Official Quant thread for CAT 2013

MBA CAT 2024
24808
@jain4444 said:


In how many ways can 7^11 be expressed as a product of 3 numbers?
78?
24809
@jain4444 said:
In how many ways can 7^11 be expressed as a product of 3 numbers?

7^11=7^a.7^b.7^c
a+b+c=11 ==>13c2=78 but these are ordered triplets.
so 6x+3y+z=78 where x denotes all three values distinct ,y denotes two values same and z denotes all three same.
here z=0 and y=6(from (0,0) to (5,5) )
so x=10....
hence unordered triplets ==>x+y+z=10+6+0=16 solutions:)

Team BV--Pratik Gauri


24810
@jain4444 said:
In how many ways can 7^11 be expressed as a product of 3 numbers?
11???

24811
@jain4444 said:
hmm did mistake chalo ye try karo A jar is filled with red, white, and blue tokens that are same but their color. The chance of randomly selecting a red token, replacing it, and then randomly selecting a white token is the same as the chance of randomly selecting a blue token. If the number of tokens of every color is a multiple of 3, what is the smallest possible total number of tokens in the jar?via - kamal/vineet sir
is it 18?
again jugaadu method. Probability mein zero hun.

i am taking total tokens as multiples of 3 starting from 9. least toh 9 hoga hi. 3 each wala case.

lets say saare 3 hi hain. total 9.
selecting a red and then white .. 1/3*1/3= 1/9 .. this is not equal to blues.. 1/3
___________________________________________

let's say saare 4 hain. total 12
selecting red and thn white.. 1/3*1/3 =1/9... this si again not equal to blue's 1/3
[isse confirmed evenly distributed nhi hoga.]

or we can make 12 , 3,3,6 or 3,6,3
For 3,3,6slcting red , then white.. 1/4*1/4=1/16.....then blue's 1/2
For 363 ... red,then white 1/4*1/2=1/8 .. then blue's 1/4

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now let's take up 15, ...
arrangments can be 3,6,6 or 6,6,3 or 3,3,9 or 9,3,3

For 3,6,6...red then white.. 1/5*2/5=2/25 ...for blues 2/5
for 6,6,3 ... red then white 2/5*2/5=4/25 for blues 1/5
for 3,3,9 ... red then white 1/5*1/5=1/25 for blues 3/5
for 9,3,3 ... red then white 3/5*1/5=3/25 for blues 1/5

___________________________________________
now let's take 18..
arrangements.. 3,6,9 or 9,6,3 or 3,9,6 or 3,3,12 or, 12,3,3 ..

for 3,6,9 ... red then white 1/6*1/3=1/18 for blues 1/2
for 9,6,3 ... red then white 1/2*1/3=1/6 for blues 1/6 [BINGO!!!]

so 9 red, 6 white and 3 blue.
total 18?

correct me if i am wrong ya if i left some arrangement. :)




24812
@bodhi_vriksha said:
7^11=7^a.7^b.7^ca+b+c=11 ==>13c2=78 but these are ordered triplets.so 6x+3y+z=78 where x denotes all three values distinct ,y denotes two values same and z denotes all three same.here z=0 and y=6(from (0,0) to (5,5) )so x=10....hence unordered triplets ==>x+y+z=10+6+0=16 solutionsTeam BV--Pratik Gauri
Could u pls explain 13c2 and 6x+3y+z=78 ??
24813
@bodhi_vriksha said:
7^11=7^a.7^b.7^ca+b+c=11 ==>13c2=78 but these are ordered triplets.so 6x+3y+z=78 where x denotes all three values distinct ,y denotes two values same and z denotes all three same.here z=0 and y=6(from (0,0) to (5,5) )so x=10....hence unordered triplets ==>x+y+z=10+6+0=16 solutionsTeam BV--Pratik Gauri
if we r not counting 1 as one of the three number than answer would be 11 na ...... forgot to count 1
24814
@Joey_Sharma said:
Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.
a+11b= 13k1
a+13b =11k2
subtract it

2b=11k1 - 13 k2
2b = 11k1 - 11k2 - 2k2
2b = 11x - 2k2 where x = k1-k2
b=11x - 2k2/2
if x=1 11 - 2k can never be even
let x= 2
b = 22- 2k2/2

k2 = 1 b = 10 a is negative

k2 = 2 b =9 a is negative
.
.
.
k2 = 6 b=5 a is 23 (first positive value)
k2=7 b=4 a is 25
so 23 + 5 = 28

second highest 25+4=29

so 28 answer

24815
In how many ways can 7^11 be expressed as a product of 3 numbers?

Please explain in detail .
24816
@Pradeep_Rss said:
There is a certain mixture of milk & water. 12 liters is drawn out from a mixture and replaced with water.after 2 such cycles the ratio of milk:water is 9:7 what is the volume of liquid with which he started .
(v-12/v)^2 = 9/16
v-12/v = 3/4
v=48 l
24817
What is the remainder when 6^83+8^83 is divided by 49 ?
24818
@Asfakul said:
In how many ways can 7^11 be expressed as a product of 3 numbers?Please explain in detail .
x=7^a
y=7^b
z=7^c

a+b+c = 11
total = 13c2 = 78

but when 2 are same -> (0,0,11);(1,1,9);(2,2,7);(3,3,5);(4,4,3);(5,5,1) ->6*3 = 18

=> 78-18 = 60

60/6 = 10

so total cases = 10+6 = 16...
24819
@catahead said:
What is the remainder when 6^83+8^83 is divided by 49 ?
6^84%49 = 1
6*x%49 = 1
x = 41

8^84%49 = 1
8*y%49 = 1
y = 43

(x+y)%49 = (41+43)%49 = 35
24820
@Logrhythm said:
x=7^ay=7^bz=7^ca+b+c = 11total = 13c2 = 78but when 2 are same -> (0,0,11);(1,1,9);(2,2,7);(3,3,5);(4,4,3);(5,5,1) ->6*3 = 18 => 78-18 = 6060/6 = 10so total cases = 10+6 = 16...
Why are you discarding the cases when power of 2 number is the same ? It's not said that the numbers needs to be different ? please explain
24821
@catahead said:
What is the remainder when 6^83+8^83 is divided by 49 ?
6^83 + 8^83 = (7-1)^83 + (7+1)^83 = N
N mod 49 = 2*83*7 mod 49 = 35
24822

solve this

24823
LCM and HCF of two numbers are 11! and 7! respectively. How many pairs of such numbers are possible?
24824
@shattereddream said:
If a bunch of positive integers adds up to 20, what is the greatest possible product of these integers?a. 1024 b. 625 c. 1458 d. 2187
1458..
24825
@The_Loser said:
LCM and HCF of two numbers are 11! and 7! respectively. How many pairs of such numbers are possible?
16 numbers
24826
@hatemonger said:
solve this
option 3..
24827
@catahead said:
What is the remainder when 6^83+8^83 is divided by 49 ?
(7-1)^83 +(7+1)^83 = 1+83*7+83*7-1 MOD 49
==>2*83*7 MOD 49 = 35 MOD 49 :)


Team BV -- Pratik Gauri