CAT 2007: Quantitative Questions a Day 51 to 100 - The Discussions

[Fermions]

Well, here's my solution:

Min LCM :6 d 6 nos. are : 6,6,3,3,3,2
Max LCM:420 d 6 nos. are : 1,2,3,5,5,7,

Positive difference of max n min LCM : M=414
which is 2 x 23 x 3^2
hence no. of factors : 2 x 3 x 2 =12 (say f1)
now M ^2 =(2^2) x (23^2) x (3^4)
hence no. of factors : 3 x 3 x 5 =45 (say f2)

Now no. of factors which divide M^2 but not M = f2-f1 =45-12 =33

Well thts my best bet .I may b wrong.
now lets wait for the actual solution by PG quant Gods...

[OperationBLACKI]

First of all sorry for such a late reply to this post ........Almost time to be replying to post 52

Quite a good sum I should say .......I did it this way

To find the min lcm of six numbers which add upto 23 , we need to start from the bottom and find out which number is fit to be the smallest lcm..Clearly 2 ,3 ,4 and 5 can not be the possible choices , as they dont set themselves as an LCM amongst the six numbers adding upto 23 ......6 becomes the obvious choice as we can have the numbers as 2 3 3 3 6 6 .....So MIN(lcm) = 6

For Max lcm take the set of three prime numbers which can be used , we cant have 11 , 13 and 17 ......neither can we have 7 , 11 , 13..... So the choice that remains is we have to use 3 , 5 and 7 which would increase the LCM and the other 3 numbers adding upto 8 should have highest powers of possible numbers , so we can have 4 , 2 and 2 as the choices for them ......remember we cant use the prime numbers ( 3 , 5 and 7 ) as they have already been used and repeating them would not increase our lcm.....So we go with 3 , 5 , 7 , 4 , 2 and 2 ......

so MAX ( lcm) = 420 ....

Diff M = 420 -6 = 414

No of factors of 414 = 12

No of factors of 414^2 = 45 ....

So the total factors that divides 414^2 but not 414 are 45 -12 = 33

Will try to post my answers before leaving for Office for the rest of the post .......

[peegeekay]

Originally Posted by apocalypse_then

peegeekay = pgk = praveen gopalakrishnan? this post is slightly impertinent to the issue at hand....sorry

Mother of God. I AM famous.

I KNEW IT!!!

So who are you? Some ghost from the NITT past?

Regards,
PeeGeeKay

[varshita]

Q)Find the last three digits of 1233^101 .

[junoonmba]

Originally Posted by varshita

Q)Find the last three digits of 1233^101 .

Last three digits 233

Moi approach......

Last three digits means 1233^101%1000
1000=125*8

1233^101 %8=233^101 %8==Remainder 1 ==>8k+1 type.....1

1233^101%125==-17^101%125
E(125)=100
therefore....remainder -17or 108===>125m+108 type.......2
solve 1 and 2--
answer comes 233.......

[marijuana_user]

Is this QQAD-52???
i doubt this????

[Jaskiran]

Originally Posted by marijuana_user

Is this QQAD-52???
i doubt this????

No its not...
It was already sent to us by mistake on Wednesday.
It is:
There is a set V of five real numbers . V(k) is the sum of all the members of V and V(k^2) is the sum of squares of all the members of V .It is known that V(k)= 8 and V(k^2) = 16. Find the largest possible value that one of the members of V can achieve .

(a)4 (b)11/5 (c)9/2 (d) 11/2 (e)None of the foregoing

[doomsayer]

52:There is a set V of five real numbers . V(k) is the sum of all the members of V and V(k^2) is the sum of squares of all the members of V .It is known that V(k)= 8 and V(k^2) = 16. Find the largest possible value that one of the members of V can achieve .

(a)4 (b)11/5 (c)9/2 (d) 11/2 (e)None of the foregoing

i go for (e)
the value will be 3.xx

[v-factor]

Originally Posted by Jaskiran

There is a set V of five real numbers . V(k) is the sum of all the members of V and V(k^2) is the sum of squares of all the members of V .It is known that V(k)= 8 and V(k^2) = 16. Find the largest possible value that one of the members of V can achieve .

(a)4 (b)11/5 (c)9/2 (d) 11/2 (e)None of the foregoing

ANS is 3 E)None of the above........set has {3,2,1,1,1}
since 3 satisfies the condition option b)is eliminated rightaway.
It can't be 4 or more as second condition wont be satisfied..(so a,c,d are eliminated).

[sastry751]

but i guess vineet and aarav mite have designed one more than that which was sent to us by mistake.it will be better if we have one becoz sat and sunday are holidays