Number System - Questions & Discussions

MBA CAT 2022
3276

acc to me its OA is 0

3277

@[386572:pathetic]
"Hey actually it is not 12 because as per the solution we wont be considering 69 and 96 so answer is 81-10 = 71"


but why to ignore 69 and 96
plz explain

3278

what is the remainder of 1!+2*2!+3*3!+...........+12*12! when divided by 13??

how to solve such questions which involve factorial??

3279
@challenger90 said: what is the remainder of 1!+2*2!+3*3!+...........+12*12! when divided by 13??how to solve such questions which involve factorial??
The answer, my friend, is to not involve factorials :)

1!+2*2!+3*3!+...........+12*12!
= 1!(1+1) +(2+1)*2!+(3+1)*3!+...........+(12+1)*12! - [1! + 2! + 3!.... + 12!]
= 2*1! + 3*2! + ... 13*12! - [1! + 2! + 3!.... + 12!]
= 2! + 3! + 4! + ..... + 13! - [1! + 2! + 3!.... + 12!]
= 13! - 1!
= 13! -1

R[(13!-1)/13] = -1 or 12 :)
3280
@ashishpai2001 said:
The answer, my friend, is to not involve factorials 1!+2*2!+3*3!+...........+12*12! = 1!(1+1) +(2+1)*2!+(3+1)*3!+...........+(12+1)*12! - [1! + 2! + 3!.... + 12!] = 2*1! + 3*2! + ... 13*12! - [1! + 2! + 3!.... + 12!]
= 2! + 3! + 4! + ..... + 13! - [1! + 2! + 3!.... + 12!] = 13! - 1!
= 13! -1R[(13!-1)/13] = -1 or 12
Instead of what you've done, will this be less complicated?

1!+2*2!+3*3!+...........+12*12! = (2-1)*1! + (3-1)*2! + (4-1)*3! + ... + (13-1)*12!
= 2! - 1! + 3! - 2! + 4! - 3! + ..... 12! - 11! + 13! - 12!

If you see, 2!, -2! are present, 3!, -3! are present and so on...

So, only! 13! - 1!, will be left.
Then just the remainder. Same way.
3281
@Omkarp said:
Instead of what you've done, will this be less complicated?1!+2*2!+3*3!+...........+12*12! = (2-1)*1! + (3-1)*2! + (4-1)*3! + ... + (13-1)*12! = 2! - 1! + 3! - 2! + 4! - 3! + ..... 12! - 11! + 13! - 12!If you see, 2!, -2! are present, 3!, -3! are present and so on...So, only! 13! - 1!, will be left. Then just the remainder. Same way.
In the end, it's completely the same. :)
n*n! = (n+1)! - n!

:D
3282
@ashishpai2001 said:
In the end, it's completely the same.
n*n! = (n+1)! - n!
I know that. I was suggesting a, possibly, cleaner method. :D
3283

hey puys... m new here,

nd i didnt join any coaching for CAT..

i have some question ..

So plzz solve these question and provide me method to solve..

1) find last two digit

101*102*103*197*198*199.

2)find the 28383term of the series. 12345678910111213...............

3)the remainder when the number 12345678910111213...............484950is divided by 16?

4)what is the remainder when 2(8!)-21(6!) divide 14(7!)+14(13!)?

5)find the remainder when 7^99 is divided by 2400?

6)find the remainder when (10^3+9)^752 is divided by 12^3?

3284

Q1 if the terms are from 101 to 199 then the last two digits are 00

if there are only 6 terms then lst two digits are 24
q2 7 ..can be find out by calculating the number upto which the total number of digits is 28383
q3 6 .....the remainder of( last four digits/16 is the final remainder
q4 7 ..can be soved out by taking 6! common
q5 343 as 7^4 =2401 so any 7^4/2400 will give rem as 1 so 7^3 will be the remainder


3285

find the last two terms

a)65*29*37*63*71*87

b)75*35*47*63*71*87*82

Is there any shortcut method to solve above question?

nd plz provide me method to solve these type of question

3286

multiply the last digit of each term ..the last two digits of the resultant will be the lasttwo digits of the whole term ,...u can PM me in case of further explanation

3287
@chaitanya1 said:

hey puys... m new here,

nd i didnt join any coaching for CAT..

i have some question ..

