Number System - Questions & Discussions

MBA CAT 2022
382
Puys,I m new to Fermat Theorem..But plz help..
Fermat theorem : for any prime number p , (p-1)times,same digit is repeated , then that number formed is exactly divisible by p ..

So if we chk for 9999(4 times),then it shud be exactly divisible by 5..but it is not the case...
Is there any mistake?


fermi little theorem is valid for prime number greater than 5
383
Hello people,

As the earlier number system thread had more than 10k replies,we have closed the old one and presenting to you a part 2 of the same

Please continue with the discussions of Number system problems here,henceforth.

Here is the link to the old thread:

http://www.pagalguy.com/discussions/number-system-25000852

Also,for getting the concepts,here is a good thread for the same:

http://www.pagalguy.com/discussions/conceptstotal-fundas-25023536

the official Quant thread for unsorted queries:

http://www.pagalguy.com/discussions/official-quant-thread-for-cat-09-ii-july-09-onwards-25042674/1605026

Hope that helps.

All the best puys

Hi

This is anosh here. Can some one t]solve the following problems

1) Find the last non zero digit of 96!

a) 2 b) 4 c) 6 d) 8

2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36

Please post the detailed answer(i.e. with the steps used for solving)
384
Hi

This is anosh here. Can some one t]solve the following problems

1) Find the last non zero digit of 96!

a) 2 b) 4 c) 6 d) 8

2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36

Please post the detailed answer(i.e. with the steps used for solving)

1.
last digit = 6
2.
option a)18
1212121212.....300 times mod 9 = 3*150 mod 9 = 0
1212121212.....300 times mod 11 = 2*150 - 150 = 150 mod 11 = 7
11k+7 = 9p
2k+7
k = 1
remainder = 18
there is a direct formula also , not able to recall.
385
Hi

This is anosh here. Can some one t]solve the following problems

1) Find the last non zero digit of 96!

a) 2 b) 4 c) 6 d) 8

2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36

Please post the detailed answer(i.e. with the steps used for solving)


1) ans is (c) 6

note: multiplying the unit digits of 1*2*3*4*6*7*8*9 = 6

multiplying 1st 90 digits excluding 5 multiples = 6*6*(9 times) = 6
from 91~96 excluding 5 multiples = 91*92*93*94*96 = 4
1st 90 digits including only 5 multiples = 1*2*3*4*6*7*...19 = 6
for 25, 50 and 75 = 1*2*3 = 6
hence 6 *4*6*6 = 4
taking 2^18, ends with 4, so 4*x= a4 , hence x= 6 or 11
in base10 rightmost non zero digits ends of n! is 2, 4, 6, 8, where n>1
ans 6
386
Hi

This is anosh here. Can some one t]solve the following problems

1) Find the last non zero digit of 96!

a) 2 b) 4 c) 6 d) 8

Please post the detailed answer(i.e. with the steps used for solving)


The Formula for the last digit is:4^n * (2n)! where n=p/10(p is the number)
Therefore find for 100!:
4^10*(20)!..This comes out to be 4...

Hence 96! * 7 * 8*9=4(Writing all the last digits)
-->96! * 4 =4

Hence 96! could be 1 or 6..
But we eliminate 1 coz there will be enough of 2's to convert the last nonzero digit to an even no..

Hence the answer is 6...
387
1.
2.
option a)18
1212121212.....300 times mod 9 = 3*150 mod 9 = 0
1212121212.....300 times mod 11 = 2*150 - 150 = 150 mod 11 = 7
11k+7 = 9p
2k+7
k = 1
remainder = 18
there is a direct formula also , not able to recall.


Plz light some candles on the underlined part...
388
hi

this is anosh here. Can some one t]solve the following problems

2) what will be the remainder when 1212121212.....300 times is divided by 99?

A)18 b) 81 c) 54 d) 36

please post the detailed answer(i.e. With the steps used for solving)


2. = 12(10^298+10^296+10^294........10^2+10^0) % 99
= 12(1^149+1^148+1^147..............1^1+1) % 99
= (12*150) % 99
= (12*51) % 99
= 612%99
= 18
389

Plz light some candles on the underlined part...

