Number System - Questions & Discussions

MBA CAT 2022
2956
For these type of questions first of all find the prime factors of 15! which are 2,3,5,7,11 & 13

=2^5 is the ans
no ways of writing a no as a product of two nos is=2^(n-1)
where n is no of prime factors

sirji ,, product of two prime there should not be anythng common ... hence cant we do it by selecting one prime i mean there are 6 prime factor set selecting 1 individual which is

6c1
then 2
6c2 .
then 3
6c3/2 .. because of repetition of same set hence

then ...it will 6+ 15+20/2= 31 ??
2957

1) Remainder of 9^36^36^24 / 88
2) Sum of digits of 5^(1!*2!*3!*4!......*13!)



I have a doubt in the above question...
plz solve this question???

2958
sirji ,, product of two prime there should not be anythng common ... hence cant we do it by selecting one prime i mean there are 6 prime factor set selecting 1 individual which is

6c1
then 2
6c2 .
then 3
6c3/2 .. because of repetition of same set hence

then ...it will 6+ 15+20/2= 31 ??


ya , i think it can be solved this way also. Only thing is we have to add 1 to the final answer to take into account the product 1*15.

Also, can you explain why did u divide 6C3 by 2 ? Which sets will repeat? I didnt get this part.
2959

Find the number of zeros in 100^1 * 99 ^2 * ..... * 1^101

2960
For these type of questions first of all find the prime factors of 15! which are 2,3,5,7,11 & 13

=2^5 is the ans
no ways of writing a no as a product of two nos is=2^(n-1)
where n is no of prime factors

Can you please explain the part in red 2^5 means the value of n =6 is it because there are 6 prime numbers before 15! ???
2961
1) Remainder of 9^36^36^24 / 88
2) Sum of digits of 5^(1!*2!*3!*4!......*13!)



I have a doubt in the above question...
plz solve this question???


1) Should be 9.
88 = 8*11.
Remainder with 8 will be 1.
9^36^36^24 mod 11.
E(11) = 10.
36^36^24 mod 10 = 6.
=> 9^6 mod 11 = (-2)^6 mod 11 = 64 mod 11 = -2 or 9.
Hence the remainder would be of the form -> 8k+1 = 11p+9. So, 9.

2) This was an Aimcat question, and it isn't "sum of digits". It would be "digital sum". And for that, take remainder with 9.
E(9) = 6.
(1! + 2! + 3! + .... + 13!) mod 6 = 1+2 = 3.
=> 5^3 mod 9 = -1 or 8.
2962
Tyranitar Says
Find the number of zeros in 100^1 * 99 ^2 * ..... * 1^101


Calculate the number of 5s in this expression.
The terms which will yield 5 are :
100^1, 95^6 , 90^11 , 85^16 , 80^21, 75^26, 70^31, 65^36, 60^41 , 55^46, 50^51, 45^56, 40^61, 35^66, 30^71, 25^76, 20^81, 15^86, 10^91, 5^96.

Note that the sum of the number and its power is 101 in each case.
Also, the number in Bold will result in two 5s.(Think why?)

So, total number of 5s = total no. of 0s = (1+1)+6+11+16+21+(26+26)+31+36+41+46+(51+51)+56+61+66+71+(76+76)+81+86+91+96 = 1124

Is it the answer. Pls confirm.
2963

guys plz help me out with these questions:
Q1how many divisors of 22400 are of the form 4n+3?
Q2how many perfect squares are there whose ten's digit =unit digit?
PS:I want to know the method and not the answer for these problems..

2964
guys plz help me out with these questions:
Q1how many divisors of 22400 are of the form 4n+3?
Q2how many perfect squares are there whose ten's digit =unit digit?
PS:I want to know the method and not the answer for these problems..

