Number System - Questions & Discussions

MBA CAT 2022
61
divishth Says
sorry for spamming...corrected my solution already...:



hmm

wht abt the next one ?

Remainder of (41!)^2 when divided by 83... ??
62
docsmile Says
Yes .Can u explain?

Because in 11!, you get the numbers 9, 10 and 11, whose product is 990 😃 And since 11 is a prime no., there can't be any no. lesser than 11! in this case.
63
docsmile Says
Yes .Can u explain?


990 can be written as (9*10*11) = (2* 3^2 *5*11)

so 11 would be least number in which all numbers would be included....
64
1) For which of the following functions of f is f(x)=f(1-x) for all x?

OA is f(x)=x^2(1-x)^2
Can some1 explain how?

2) If n is positive integer and product of all integers from 1 to n inclusive is multiple of 990 ,then what is the least possible value of n?

For the 1st question, it's pretty easy if you just put 1-x in place of x and check yourself.

Original: f(x)=x^2(1-x)^2
Substituting x with (1-x): f(1-x) = (1-x)^2(1-1+x)^2 = (1-x)^2.x^2

Thus, f(x) = f(1-x)
65

1)distribution of test scores for a group of trainees is as follows.what score interval contains median of 73 scores?
score interval are 50-59,60-69,70-79,80-89,90-99
corresponding number of scores for each interval are 2,10,16,27,18

66
1)distribution of test scores for a group of trainees is as follows.what score interval contains median of 73 scores?
score interval are 50-59,60-69,70-79,80-89,90-99
corresponding number of scores for each interval are 2,10,16,27,18

i think the interval 80-89....correct me if i am wrong
67
Simple number problems ..

2. when 41!^2 divided by 83.


Thanks


We know tht--frm Wilson's Theorem
(p-1)!%p = (p-1) or -1(iff p is prime)
so, we can deduce
82!%83=-1 or 82---------------(1)
frm Wilson's Corollary
(p-2)!%p = 1
so, we can deduce here
81!%83=1-----------------(2)
Now 81!=81.80.79.-------.42.41!----------(3)
from 2 and 3
(81.80.79.--------.42.41!)%83=1
-2*-3*-4*-5....-41*41! = 41!*41!%83 = (41!)^2%83 = 1
68
after 5!,,each term will end with 0..so we'll add 1!+2!+3!+4!= 1+2+6+24=33
5789^33/10
(5780+9)^33/10
9^33/10=9 remainder....


In above solution:
1!+2!+3!+4!= 1+2+6+24=33
But (5!+6!+7!+...+1000!) will have 0 as unit digit, I understand easily.

But you are taking it granted that (5!+6!+7!+...+1000!) will have 0 also TENS DIGIT so that (1!+2!+.....+1000!) has 33 as last 2 digits.

My doubt is how (5!+6!+7!+...+1000!) has 0 as TENS DIGIT?
69
divishth Says
i think the interval 80-89....correct me if i am wrong

Correct.plz explain....
70
There is a 20 X 20 array. In Each row , the tallest person is called and among them, the tallest person is A. In Each column, the shortest person is called and among them, the shortest person is B. Who is taller
71
If its 1!+2!+...+1000! then the ten's digit is 0.From 10! onwards no number contributes to 10's digit.
If its 5!+6!+7!+.....+1000! then the ten's digit is 8.


If 10! onwards all the terms DOES NOT CONTRIBUTE to TENS DIGIT then how (1!+2!+3!+....+1000!) will have TENS DIGIT as 0?

as (1!+2!+....+9!) = 409113 has TENS DIGIT as 1.
72
In above solution:
1!+2!+3!+4!= 1+2+6+24=33
But (5!+6!+7!+...+1000!) will have 0 as unit digit, I understand easily.
But you are taking it granted that (5!+6!+7!+...+1000!) will have 0 also TENS DIGIT so that (1!+2!+.....+1000!) has 33 as last 2 digits.
My doubt is how (5!+6!+7!+...+1000!) has 0 as TENS DIGIT?


From 10! onwards the ten's digit is 0
whereas we can see that
5!=120,6!=720,7!=5040,8!=40320 9!=362880
so 5!+6!+7!+8!+9! = 20+20+40+20+80=80(last two digits)
therefore tens digit shd be 8 and not 0.

Correct me if i am wrong
73
docsmile Says
Correct.plz explain....


common sense....maximum number of elements jismein...woh answer..

mathematical solution
here we have...(2*50 + 10*60 + 16*70 + 27*80 + 18*90)/73

would give approx > 80
so answer 80-89
74
From 10! onwards the ten's digit is 0
whereas we can see that
5!=120,6!=720,7!=5040,8!=40320 9!=362880
so 5!+6!+7!+8!+9! = 20+20+40+20+80=80(last two digits)
therefore tens digit shd be 8 and not 0.

Correct me if i am wrong


5!+6!+7!+8!+9! = 20+20+40+20+80=80

you are adding from 5! TO 9!...but forgot to add ten's digit of 1! to 4!..which is 33
so 80+33 = x13.........so ten's digit = 1...and not 8..
75
From 10! onwards the ten's digit is 0
whereas we can see that
5!=120,6!=720,7!=5040,8!=40320 9!=362880
so 5!+6!+7!+8!+9! = 20+20+40+20+80=80(last two digits)
therefore tens digit shd be 8 and not 0.

Correct me if i am wrong


To clear all doubts let us find:
What is the TENS digit of (1!+2!+3!+....+1000!)?
76

It is nothing but 1
1!+...4! = 33
5!+........+9! = 80
33+80=....13. so 1 is tens digit.

77
To clear all doubts let us find:
What is the TENS digit of (1!+2!+3!+....+1000!)?


It's 1...for all practical purposes...
78

series 22 , 34 , 48 , ? , 79, 96
what is d ans.
fast
.

79
hmm

wht abt the next one ?

Remainder of (41!)^2 when divided by 83... ??

The answer is (41!)^2 mod 83 = 1
81! mod 83 = 1
81*80*79*78*....*42*41! mod 83 = 1
=> -2*-3*-4*....*-41*41! mod 83 = 1
=> (41!)^2 mod 83 = 1
80
The answer is (41!)^2 mod 83 = 1
81! mod 83 = 1
81*80*79*78*....*42*41! mod 83 = 1
=> -2*-3*-4*....*-41*41! mod 83 = 1
=> (41!)^2 mod 83 = 1

@Ashu,
plz explain this part: "-2*-3*-4*....*-41*41! mod 83 = 1"

What happened to (-2*-3*-4*-5*.....*-1) part (here -1 is extracted from -41)?