=========================================================
option a)18
1212121212.....300 times mod 9 = 3*150 mod 9 = 0
1212121212.....300 times mod 11 = 2*150 - 150 = 150 mod 11 = 7
11k+7 = 9p
2k+7
k = 1
remainder = 18
there is a direct formula also , not able to recall.
=========================================================


divisibility of 9, a number is divisible by 9 if the sum of its digit is divisible by 9, hence 12=3, since 12 comes 300 times its as though 3 comes 150times, so 3*150 = 450 which is divisible by 9, hence 450 % 9 = 0.

divisibility of 11, a number is divisible by 11 if the difference between the sum of its even digits to odd digits is divsible by 11, here we have 150 (2s) and 150 (1s), so 300 - 150 =150%11 is 7

hope this helps
390
Hi

This is anosh here. Can some one t]solve the following problems

1) Find the last non zero digit of 96!

a) 2 b) 4 c) 6 d) 8

2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36

Please post the detailed answer(i.e. with the steps used for solving)

last non zero digit 96!

prime factors in 96-2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89

96!/2=48,96!/4=24,96!/8= 12 ,96!/16= 6 ,96!/32=3 ,96!/64= 1
96!/3=32,96!/9=10,96!/27=3,96!/81=1
96!/5=19,96!/25=3
96!/7=13,96!/49=1
96!/11=8
96!/13=7
96!/17=5
96!/19=5
96!/23=4
96!/29=3
96!/31=3
96!/37=2
96!/41=2
96!/43=2
96!/47=2
96!/53=1
now after 53 onwards till 89..u get the quotient 1..

now 96!= 2^94*3^46*5^22*7^14*11^8*13^7*17^5*----------------83^1*89^1
2^72*3^46*7^14*11^8*13^7*17^5*----------------83^1*89^1...
any power of 2 dats even will give u last two digits 76

6*9*9*1*7*7*9*1*1*9*9*9*1*1*3*9*7*1*3*9*3*9
=6 ans........ very lenghty :(
391
Hi

This is anosh here. Can some one t]solve the following problems

1) Find the last non zero digit of 96!

a) 2 b) 4 c) 6 d) 8

2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36

Please post the detailed answer(i.e. with the steps used for solving)

What will be the remainder when 1212121212.....300 times is divided by 99??

easiest way to solve dis is

99=100-1or 10^2-1 start making pairs of 2 frm the right end
12+12+12+12+-----------15o times

12*150=1800

1800/99=18 remainder 😃
392

My 1 bit operating system based mind said that......
1) Find the last non zero digit of 96!

a) 2 b) 4 c) 6 d) 8
see 96! = 1.2.3.4......96
now 1.2.3....10 its unit digit will be the unit digit of (2^9).(3^4).7 = unit digit of 2.1.7 = 4
for the terms 11.12...20 again the unit digit will be (2^10).(3^4).13.7.17.19 = unit digit of 4.1.3.7.7.9 i.e. 2....
this is how one can proceed... i'm not giving the answers due to time constraints. sorry for this.........

2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36

Please post the detailed answer(i.e. with the steps used for solving)
What i'm when my mentor failed to solve this. don't worry i'll be back with its solution next time........


"IT'S BETTER TO FAIL IN DOING SOMETHING THAN TO EXCEL IN DOING NOTHING"

393
My 1 bit operating system based mind said that......
1) Find the last non zero digit of 96!

a) 2 b) 4 c) 6 d) 8
see 96! = 1.2.3.4......96
now 1.2.3....10 its unit digit will be the unit digit of (2^9).(3^4).7 = unit digit of 2.1.7 = 4
for the terms 11.12...20 again the unit digit will be (2^10).(3^4).13.7.17.19 = unit digit of 4.1.3.7.7.9 i.e. 2....
this is how one can proceed... i'm not giving the answers due to time constraints. sorry for this.........


