Number System - Questions & Discussions

MBA CAT 2022
482
Last three digits is nothing but finding the remainder when the number is divided by 1000

3^1994 mod 1000= 3^394mod 1000= 3^4 *(3^10)^39 mod 1000

NOw 3^10mod1000 = 049

(3^10)^39 mod 1000 = 49^39 mod 1000 =7^78 mod 1000

Closest multiple of 4 near 78 is 76

7^76 mod 1000

= 2401^19 mod 1000= (2400+1)^19 mod 1000= (1 + 19C1 *2400)mod 1000 = 45601mod1000 =601

So 3^4*7^2*601 mod 1000= 81*49*601 mod 1000 =369



**
I didn't understood this problem !! Can u explain again :shocked:
483
**
I didn't understood this problem !! Can u explain again :shocked:


Pl PM me and tell me the parts you didnt understand.Let us not clutter this thread unnecesarily.
484

i somewhere read this ( trick )

3^1994 mod 1000= 9^997= 9*81^498
now last two digit of 81^498= 1 will be surely the last digit
second last digit will be 8*8=64----------------4

so inal last two digit is 41*9=369

so last two digit will be surely 69

m i wrong somewhere?
pls tell

485

@ viveknitw

3^1994 mod 1000= 9^997= 9*81^498
now last two digit of 81^498= 1 will be surely the last digit
second last digit will be 8*8=64----------------4

so inal last two digit is 41*9=369

so last two digit will be surely 69

m i wrong somewhere?
pls tell


dere's no flaw in ur method...69 are the last two digits....

486

how can i download any thread?

487
sagartule Says
how can i download any thread?

:) how can we download a thread..???but yes we can download the attachments in it..if u are new to PG..kindly refer PAGAL VISUAL GUIDE...
488
Mani_23 Says
:) how can we download a thread..???but yes we can download the attachments in it..if u are new to PG..kindly refer PAGAL VISUAL GUIDE...



well u can download the thread and view it later but it will take some time.
dowload idm.
then integrate it with ur browser.
then download the thread page wise.
if u want the full thread which has 100 pages , u have to download 100 times each page.
it will take time and later u can these pages saved as html.

(note - it will take a lot of time though )
489

is the answer is 0??

u can try this way

take the ap as

4,2,0,-2........


thus t(1) *1=4*1=4
t(2)*2=4
hence the t(3) is 0

490
guys hav a question here: Progressions.

If a times ath term of an AP equals b times bth term. find the (a+b)th term.

- What i wanna ask is, is there a non-algebraic way of cracking this? A short-cut... u go thru the formulae, things work out fine.. but takes a lil time...

Thanks much!




is the answer is 0??

u can try this way

take the ap as

4,2,0,-2........


thus t(1) *1=4*1=4
t(2)*2=4
hence the t(3) is 0
491

what is the remainder when 28! is divided by 29 and 39! is divided by 41? plzz explain the answer for the same.

492
destiny1400 Says
what is the remainder when 28! is divided by 29 and 39! is divided by 41? plzz explain the answer for the same.


28! mod 29 =-1~28

39!mod41 =1
493
destiny1400 Says
what is the remainder when 28! is divided by 29 and 39! is divided by 41? plzz explain the answer for the same.



by wilsons theorm

(p-1)!+1/p= remainder 0
thus 28!+1/29= 0
so -1 is the remainder
thus 29-1= 28 reaminder

similarly
41!/39 = remainder 1
494
There are 8436 steel balls, each with a radius of 1 cm, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 balls in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is :

a) 34 b) 38 c) 32 d) 36 e) 40

Someone explain this problem
495
sachy24 Says
Someone explain this problem

Balls in row 1 = 1
Balls in row 2 = 3 = 1 + 2
Balls in row 3 = 6 = 1 + 2 + 3

So, balls in row n , T(n) = summation of natural numbers upto n = /2

total sum = summation of T(n) within limits
496
sachy24 Says
Someone explain this problem


1+3+6+10+...=8436

Summation(n(n+1)/2) = 8436

(1/2)(Summation(n^2+n) = 8436

Just solve it and find the answer.:shocked:
Or use options.
497
GOOD Morning :-P
1. Find 28383rd term in series 1234567891011..
a.3 b.4 c.9 d.7
2. sum of 1/1*5+1/5*9+1/9*13+..+1/221*225
a.28/221 b.56/221, c.56/225 d. none
3. sum of 1/(2^1/2+1^1/2)+1/(2^1/2+3^1/2)+..+1/(120^1/2+121^1/2)
a. 10 b.11 c.12 d.none
4. sum of infinite term 1/1+1/3+1/6+1/10+1/15.
a.2 b.2.25 c.3 d.4
5. sum of +1/6+1/12+1/20+..+1/156+1/182
6.how many 3 digit no:have the property tht their digit taken from left to right form an AP or GP
498
2. sum of 1/1*5+1/5*9+1/9*13+..+1/221*225

ans-56/ 225
499
4. sum of infinite term 1/1+1/3+1/6+1/10+1/15.
a.2 b.2.25 c.3 d.4

is ans - a.2 ?
500
5. sum of +1/6+1/12+1/20+..+1/156+1/182

select the one near to 1(sum )
501
GOOD Morning :-P
1. Find 28383rd term in series 1234567891011..
a.3 b.4 c.9 d.7
2. sum of 1/1*5+1/5*9+1/9*13+..+1/221*225
a.28/221 b.56/221, c.56/225 d. none
3. sum of 1/(2^1/2+1^1/2)+1/(2^1/2+3^1/2)+..+1/(120^1/2+121^1/2)
a. 10 b.11 c.12 d.none
4. sum of infinite term 1/1+1/3+1/6+1/10+1/15.
a.2 b.2.25 c.3 d.4
5. sum of +1/6+1/12+1/20+..+1/156+1/182
6.how many 3 digit no:have the property tht their digit taken from left to right form an AP or GP


answer of q1 is d) 7
q2 is 56/225
q4 is 2