Official Quant thread for CAT 2013

@hiteshkhurana82 said:
abc a three digit number such that a>= b>= >=c. hw many no feasible?
For all digits diff. --> 10C3 cases
2 digits same --> 10C2 * 2
All Digits same--> 9C1 (except 0)

10C3 + 10C2*2 + 9C1 = 120+ 45*2 + 9 = 219
@hiteshkhurana82 said:
abc a three digit number such that a>= b>= >=c. hw many no feasible?
220 ??
@ScareCrow28 said:
198??3^5 - 3*(1+2+3+4+5) = 198
options - 1,2,3,4..
@hiteshkhurana82 said:
a cube is to be painted wid 2 colours, each colour being used for 3 faces of cube. hw many diff cubes painted possible?
2?
@ScareCrow28 said:
220 ??
sir approach. it's OA i dn have
@hiteshkhurana82 said:
options - 1,2,3,4..
Damn!! I misread it as red, blue or green!!
Ans should be 2??
@adwaitjw said:
2?
approach?
@hiteshkhurana82 said:
approach?
Faces with same colour can be either 3 adjacent or 2 opposite faces n one connecting both of them... Only Two possible cases I can see..
@hiteshkhurana82 said:
sir approach. it's OA i dn have
____a_b_c___
For a=9
b= 9, c ranges from 9-0 = 10 values
for b= 8, c ranges from 8-0 = 9 values
........
....
......
for b=0, 1 value

So, Total 1+2+...+10 = 55 values for a=9

Similarly, for a=8, total 1+2+..+9 = 36
So, Total = 1+ (1+2) + (1+2+3) +....+ (1+2+...+10) = 220
@hiteshkhurana82 said:
Using colour red blue & green, hw many diff ways can we paint edges of a regular pentagon sch tat no 2 edges meeting at a common vertex are of same colour ?
Without counting rotations and reflections to be same

Let make edges linear
_ _ _ _ _

Possible combinations,

For R any R,G,B can appear; First & last and consecutive edges can't be same

R _ R _ _ ways=3*2*1*2*1
R _ _ R _ ways=3*2*1*1*2
R B G B G ways 3*1*1*1*1
R G B G B ways 3*1*1*1*1

Add all 30?
@ScareCrow28 said:
____a_b_c___For a=9b= 9, c ranges from 9-0 = 10 valuesfor b= 8, c ranges from 8-0 = 9 values..................for b=0, 1 valueSo, Total 1+2+...+10 = 55 values for a=9Similarly, for a=8, total 1+2+..+9 = 36So, Total = 1+ (1+2) + (1+2+3) +....+ (1+2+...+10) = 220
Underlined one shouldnt be there in total... As it is the case of 000... It wont b a 3 digit no. in that case...
@adwaitjw said:
Underlined one shouldnt be there in total... As it is the case of 000... It wont b a 3 digit no. in that case...
Well. that's something that has always intrigued me. I have seen a few questions where 000 is included as a 3 digit number. I know 0 is not a 3 digit number. As @hiteshkhurana82 said OA is 220, so I wonder what could the reason be.
In how many ways can the number 105 be written as a sum of two or more consecutive positive integers?


If your answer is No. of odd factors -1 i.e 7 then kindly show that 7 No.'s in detail

@ScareCrow28 said:
Well. that's something that has always intrigued me. I have seen a few questions where 000 is included as a 3 digit number. I know 0 is not a 3 digit number. As @hiteshkhurana82 said OA is 220, so I wonder what could the reason be.
I guess he said he dnt hav OA...:P Right @hiteshkhurana82 ? Otherwise Explaination is same in my post also bt diff way... N excluding 000...:P
@Aman.Malhotra said:
In how many ways can the number 105 be written as a sum of two or more consecutive positive integers?If your answer is No. of odd factors -1 i.e 7 then kindly show that 7 No.'s in detail
4?

105 = 3*5*7

105 = 52 + 53
105 = 34 + 35 + 36
105 = 19 + 20 + 21 + 22 + 23
105 = 12 + 13 + 14 + 15 + 16 + 17 + 18
The other series won't be considered because the question asks for two or more consecutive "positive" integers.
@grkkrg said:
4?105 = 3*5*7105 = 52 + 53105 = 34 + 35 + 36105 = 19 + 20 + 21 + 22 + 23105 = 12 + 13 + 14 + 15 + 16 + 17 + 18The other series won't be considered because the question asks for two or more consecutive "positive" integers.
1+2+3+4+5+6+7+8+9+10+11+12+13+14 = 105
@Aman.Malhotra said:
In how many ways can the number 105 be written as a sum of two or more consecutive positive integers?If your answer is No. of odd factors -1 i.e 7 then kindly show that 7 No.'s in detail
52+53
34+35+36
19+20+21+22+23
15+16+17+18+19+20
12+13+14+15+16+17+18
6+7+8+9+10+11+12+13+14+15
1+2+3+4+5+6+7+8+9+10+11+12+13+14



@ScareCrow28 how to get this series 1+2+3+4+5+6+7+8+9+10+11+12+13+14 = 105

by any formula ??
@Aman.Malhotra said:
In how many ways can the number 105 be written as a sum of two or more consecutive positive integers?If your answer is No. of odd factors -1 i.e 7 then kindly show that 7 No.'s in detail
2k+1 = 105--possible

3k+3 = 105---Possible
4k+7 = 105---NOT
5k+11 = 105---Possible
6k+16 = 105---NOT
7k+22 = 105---Possible
8k+29 = 105---NOT
9k+35 = 105---NOT
10k+9..
I won't write others but I have found 7 such conditions
Ignore the typos Confused and tedious to edit....
Method is same..
@Aman.Malhotra said:
@ScareCrow28 how to get this series 1+2+3+4+5+6+7+8+9+10+11+12+13+14 = 105by any formula ??
if an odd number divides it completely then it can be expressed as sum around it

if an even number divides it with q=xxx.5 then it can be expressed as sum around it