Hi,
For Q1
When it speaks of divisors the 1st thing that comes to my mind is Factors->Factorization
22400= 2^7 x 5^2 x 7
Now a little introspection tells me (keep aside 2^7 now) that 7, 35,175 are in form 4n+3 ....Do the following multiplications 7x5, 7x25 etc.
Now we can't take 2's power more than 1 else it'll be iv. by 4 so check 7x2, 35x2,25x2 etc.
After all this I came up with 3 values 7,35,175
Q2...Will think

Regards,
Never Back Down
2965
guys plz help me out with these questions:
Q1how many divisors of 22400 are of the form 4n+3?
Q2how many perfect squares are there whose ten's digit =unit digit?
PS:I want to know the method and not the answer for these problems..


22400 = 2^7 * 7 * 5^2....

divisors=(1,2,4,8,16,32,64,12(1,7)(1,5,25)

divisors of the form 4n+3= (7)(1,5,25)= 7*5,7*35, 7*1 =>3 values

question is incomplete unless u mention the range
1st
such square is 10^2 and then 12^2=144.

subsequent squares with 00 and 44 will occur at mutilpes of 10, viz 20,30,...etc
and ending with 44 = (50k+-12) , viz 38,62,88,112,138,162....so on
2966

hey guys please help me with this question
how many zeros will be there at the end for the expression:
1*2^2*3^3*4^4..............100^100

2967
hey guys please help me with this question
how many zeros will be there at the end for the expression:
1*2^2*3^3*4^4..............100^100


Is the answer 1300???.........
2968
incognito7 Says
Is the answer 1300???.........

1300 it is.....
Concept:- Just check powers of 5 at 25,50,75,100.....

Regards,
Never Back Down
2969
guys plz help me out with these questions:
Q1how many divisors of 22400 are of the form 4n+3?
Q2how many perfect squares are there whose ten's digit =unit digit?
PS:I want to know the method and not the answer for these problems..


1) 3.
22400 = 2^7*5^2*7.
7*1, 7*5, 7*25 would be of 4n+3 form.

2) CBD.
We need an outer limit for such cases.
But you can memorize that Squares of numbers ending in 0 would always have their units and tens digit same,ie, 0.
And, squares of number of the form 50k+/-12 will always end in 44.
2970
But you can memorize that Squares of numbers ending in 0 would always have their units and tens digit same,ie, 0.
And, squares of number of the form 50k+/-12 will always end in 44.

Seriously Bhai,
Where do you get these Fundas...

Regards,
Naver Back Down

2971
Seriously Bhai,
Where do you get these Fundas...

Regards,
Naver Back Down



Bhai no where special. PG.

Another concept -> Squares of numbers of the form 500k +/- 38 will always end in 444. :).
These are not axioms. They can be logically proven with binomial expansion.
2972

hey thnx...but i didnt understood E(11) = 10 and E(9)=6????????

hw it cm can u plz tell me

2973
hey thnx...but i didnt understood E(11) = 10 and E(9)=6????????

hw it cm can u plz tell me


11's euler number is 10, E(11) = 10.
and E(9) = 6.

Read the post on this link and you will understand. :)

http://www.pagalguy.com/forum/quantitative-questions-and-answers/55747-cat-2010-concepts-fundas-tips-13.html#post2629058
2974

hey thnx yaar....
bt i didnt understood the solution given to the second question y u took ----- take remainder with 9????
can u plz explain....

and by using the logic for the first question like i am tryin to solve the question 30^72^87/11 but i am geting wrong answer....i am answer as 9....but in arun sharma its given as 5
question is on page number 52(arun sharma)

2975
hey thnx yaar....
bt i didnt understood the solution given to the second question y u took ----- take remainder with 9????
can u plz explain....

and by using the logic for the first question like i am tryin to solve the question 30^72^87/11 but i am geting wrong answer....i am answer as 9....but in arun sharma its given as 5
question is on page number 52(arun sharma)


Yar to find the digital sum we take remainder with 9.


30^72^87 mod 11.
E(11) = 10.
=> 72^87 mod 10 = 2^87 mod 10 = 8.
=> 30^8 mod 11 = (-3)^8 mod 11 = 9^4 mod 11 = (-2)^4 mod 11 = 16 mod 11 = 5.

5 hi ayega yaar.