So plzz solve these question and provide me method to solve..

1) find last two digit

101*102*103*197*198*199.

2)find the 28383term of the series. 12345678910111213...............

3)the remainder when the number 12345678910111213...............484950is divided by 16?

4)what is the remainder when 2(8!)-21(6!) divide 14(7!)+14(13!)?

5)find the remainder when 7^99 is divided by 2400?

6)find the remainder when (10^3+9)^752 is divided by 12^3?

1) Should be 64.

Just take remainder by 100, for each term.
=> 1*2*3*-1*-2*-3 = -36 or 64.

2) My take is 3.

we need to find 28383 digits.
1 digit numbers - 9.
2 digit numbers - 90. Hence number of digits = 90*2 = 180.
3 digit numbers - 900. Hence number of digits = 900*3 = 2700.
Total digits till now = 9+180+2700 = 2889.
We need 28383 - 2889 = 25494 more digits.
Hence 25494/4 = 6373*4 + 2.
So, last few digits would be 637263. Ends in 3.

3) Should be 6.

Just take remainder with last 4 digits.
=> 4950 mod 16 = 6.

4) Should be 7!.

2(8!) - 21(6!) = 16(7!) - 3(7!) = 13(7!).
14(7!) + 14(13!) = (1+13)7! + 14(13!) = 7! + 13(7!) + 14(13!). Here the last two terms are perfectly divisible, and the first is not. Hence, 7!.

5) Should be 343.

2400 = 32*3*25.
7^99 mod 32 = 23.
7^99 mod 3 = 1.
7^99 mod 25 = 18.
Hence, remainder would be of the form: 32x+23 = 3y+1 = 25z+18 or 343.

6) Do it the same way.

3288

what is the remainder when (1!)^3+(2!)^3+ (3!)^3+.......+(1152!)^3 is divided by 1152

3289
@chaitanya1 said:what is the remainder when (1!)^3+(2!)^3+ (3!)^3+.......+(1152!)^3 is divided by 1152
is it 225..???
1152=2^7*3^2
onwards 4! every term contains this..
(4!)^3=2^9*3^3
so
remainder is (1!)^3+(2!)^3+(3!)^3
3290
@janvats said: find the remainder when (38^16!)^1777 is divided by 17a. 1 b. 16 c. 8 d. 13
a) 1
3291
@chaitanya1 said:

hey puys... m new here,

nd i didnt join any coaching for CAT..

i have some question ..

So plzz solve these question and provide me method to solve..

1) find last two digit

101*102*103*197*198*199.

2)find the 28383term of the series. 12345678910111213...............

3)the remainder when the number 12345678910111213...............484950is divided by 16?

4)what is the remainder when 2(8!)-21(6!) divide 14(7!)+14(13!)?

5)find the remainder when 7^99 is divided by 2400?

6)find the remainder when (10^3+9)^752 is divided by 12^3?

for Q5) 7^4=2401
(2401^24)x7^3/2400
= 1x343=343 remainder

by using same method for Q6) my take 490
3292

the mean proportion between 2 numbers is 9 and third proportion of 2 numbers is 243.find the larger of 2 numbers??

3293

A 3-digit number has it's first two digit equal.find the maximum possible ratio of number and sum of digits, if number is not divisible by 10??

3294
@challenger90 said: the mean proportion between 2 numbers is 9 and third proportion of 2 numbers is 243.find the larger of 2 numbers??
Answer is 27

3295
@challenger90 said: A 3-digit number has it's first two digit equal.find the maximum possible ratio of number and sum of digits, if number is not divisible by 10??
not getting this question ,please anyone solve this question.