[saloni30]

Originally Posted by warrior

3

last term of (1!+3!+5!+........999!) = 7 just add the 1! and 3! term rest all will have one zero

also last two terms of (1!+3!+5!+........999!) = 47

which when divided by 4 give 3 as the ans

so 7^3 ends in 3

hi,

i understand that from 5! onwards there will be one zero. but why havent u taken 2! and 4!?

i havent quite understood the expalantion you have given... could you be a bit more explicit.

thanks.

[neeraj.acc]

Originally Posted by icarus

Hi

Juz found some number related probs in CAT'99 Paper. Most of the questios are sitters. I dunno if this would help the Junta

1. The difference between the biggest and the smallest number formed by using 0, 1, 2, 3, 4 exactly once is :

(1) 28000 (2) 32591 (3) 27143 (4) 41976

43210-01234

2. If (AB) x (AB) = XYA, then X stands for :

(1) 3 (2) 1 (3) 6 (4) 9


3. A number divided by 899 gives remainder of 63. What will be the remainder when it is divided by 29 ?

(1) 5 (2) 7 (3) 9 (4) 15

Let the no. be x
Then x=899y+63=29*31y+63=29*31y+(29*2+5)
Thus the remainder =5


4. How many numbers can be formed by using 8,7,5,3,2 such that they are divisible by 125 ? Digits to be used exactly once.

(1) 20 (2) 4 (3) 2 (4) 1

For the no. to be divisible by 125 the last three digigits must be divisible by 125
Here the last three digits can be 375 or 875 (i.e in 2 ways). And for both these values the first two digits can be arranges in 2 ways.
Thus the number of req. nos. formes = 2*2=4

5. If n x n x n is odd, then which is true ?
(I) n x n is even
(II) n x n is odd
(III) n is odd

(1) (I) alone (2) (I) and (II) (3) (II) and (III) (4) (II) alone

n x n x n is odd => n is odd (n x n x n can not be odd for even nos.)
and if n is odd then n x n is also odd.

6. The unit's digit in the expansion of 2 raised to power 51 is :

(1) 2 (2) 4 (3) 6 (4) 8

Cyclicity of 2 is 4 i.e after every 4th power of two the unit digit will start repeating.
51=4x12 + 3. Therefore the unit digit of 2^51 will be the same as the unit digit for 2^3. Hence the answer 8

7. Take a 99 digit number. It is created by using first 54 natural numbers. The number thus is 1234567891011.....5354. This number is now divided by 8. The remainder is :

(1) 1 (2) 2 (3) 4 (4) 8
:(

8. How many 5 digit numbers can be made using 1, 2, 3, 4, 5 such that the digit in the unit place is greater than the digit in the hundred's place ?

(1) 40 (2) 120 (3) 60 (4) 720

And one or two more similar, easy questions also appeared.

I am too bad at EU and RC . Can any one help

[warrior]

Originally Posted by neeraj.acc

I am too bad at EU and RC . Can any one help

abe ye EU RC thode hi hai pagale

ise tu hum Qunat kahte hai

[beldcat]

Originally Posted by saloni30

hi,

i understand that from 5! onwards there will be one zero. but why havent u taken 2! and 4!?

i havent quite understood the expalantion you have given... could you be a bit more explicit.

thanks.

its simple ...as u have asked y not 2! and 4 ! just read the quest its written
1!+3!.... and so on...

[anupam_khanna]

@neeraj.acc

7. Take a 99 digit number. It is created by using first 54 natural numbers. The number thus is 1234567891011.....5354. This number is now divided by 8. The remainder is :

(1) 1 (2) 2 (3) 4 (4) 8

the answer for this is 4

we just need to divide the last 3 digits... 354 by 8 to get the remainder...

[nduhan]

In a number system product of 114 and 21 is 2444....what is base ? any method to do it fast ??

[viccy_776]

Originally Posted by anupam_khanna

@neeraj.acc

7. Take a 99 digit number. It is created by using first 54 natural numbers. The number thus is 1234567891011.....5354. This number is now divided by 8. The remainder is :

(1) 1 (2) 2 (3) 4 (4) 8

the answer for this is 4

we just need to divide the last 3 digits... 354 by 8 to get the remainder...

bhaiya 354 ko 8 se divide karoge to 2 bachega

354=8*44+2

[viccy_776]

Originally Posted by nduhan

In a number system product of 114 and 21 is 2444....what is base ? any method to do it fast ??

let the base be x

114 = 1*x^2+1*x+4

21= 2*x+1

product = 2x^3+3x^2+9x+4

comparing with RHS i.e. 2x^3+4^x2+4x+4 u get

3x^2+9x= 4x^2+4x which gives x=5

[warrior]

Originally Posted by nduhan

In a number system product of 114 and 21 is 2444....what is base ? any method to do it fast ??

its 5


simple multiply kar

114
*21
------
114
330
-----
444

now 2*4 will give 3 when base is 5

[esjay]

ive wasted whole of my evening on this one...i know the ans...sm1 pls post the soln
Find the largest integer which definitely divides a^(4*b+1) – a
a. 10 b. 5 c. 20 d.30