Official Quant thread for CAT 2013

MBA CAT 2024
25148
@ChirpiBird said:
total codes = 9*10= 90digits that can cause confusion 0,1,6,8,95C2 = 10ways.and they can be mirrored so .... 10*2 = 20but here 69 when mirrored is 69 only.90-20 +1 = 71 numbers?EDIT : 11, 88 should also be considered just like 69.. total 73?
Bhai only distinct numbers are allowed so 11, 88 shouldn't be there
25149
@albiesriram said:
Consider ordered quadruple of integers (a,b,c,d) such that 1b+c. Find the number of ordered quadruples possible.
b + c 1 + 1 quadruples
1 + 1
1 + 1

for b+ c= 2 , I am getting 90 quadruples

what should be the method ?

@bodhi_vriksha sir ji

25150
@scrabbler said:
I get 135.27, 26, 25, 23, 20, 14 seems to satisfy.regardsscrabbler
sir agar aapke solution me 14 ko 15 se replace kar de..........
it works i think...
which adds upto 136(IMO)
can u plzz check...
25151

@bodhi_vriksha said:
Now that you have, go for it!Team BV - Vineet
i tried to scale it down to 10 integers....
so the solution set that i got was 10 9 8 6...... basically we need to check till the N/2 .... where N is the highest number...
now did the same for original set....
27 26 25 23 20 15.... total 136... none of these...()

25152
@Dexian said:
sir agar aapke solution me 14 ko 15 se replace kar de..........it works i think...which adds upto 136(IMO)can u plzz check...
But 27 + 26 + 15 = 25 + 23 + 20 so two subsets still equal...

regards
scrabbler
25153
@scrabbler said:
But 27 + 26 + 15 = 25 + 23 + 20 so two subsets still equal...regardsscrabbler
I QUIT......
i missed this .......... so if 15 is in my sol set then it can not have 14.. that why i thought ki for once i have found a BUG in ur sol......
25154
@Dexian said:
I QUIT......i missed this .......... so if 15 is in my sol set then it can not have 14.. that why i thought ki for once i have found a BUG in ur sol......
Waisa mera abhi tak confirmed nahin hai 😃

And I make lots of mistakes (ask the people who were here is Sep/Oct, :splat: was my favourite smiley); speed is my forte, not accuracy (especially on the forum, I challenge myself to solve orally first and only write if it is necessary....this was one where it was necessary 😞 :banghead: )

regards
scrabbler
25155
@ScareCrow28 said:
1001 = 7*11*13 3^1001 mod 7 = 53^1001 mod 11 = 33^1001 mod 13 = 9No of the form = 7k + 5 = 11l + 3 = 13m + 9No 971 satisfies! Hence, remainder is 971 ( Edited )
Bhai how to find 3^1001 mod 7 = 5 ?
25156
@dushyantagarwal said:
Bhai how to find 3^1001 mod 7 = 5 ?

Take powers of 3 and check remainders with 7
3 = 3
3^2 = 3 * 3 = 9 = 2
3^3 = 2 * 3 = 6
3^4 = 6 * 3 = 18 = 4
3^5 = 4 * 3 = 12 = 5
3^6 = 5 * 3 = 15 = 1

Thus 3^6 mod 7 = 1.

So write 3^1001 as (3^6)^166 * 3^5 and the remainder will be 1 * 5 = 5

regards
scrabbler
25157
@sa1m007 said:
Can anyone try to solve this cryptarithmetic sum ....... UMA SPY-------- MAMM AYTL*AXUY**-----------ASYLUM
543
678
__________
4344
3801 *
3258**
___________
368154


25158
@amresh_maverick said:
Min value of (x+y^2+z^3) (1/x+1/y^2+1/z^3) for +ve x,y,z
(x+1/x)+(y^2+1/y^2)+(z^3+1/z^3)

Minimum value of (variable)+1/(variable)>=2

=2+2+2=6

25159

@Dexian said:
i tried to scale it down to 10 integers....so the solution set that i got was 10 9 8 6...... basically we need to check till the N/2 .... where N is the highest number...now did the same for original set....27 26 25 23 20 15.... total 136... none of these...()
27+15+26=23+25+20

Team BV - Vineet
25160
@amresh_maverick said:

what should be the method ?@bodhi_vriksha sir ji

sorry the condition should be 1

25161

Q) there are 10 stations between station A and Station B. 4 new stations have been added now. what are the number of ways in which new tickets can be issued considering only 1 way of the journey ?

25162
@ayushbhalotia said:
Q) there are 10 stations between station A and Station B. 4 new stations have been added now. what are the number of ways in which new tickets can be issued considering only 1 way of the journey ?
46?
25163
@ayushbhalotia said:
Q) there are 10 stations between station A and Station B. 4 new stations have been added now. what are the number of ways in which new tickets can be issued considering only 1 way of the journey ?
120 -66 = 54
25164
Q) Figures 0,1,2 3 consist of 1,5 ,13 ,25 non overlapping unit squares, respectively. If the pattern were continued, how many non overlapping unit squares would there be in figure 100?


25165

I went through few articles on using modular arithmetic for Remainder Problems , but none was clear enough..

someone pl help me with this concept , or provide a link where i can go through it.

Thanks

25166
@albiesriram said:
Consider ordered quadruple of integers (a,b,c,d) such that 1b+c. Find the number of ordered quadruples possible.
when , a + d = 20 (1 solution)
for b + c , we have 2 + 3 + 4 ......... 19 = 189 solutions
=> total = 189*1

when , a + d = 19 (2 sol.)
for b + c we have 3 + 4 + 5.......19 = 187 sol.
=> total = 187*2

when , a + d = 18 (3 sol.)
for b + c we have 4 + 5 + 6.......19 = 184 sol.
=> total = 184*3
.
.
.

anybody have easy approach for this one ?
25167
@amresh_maverick said:
120 -66 = 54
@albiesriram said:
46?
nope...