Official Quant thread for CAT 2013

MBA CAT 2024
13843
@vbhvgupta said:
How many natural number not exceeding 4321 can be formed with the digits 1,2,3 and 4 if the digits can repeat.
3.4^3 + 2.4^2 + 4 + 1 = 229 ? (Considering 4 digit numbers only )
13844
@Budokai001 said:
Editing ...4 Digit nos= 2291xxx -> 64 nos2xxx -> 643xxx-> 6441xx ->1642xx ->16431x ->44321 ->1
shd we have to consider 1,2 and 3 digit no also??

Got it...
13845
@Enigma001 said:
3.4^3 + 2.4^2 + 4 + 1 = 229 ?
Buddy u didnt consider 1,2 and 3 digit nos.....
13846
@vbhvgupta said:
Buddy u didnt consider 1,2 and 3 digit nos.....
yup..
considering 1,2,3 and 4 digit numbers : 4 + 4^2 + 4^3 +229 =313
13847

One from my side

A positive integer is the nth multiple of p (n>1) and its reverse is (n^2-1)th multiple of (p+1).Which of the following is the number ?

a)1008

b)7201

c)4012
d)2004

13848
@Budokai001 said:
One from my sideA positive integer is the nth multiple of p (n>1) and its reverse is (n^2-1)th multiple of (p+1).Which of the following is the number ?a)1008b)7201c)4012d)2004
1008 :P
13849
@Budokai001 said:

One from my side

A positive integer is the nth multiple of p (n>1) and its reverse is (n^2-1)th multiple of (p+1).Which of the following is the number ?

a)1008

b)7201

c)4012d)2004

a)1008
n=8,p=126
13850
@vbhvgupta said:
How many natural number not exceeding 4321 can be formed with the digits 1,2,3 and 4 if the digits can repeat.
All possible numbers=4+4^2+4^3+4^4
Numbers exceeding 4321=3+2*4+4^2=27
So ANS=340-27=313
13851
@vbhvgupta said:
How many natural number not exceeding 4321 can be formed with the digits 1,2,3 and 4 if the digits can repeat.
All possible numbers=4+4^2+4^3+4^4
Numbers exceeding 4321=3+2*4+4^2=27
So ANS=340-27=313
13852
@Budokai001 said:
One from my sideA positive integer is the nth multiple of p (n>1) and its reverse is (n^2-1)th multiple of (p+1).Which of the following is the number ?a)1008b)7201c)4012d)2004
1008?
13853
@bs0409 (2^5002)+1 is divisible by 5.......Am I right???
13854

@vbhvgupta yes it is divisible

13855

(16!+91 ) %323 equals ?

13856
@vbhvgupta said:
@bs0409(2^5002)+1 is divisible by 5.......Am I right???
yes it is divisible by 5
13857
@Budokai001 said:
(16!+91 ) %323 equals ?
5?
13858
@vbhvgupta said:
@bs0409(2^5002)+1 is divisible by 5.......Am I right???
yup
13859
@Budokai001 said:
(16!+91 ) %323 equals ?
Is it 3??
13860
@pirateiim478 will be 5 mate.. post your approach ..will correct it
13861
😛 Edited!!

x+y=8 and P= 5x^2+11y^2 where x,y>0. What is minimum possible value of P?


13862
@Budokai001 said:

(16!+91 ) %323 equals ?

Lets find 16!mod 323
Now 323=17*19
apply CRT
16!mod17=16
We have 18!mod19=18
or, 16! *17 *18 mod 19=-1 mod19
(16!mod19) *(-2)*(-1)= -1 mod19
or, 16!mod19=9
17a+16=19b+9=237.......
ANS=237+91 mod323=5