Official Quant thread for CAT 2013

MBA CAT 2024
15123
@getupsid said:
find remainder when 7+77+777+7777+......100 times is divided by 8?? find remainder when 2222^5555 and 5555^2222 is dividived by 7??
1) since the remainder from 8 depends on only last 3 digits .. so from 777 onwards the remainder of all terms will be same as that of 777.
so remainder is {7 + 5 + 98 (1)} mod 8 == 6
15124

@Estallar12 said:
1.)Last non-zero digit of p! = Last non-zero digit of 4^n*(2n)! where n = P/10.So, For 100!=> Last Digit of = 4^10 * 20! = 6 * 4^2 * 4! = 6*16*24 = 96*24 = 4.So, 100! = 100*99*98*97 * 96! = 4=> Last Digit of 4 * 96! = 4So, 96! Last Digit = 6. 2.) What's this ?
an interesting number is a 10 digit number in which all digits are distinct and a multiple of 11111.
15125
@The_Loser said:
kuch samaj nhi aa rha ya.
Here is the method to find the last non zero and the last 2 non zero with example.
R(n!) = Last Digit of [ 2^a x R(a!) x R(b!) ]
where n = 5a + b
Example: What is the rightmost non-zero digit of 25! ?
†' R (25!) = Last Digit of [ 2^5 x R (5!) x R (0!) ]
†' R (25!) = Last Digit of [ 2 x 2 x 1 ] = 4
OR
Last two digits of a fact. (!)
Z(ab) = 44 * Z(ab/5) * Z(b!)
Z(ab/5) = highest multiple of 5 in ab
Z(b!) = last two non-zero digit in b! ... Z(90!) = 44 * Z(18!) * z(0!)
Z(18!) = 44 * Z(3!) * Z(8!) =>Z(18!) = 44 * 6 * 32 = 48
Z(90!) = 44 * 48 * 1 = xx12
NOTE : if ten's digit of 'ab' was even then we will take 66 instead of 44 and if we were asked to find only unit digit then 4 and 6
R(n!) i.e. right-most non-zero digit of n! is found out like this:
Step 1: Express n as 5a+bSay, 37 = 37 = 35 + 2 = 5*7 + 2 => (a, b) = (7, 2)
Step 2:R(n!) = R(2^a) * R(a!) * R(b!) So, for 37 => R(2^7) * R(7!) * R(2!)
Note: We can repeat process for R(a!) if a is on the higher sideHere, R(37!) = R(128 ) * R(5040) * R(2)= 8*4*2 = 64 => 4
15126
@Cat.Aspirant123 said:
three unbiased dice are thrown together, probablity that all 3 show multiple of 3, but do not show the same multiple of 3 ?
336 or 663.
Arrange dose in 3!/2 * 3!/2 ways
= 6
sample space = 6*6*6
requd prob = 1/ 36 ?
15127
@ravi.theja said:
1/36?? "taking that all three cannot have same multiple ,but two can have same multiple"
where its mentioned that two have same multiple
15128
@The_Loser said:
kuch samaj nhi aa rha ya.
He meant to say that for finding Last Digits of any number N! , you need to write N in the form of 5a + b and find the last digits of 2^a * a! * b!

Here, N = 96 = 5*19 + 1
So, a = 19 and b = 1.
Now, Last Digit of ( 2^19 * 19! * 1 ).

2^19 ends in 8 as 19 mod 4 = 3 and 2^3 = 8.
For 19! => 19 = 5*3 + 4. So, b = 4 and a = 3. => LD of 2^3 * 3! * 4! = 8*6*24 = 2
So, LD of ( 2^19 * 19! * 1 ) = 8*2*1 = 6. :)

15129
@getupsid said:
an interesting number is a 10 digit number in which all digits are distinct and a multiple of 11111.
Oh. If it is so then --

Since all are distinct digits => All digits have to be present in the number.

So, Sum of Digits SOD = 45. => SOD or the number is divisible by 9.
Or, By 99999 too.

Let Number be abcdefghij.

abcdefghij = 100000abcde + fghij = 99999abcde + (abcde + fghij)

=> (abcde + fghij) should also be divisible by 99999.

=> abcdef + fghij = 99999 {As 0 or any multiple of 99999 is not possible}.

Thus, a + f = b + g = c + h = d + i = e + j = 9

Now, we have 5 sets which give sum as 9 = (0, 9), (1, 8 ), (2, 7), (3, 6) and (4, 5).

Thus, 9 ways for a, 8 for b, 6 for c, 4 for d and 2 for e.

So, 9*8*6*4 *2 = 3456 such numbers.
15130
@Cat.Aspirant123 said:
where its mentioned that two have same multiple
if its consider 2 can have same multiple den its 1/36 else probability = 0.
15131
@The_Loser said:
kuch samaj nhi aa rha ya.
@The_Loser we have to find the last non zero digit of 96!..
express the given num in the form of 5a + b
for 96 = 5*19 + 1
here the formula for rightmost digit of a number is : unit digit of 2^a* {rightmost digit of a! * rightmost digit of b!}
here a= 19 and b= 1
r(96!) = U(2^19){r(19!)*r(1!)}
r(96!) = 8*(19!)*1) ---- (a)
now 19= 5*3+ 4
r(19!)= U(2^3){r(3!)*r(4!)}
== 8*6*4
==> r(19!)= 2 (unit digit)
putting above in (a)
we get
r(96!)= 8*2*1 = 6
15132
@The_Loser said:
kuch samaj nhi aa rha ya.
@The_Loser we have to find the last non zero digit of 96!..
express the given num in the form of 5a + b
for 96 = 5*19 + 1
here the formula for rightmost digit of a number is : unit digit of 2^a* {rightmost digit of a! * rightmost digit of b!}
here a= 19 and b= 1
r(96!) = U(2^19){r(19!)*r(1!)}
r(96!) = 8*(19!)*1) ---- (a)
now 19= 5*3+ 4
r(19!)= U(2^3){r(3!)*r(4!)}
== 8*6*4
==> r(19!)= 2 (unit digit)
putting above in (a)
we get
r(96!)= 8*2*1 = 6
15133

ab no one is posting any ques, so take an easy one.
dre are 100 articles n1, n2, n3... n100.
They are arranged in all possible order.
hw many arrangements wud be dre in which n28 wud always be before n29.

15134
@getupsid said:
1) find the last non zero digit of 96! ??



76
15135
P, Q and R are in arithmetic progression but not in geometric progression. Their squares are in harmonic progression. 2Q^2 + PR equal ??

@Estallar12 bhai solve dis
15136
@joyjitpal said:
99!
par kuch kami hai abhi bhi.
15137
@The_Loser said:
par kuch kami hai abhi bhi.
100!/2 hoga kya
15138
@joyjitpal said:
100!/2 hoga kya
yes
exactly in half of the cases required arrangements wud be there
15139
@The_Loser said:
yesexactly in half of the cases required arrangements wud be there



pichli bar just before assume kar ke kiye the

15140
@ravi.theja said:
P, Q and R are in arithmetic progression but not in geometric progression. Their squares are in harmonic progression. 2Q^2 + PR equal ??@Estallar12 bhai solve dis
i think [Q(P^2+R^2)]/2 +2Q
15141
@The_Loser said:
ab no one is posting any ques, so take an easy one.dre are 100 articles n1, n2, n3... n100.They are arranged in all possible order.hw many arrangements wud be dre in which n28 wud always be before n29.
100c2*98!
15142
suppose 1st march 1987 ko friday hai.... ko 17th august 1988 ko kya hoga(day)