Hedonist
Originally Posted by Rajasekaran Q#64:A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural number starting with 1. How many triangular > 1000, have the property that they are the difference of squares of two consecutive natural number?
Try this guys
hmmm.... taking it as lesser than 1000......
triangle number = n(n+1)/2
so we need the case where... n(n+1)/2 = (p+1)^2p^2
that is n(n+1)/2 = 2p+1
now rhs is always odd.... while lhs needs to be odd aswell...
only possible when neither n or n+1 are mult of 4
so for n=44 we get a triangular number lesser than 1000
so all numbers from n=2 to n=44 subtract the mult of 4
that is 11*2 since either n or n+1 can be a multp of 4
so 4322 = 21 

Ques. 68. What is the remainder when 2(8!)21(6!) divides 14(7!) + 14(13!) ?
A. 1
B. 120
C. 720
D. 5040
E. None of these


Ques.69. What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ...
......... + (1152!)^3 is divided by 1152
A. 125
B. 225
C. 325
D. 205
E. 305


Ques. 70. N = 98765432109876543210....1000 digits.What is the least positive value of n such that N + n is divisible by 11
A. 5
B. 6
C. 7
D. 8
E. None of these


Virtuoso
Originally Posted by anjalig Ques. 68. What is the remainder when 2(8!)21(6!) divides 14(7!) + 14(13!) ?
A. 1
B. 120
C. 720
D. 5040
E. None of these option A. reminder will be 1


Virtuoso
Originally Posted by anjalig Ques.69. What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ...
......... + (1152!)^3 is divided by 1152
A. 125
B. 225
C. 325
D. 205
E. 305 option B. 225


Virtuoso
Originally Posted by anjalig Ques. 70. N = 98765432109876543210....1000 digits.What is the least positive value of n such that N + n is divisible by 11
A. 5
B. 6
C. 7
D. 8
E. None of these Option B 6.


Originally Posted by anjalig ques. 68. What is the remainder when 2(8!)21(6!) divides 14(7!) + 14(13!) ?
A. 1
b. 120
c. 720
d. 5040
e. None of these d.5040 CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by anjalig ques.69. What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ...
......... + (1152!)^3 is divided by 1152
a. 125
b. 225
c. 325
d. 205
e. 305 b.225 CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by anjalig Ques. 70. N = 98765432109876543210....1000 digits.What is the least positive value of n such that N + n is divisible by 11
A. 5
B. 6
C. 7
D. 8
E. None of these B.6 CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by anjalig P and Q are two natural numbers, both greater than 1 such that P * Q = 5!.If Q can be represented as xy where (x + y) is minimum and P is not a multiple of x,what is the value of P ?
A. 10
B. 30
C. 40
D. 60
Please explain it Take x, y = 3,1
x+y = 4 (It cant be less as we have to take 1 of dem as 1 and also the other cannot be 2 so its 3)
hence p = 40.
Please let me know if m wrong.
All the best guys..! 

Q 71
Mc donald's ran a campaign in which it gave game cards to its customers.These game cards made it possible for customers to win hamburgers,french fries,soft drinks and other fast food items,as well as cash prizes.Each card had 10 covered spots that could be uncovered by rubbing them with a coin.Beneath three of these spots were "no prizes sign".Beneath the other seven spots were names of prizes,two of which were identical.for ex one card might have two pictures of a hamburger,one picture of a coke,one of french fries,one of a milk shake ,one of $5,one of $1000 and three "no prizes "sign.for this prize the customer could win a hamburger.to win on any card the customer had to uncover the two matching spots before uncovering a "no prize";any card with a "no prize" uncovered was automatically void.Assuming that the two matches and the three "no prize" signs were arranged randomly on the cards what is the probability of a customer winning?


Originally Posted by anjalig What is the remainder when 1 + (11)^11 + (111)^111 + (1111)^1111 + .....
+ (111....111)^111...111
 
10 digits 10 digits
is divided by 100 ?
A. 40
B. 30
C. 10
D. 0 option D
All the best guys..! 

Originally Posted by Rajasekaran 123123......300 digits when divided by 9 gives a remainder 6
When divided by 11 the remainder will be Zero.
So this number can be represented in terms of two AP series,
9A+6 = 11B
Write down the first series values, 6, 15, 24, 33, 42, 51,...
Second series values 11, 22, 33, 44, ....
The first common terms of the series will be remainder. 123123123....300digits = 123 * 100
Now we can easily find the remainder.
All the best guys..! 

Q 72
2  (6407522209/3600840049) is equal to
a) 0.666039
b) 0.666029
c) 0.666009
d) none of these


68. 5040 option D 69. 225 option B 70. 6 option B
Plz let me know if I am correct. Den wil post d soln... All the best guys..! 

Originally Posted by quixotic_me 2  (6407522209/3600840049) is equal to
a) 0.666039
b) 0.666029
c) 0.666009
d) none of these option D
The number inside squareroot is >1 so its squareroot is also > 1. Hence none of theoptions is correct.
All the best guys..! 

Virtuoso
Originally Posted by anjalig Ques. 68. What is the remainder when 2(8!)21(6!) *divides 14(7!) + 14(13!) ?*
A. 1
B. 120
C. 720
D. 5040
E. None of these 14(7!) + 14(13!) = 14*7![1 + 8*9*10*11*12*13]
2(8!)21(6!) = 6![2*7*8  7*3] = 7![2*83]
14*7![1 + 8*9*10*11*12*13]

7![2*83]
7![14 + 8*9*10*11*12*13]

13*7!
Now the second term is completely divisible by 13 and hence the remainder will be 7!
Hope i am correct.

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