well bro..we have a formula for calculating last non-zero unit digit of number which is divisible by 10..
here 90!'s last non-zero unit digit = (4^9) * (18!) ...formula is (4^k)*((2k)!)..where k = number/10..in our case 9...

now 90!'s last non-zero unit digit = 2
96! = 2*91*92*93*94*95*96 = 2*(13*7)*(23*2^2)*(93)*(2*47)*(19*5)*(2^5*3)
= 10 * 2*(13*7)*(23*2)*(93)*(2*47)*(19)*(2^5*3)
= 6....which is the answer....
394


2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36

Please post the detailed answer(i.e. with the steps used for solving)
What i'm when my mentor failed to solve this. don't worry i'll be back with its solution next time........


12121212....300 digits mod 99 = mod 99

now 100 mod 99 = 1....
so 12*100^298 mod 99 = 12
so remainder 12 will come 150 times...so 12*150 mod 99 = 18..wats the answer ??
395
last non zero digit 96!

prime factors in 96-2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89

96!/2=48,96!/4=24,96!/8= 12 ,96!/16= 6 ,96!/32=3 ,96!/64= 1
96!/3=32,96!/9=10,96!/27=3,96!/81=1
96!/5=19,96!/25=3
96!/7=13,96!/49=1
96!/11=8
96!/13=7
96!/17=5
96!/19=5
96!/23=4
96!/29=3
96!/31=3
96!/37=2
96!/41=2
96!/43=2
96!/47=2
96!/53=1
now after 53 onwards till 89..u get the quotient 1..

now 96!= 2^94*3^46*5^22*7^14*11^8*13^7*17^5*----------------83^1*89^1
2^72*3^46*7^14*11^8*13^7*17^5*----------------83^1*89^1...
any power of 2 dats even will give u last two digits 76

6*9*9*1*7*7*9*1*1*9*9*9*1*1*3*9*7*1*3*9*3*9
=6 ans........ very lenghty :(


never go by this method....always remember whenever a number is divisible by 10...it's last non-zero digit can be found out by....(4^k)*((2k)!)....where k = N/10...where N is the number....
396
The prime factor of (2^31) - 1 lying between 200 and 300 is :
a)227 b)223 c)251 d)271
397
well bro..we have a formula for calculating last non-zero unit digit of number which is divisible by 10..
here 90!'s last non-zero unit digit = (4^9) * (18!) ...formula is (4^k)*((2k)!)..where k = number/10..in our case 9...

now 90!'s last non-zero unit digit = 2
96! = 2*91*92*93*94*95*96 = 2*(13*7)*(23*2^2)*(93)*(2*47)*(19*5)*(2^5*3)
= 10 * 2*(13*7)*(23*2)*(93)*(2*47)*(19)*(2^5*3)
= 6....which is the answer....



well i have one ore thing to add along with this --the pattern if u see suggests

1*3*4*6*7*8*9=8
8 will come 9 times=8 will be the unit digit--1
0-10,10-20....till 80-90
now for 90-96 we kan again do the same
1*3*4*6=2(unit digit)--2

kombine the two we get 6 as the unit digit
398

2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36



12 +12 =24
24*75=1800
1800/99=18 remainder ( the method is of naga 25 french--finally i'm using it )

399
The prime factor of (2^31) - 1 lying between 200 and 300 is :
a)227 b)223 c)251 d)271



answer is none of these

(2^31) - 1 by itself is a prime number .. i dont think there can be any prime factors

correct me if am wrong
400
2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36



12 +12 =24
24*75=1800
1800/99=18 remainder ( the method is of naga 25 french--finally i'm using it )



which method r u talkin abt
can u pls explain again?
401

2) What will be the remainder when 1212121212.....300 times is divided by 99?

a)18 b) 81 c) 54 d) 36

well simply divide by 9, we find it is easily divisible by it,
dividing by 11 remainder-7
thru chinese remainder theorm
answer will be 18