Number System - Questions & Discussions

MBA CAT 2022
2776
bhow bhow Says
what is the remainder when 44! is divided by 47


according to wilson theorem (n-1)! / n

gives remainder n-1 if n is a prime no .

so

46!/47 remainder = 46

46 * 45! /47 remainder = 46

so 45! /47 remainder is 1

45 * 44! /47 remainder = 1

-2 * 44 ! /47 reminder =1
so

when 44 ! /47 it should be 23

ans is 23
2777

find out the last three digit of 3^1001???

2778
ajeetaryans Says
find out the last three digit of 3^1001???

My take is 003.

1000 = 125*8.
E(8 ) = 4.
=> 1001 mod 4 = 1.
=> 3 mod 8 = 3.
E(125) = 100.
=> 1001 mod 100 = 1.
=> 3 mod 125 = 2.
So, we need to find the smallest 3 digit number which is of the form: 8p+3 = 125k+3, here it is 003 simply.
2779
ajeetaryans Says
find out the last three digit of 3^1001???


answer is 3 by wilson's rule
2780
bhow bhow Says
answer is 3 by wilson's rule


Wilson's rule?? Here??
Can u elaborate on that a little!
2781
Wilson's rule?? Here??
Can u elaborate on that a little!

lol sorry by mistake...its by euler's rule.....i think name of the doesnt matter :-P my bad
2782

hey freinds,
can't find d solution of:
1)what is the reminder when 1923^1924^1925 is divided by 1924?
2)n=7777...(101 digits) .when n is divided by 440,the reminder is?
3)n=171717...(101 digits).when n is divided by 625,the reminder is?

plz help me frends..:banghead:

2783
hey freinds,
can't find d solution of:
1)what is the reminder when 1923^1924^1925 is divided by 1924?
2)n=7777...(101 digits) .when n is divided by 440,the reminder is?
3)n=171717...(101 digits).when n is divided by 625,the reminder is?

plz help me frends..:banghead:


1)1924^1925/1924 rem=0
1923^0=1

2) using crt 8x+1=5y+2=11z+7
on solving 337

3)we need to check the remainder by last four digits so the last four digits are 7171/625
rem=296
2784

thank you..
can u elaborate 2nd problem ..
i am nt aware of how to use crt theorem...:)

2785
thank you..
can u elaborate 2nd problem ..
i am nt aware of how to use crt theorem...:)


k crt is like write 440 as 8 *5*11

and now check rem by each individually so
with 8 its 1 and 5 its 2 and 11its 7
write 8x+1=5y+2=11z+7
so when y=3
x=2
lcm(8,5)m+17=11z+7

solve similarly to find values of z and m ..
it will come 337 ..btw what are the ans
2786
hey freinds,
can't find d solution of:

2)n=7777...(101 digits) .when n is divided by 440,the reminder is?

plz help me frends..:banghead:


2. 440 = 11*8*5

divisibility by 11, up to 100 digit remainder is 0 so remainder is 7.
divisibility by 8, take last 3 digit and get remainder 777 % 8 = 1
by 5 => 2

so we have to find number when divided by 5 gives remainder 2, by 11 gives 7 and by 8 gives 1.

so the number is 337.
2787
mbajamesbond Says
1- Find the number of divisors 544 which are greater than 3.


544 = 2^(5)*17
6*2 = 12
Factors less than 3 would be 1,2. So there are 10 factors greater than 3.


2- Find the number of divisors 544 excluding 1 and 544.

544 has 12 factors, so removing 1 and 544, we have 10 factors.


3- Find the number of divisors 544 which are perfect squares.

4,16 - So just 2.


4- Find the number of zeros in 100^1*99^2*98^3*97^4*...................*1^100.


(20+4)*(19+3)^(2+3+4+5)*18^(some addition) ... Blank, can't think of a good method atm, will come back.


5- Find the max. value of n such that 477x42x37x57x30x90x70x2400x2402x243x343 is perfectly divisible by 21^n.


Count the number of 7s.
477 = Not divisible by 7(No7)
42 = 7*6
37 = No7
57 = No7
30 = No7
90 = No7
70 = 7*10
2400= No7
2402= No7
243 = No7
343 = 7^3
Number of 7s = 5.So, n = 5.*


6- 100! + 200! =

7- 57*60*30*15625*4096*625*875*975=

8- 1!*2!*3!*4!*..................50! =


6. No idea.
7. I can see a 16^3 and some powers of 25, so must be something like that, too lazy. :lookround:
8. No idea


9- Find the remainder when 51^203 is divided by 7?


