Number System - Questions & Discussions

MBA CAT 2022
1370
Hi Guys,

This question is posted on the second page of the thread but I have posted here for my doubt.
My way for this question-
9^odd=9 and 9^even=1
so the expression can be written as-
9^!1+9!2............... (Acc. to my mathematics,it shud be 9^1! * 9^2! *....chk this)
=9+1+9+1+9+1......................
=9*500+1*500, as there will be 500 even and 500 odd nos
=....oo
so isnt the last digit zero?


see the question reduces to 9^(1! + 2! + 3!)/4 [As further factorials will be divisible by 4,so no need to consider them)
now 1! + 2! + 3! = 9%4 =1
Hence remainder is 9^1 =9 ...
1371

Someone plz explain ----

gcd of (2^100-1,2^120-1) = (2^20-1)

1372
Someone plz explain ----

gcd of (2^100-1,2^120-1) = (2^20-1)


gcd(a,b)=gcd(a,b-a)
thus
gcd of (2^100-1,2^120-1) = gcd
=>gcd
=>gcd=(2^20 -1)
i hope ...it is clear now bro
1373
gcd(a,b)=gcd(a,b-a)
thus
gcd of (2^100-1,2^120-1) = gcd
=>gcd
=>gcd=(2^20 -1)
i hope ...it is clear now bro


How cum (2^(5x20)-1) became (2^20-1)k ???
It cant be this way dude....

Well, after applying on various examples,i m able to deduce that for HCF and LCM,just take the HCF/LCM of power and u ll get the answer...
In this case,HCF(100,120)= 20...Hence answer is 2^20-1
1374

since
(a^5-1)=(a-1)(a^4 +.............+1)
where a =2^20
i think... i have driven my point home

1375
since
(a^5-1)=(a-1)(a^4 +.............+1)
where a =2^20
i think... i have driven my point home


well, k in most general sense denotes a variable n not a bracket..hence...
1376

Just another way :

we have to find GCD of 2^100-1 and 2^120-1
or we can say we have to find a common Divisor which when divides 2^100 and 2^120.......leaves remainder 1. so that , 1-1=0

this means it will evenly divide 2 ^120 - 2^100

{ example : 2 leaves remainder 1 when it divides 5 and 7 ....but evenly divides 7-5 =2 }

= 2^120 - 2^100
= 2^100(2^20-1)

hence GCD is 2^20-1

1377

himanshu ....
would you mind elaborating this method ?

1378
Someone plz explain ----

gcd of (2^100-1,2^120-1) = (2^20-1)

general rule...
gcd(2^a - 1 , 2^b - 1) = 2^gcd(a,b) - 1
hence
hence gcd(120,100) = 20
hence ans = 2^20 - 1
1379

@hsemu:
please refer this link .....and understand the concept Type#8 :

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

mate here same funda applies ...just question is molded in form that " Find the largest no. that leaves remainder 1 when it divides 2^120 and 2^100"

=2^120-2^100
=2^100(2^20-1)
clearly 2^100 will never leave remainder 1 ..hence its 2^20-1

now since remainder left with 2^120 is 1 .....now 2^120 - 1 will turn out to be 1-1 ..hence 0.
same is the case with 2^100-1 .

same approach i have followed as described by Maxximus sir.
Please Do correct me ..if i am wrong ..as i am new here 😛

1380

thanx puys.....
and yes himanshu u r correct....

1381

Find the sum of all the possible whole number divisors of 720.
can u form a series for this one?

1382

solution:
720= 2^4 * 3^2 *5

sum= (2^0 + 2^1 + ..... + 2^4)*(3^0 + 3^1 + 3^2)*(5^0 + 5^1)
= 93*26
=2418

1383

if 2 ^ 2010 is a 606 digit no with the first digit being 1. then how many no in the following set has 4 as first digit ?
S = { 2 ^ 0, 2 ^1, 2^2,...........,2^2009}

Please also give the procedure applied to solve the question..

Note: Sorry, I am new to this forum. Earlier posted this question as an independent question.

1384
if 2 ^ 2010 is a 606 digit no with the first digit being 1. then how many no in the following set has 4 as first digit ?
S = { 2 ^ 0, 2 ^1, 2^2,...........,2^2009}

Please also give the procedure applied to solve the question..

Note: Sorry, I am new to this forum. Earlier posted this question as an independent question.

see the form :-
2^2=4

2^12=4096

2^22=4....

2^n where n equals 2,12,22,32,.........,2002

so total 201 digits..
please correct me if iam wrong..:)
1385
see the form :-
2^2=4

2^12=4096

2^22=4....

2^n where n equals 2,12,22,32,.........,2002

so total 201 digits..
please correct me if iam wrong..:)


Answer given is 195...
1386

how can i subscribe to this thread?

1387

find the last digit of the lcm of and

1388
notinuse Says
find the last digit of the lcm of and

What's the logic behind this ? Also how can I find the last digit. plz explain
1389
notinuse Says
find the last digit of the lcm of and

LCM = * / 2 --> dividing by 2 becus both are consecutive even nos. hence will have 2 as HCF
to find the last digit:
3^2003 -1 mod100 = 26
3^2003 + 1 mod100 = 28
hence 26*28 / 2 =**364.
hence the last digit shud be 4(if i have not done any mistake)