(51^203)%7 = (2^203)%7.

(2^6)%7 = 1
(2^203)%7 = (2^5)%7 = 4.
So, remainder - 4.


10- Find the remainder when 67^99 is divided by 7?



(67^99)%7 = (4^99)%7.
(4^3)%7=1
So, (4^99)%7 = 1
So, remainder = 1.


11- Find the remainder when 75^80 is divided by 7?



(75^80)%7 = (5^80)%7 = (-2)^80%7 = (2^80)%7 = 4
So, remainder - 4.


12- Find the remainder when 54^124 is divided by 17?



(54^124)%17 = (3^124)%17
9^4 = 6561
3^8 = 6561
6561%17 = -1

(3^124)%17 = (3^4*(-1)^15)%17

= (81*(-1))%17 = -13 = 4.
So, remainder = 4.


13- Find the remainder when 25^102 is divided by 17?


(25^102)%17 = (8^102)%17 = (2^306)%17
(2^4)%17 = -1
Therefore, (2^306)%17 = -4.

So, remainder is 13.


14- Find the units digit in each of the following case-


1^1*2^2*3^3........................*100^100



1+4+7+6+5+6+3+6+9 = 7
So unit's digit is 7.


173^45x152^77x777^999


3*2*3 = 8
So, unit's digit is 8.


82^43*83^44*84^97*86^98*87^105*88^94


8*1*4*6*7*4 = 6
So, unit's digit is 6.


15- Find number of numbers between 300-400 both included that are not divisible by 2,3,4 and 5.


Numbers divisible by 2 = 51.
Numbers divisible by 3 = 34, but remove even multiples so, 17.
Numbers divisible by 4 are included in 2.
Numbers divisible by 5 = 21, but remove multiple of 3, so, 7.
So, total numbers divisible by 2,3,4,5 = 75.
So, total numbers not divisible by 2,3,4,5 = 26.
2788

on dividing a certain no. by 5,7,8 successively ,the reminders are 2,3,4 respectively.What would be reminder if the order of division is reversed?

2789
newavtar Says
on dividing a certain no. by 5,7,8 successively ,the reminders are 2,3,4 respectively.What would be reminder if the order of division is reversed?


5 7 8
2 3 4

the number is smallest such number 157 so when its divided by 8 ,7 ,5 the rem will be 5,5,2
2790

1-Find the arithmetic mean of the AP with 41 terms whose 1st term is 2.5 and common difference is 0.75.

2-Find the 15th term of the sequence 20,15,10.........

2791
1-Find the arithmetic mean of the AP with 41 terms whose 1st term is 2.5 and common difference is 0.75.

2-Find the 15th term of the sequence 20,15,10.........


1.

a = 2.5
d = 0.75
n = 41

AM = {a + a + (n-1)*d} / 2 = {5 + 40*(3/4)}/2 = 35/2 = 17.5

Another way would be to find the value of the 21st term.
In 41 terms, 21st term would be right in the middle, which would be the mean.
If it'd have been an even number, say 40 terms, then take the average of the 20th term and the 21st, but in that case, how I initially solved it would be a better way to go.

2.

a = 20
d = -5
n = 15.

T(n) = a + (n-1)d = 20 - 14*5 = -50.
2792
1.


2.

a = 20
d = -5
n = 15.

T(n) = a + (n-1)d = 20 - 14*5 = -50.

(why d= -5 and not 5 , what if the ap was mix,increasing and dcreasing)
2793

1-Divide 124 into 4 parts which are in AP such that the product of 1st and 4th part is 128 less than the product of the 2nd and 3rd part.

Sol-The basic formulae is same here... but why those 4 numbers cant be a,2a,3a and 4a.

As in book its a-3, a-1, a+1 , a+3

2-The 63rd and 6th terms of an AP are -77 and 37. find the 17th term.

2794

1.Numbers are :- 19,27,35,43
2.17th term :- 15

2795
1-Divide 124 into 4 parts which are in AP such that the product of 1st and 4th part is 128 less than the product of the 2nd and 3rd part.

Sol-The basic formulae is same here... but why those 4 numbers cant be a,2a,3a and 4a.

As in book its a-3, a-1, a+1 , a+3
coz if u wud add the the nos.(a+2a+3a+4a) it will give 10a,which is not multiple of 124.
2-The 63rd and 6th terms of an AP are -77 and 37. find the 17th term.
t63=a+62d
t6=a+5d
d= -2
so a=47
therefore,t17=15:):)