can u explain how did u get 1 as remainder?
2) What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(13!)?
a) 1 b)7! c)8! d)9!
My Answer:
The remainder will be 1 !
{14(7!) + 14(13!)} / {2(8!) -21(6!)}
take common 7! in both part and cancel and then remaining part is
{ 14 +(14*13*12*11*10*9*8 )} / 13
remainder will be 1
but we cancel 7!
so multiply result by 7!
so answer is 7!
pls provde detailed solution for these questions
1)How many four-digit numbers can be formed using the digits 1, 2, 4, 6, 7 and 9 such that each number is divisible by 3 but not by 9? (Repetition of digits is not allowed) ?
2)What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 11!?
3)Find the reminders when 10^ 2 + 11^ 2 + 12^ 2 + . . . . 28 ^2 is divided by 19
4)The average of 25 distinct positive integers is 25. What is the greatest possible value of any one of these 25 numbers?
5)What is the remainder when 25 x (1331)2196 is divisible by 13?
:
:Q1 :->How many four-digit numbers can be formed using the digits 1, 2, 4, 6, 7 and 9 such that each number is divisible by 3 but not by 9? (Repetition of digits is not allowed) ?
select 4 digits from 1,2,4,6,7 and 9 such that no is divisible by 3 but not 9
there two possible cases
case I :->
2,6,7 ,9
sum of digits 24(which is divisible by 3 but 9 ) and no's
form through these digits are 4! = 24
case II :- 1,4,7,9 sum of digits (1+4+7+9 ) = 21
so again total no 4! = 24
so total 24+24 = 48 no's
2)What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 11!?
ans will be
1!+2!+ .............+10! are not divisible by 11! so remainder is also this
so if i want to total
remainder
1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880 + 3628800
3)Find the reminders when 10^ 2 + 11^ 2 + 12^ 2 + . . . . 28 ^2 is divided by 19
remainder will be 0
= 10^ 2 + 11^ 2 + 12^ 2 + . . . . 28 ^2
= {1^2 + 2^2 +3^2 .................+28^2} - {1^2 + 2^2 ......... + 9^2}
={28 *29*57}/6 - {9*10 *19}/6
= 19{14*29 -15}
and this a multiple of 19 so ans is zero
4)The average of 25 distinct positive integers is 25. What is the greatest possible value of any one of these 25 numbers?
1+2+ ..........+24 = 24*25/2
= 12 *25
= 300
so total is 625
maximum possible value of 25th element = 325
5)What is the remainder when 25 x (1331)^2196 is divisible by 13?
25/13 = remainder =(-1)
1331 /13 remainder is = 5
(1331)^2196/13
or
( 5 ) ^ 2196/13
or
(25)^ 1098 /13
(-1) ^ 1098 = 1
(-1) *1= -1
so remainder will be 12
Block 1 - Number system
Q1-For the no. 7200 find..
A-The sum and no. of factors divisible by 150
B-The sum and no. of factors divisible by 15
C-The sum and no. of factors not divisible by 75
D-The sum and no. of factors not divisible by 24
Q2-Find the No. of divisors 544 which are greater than 3?
Q3-Find the No. of divisors 544 excluding 1 and 544?
Q4-Find the No. of divisors 544 which are perfect squares?
Q5-Find the number of zeroes in
100^1* 99^2* 98^3 * 97^4.....................* 1^100
Q1-For the no. 7200 find..
A-The sum and no. of factors divisible by 150
B-The sum and no. of factors divisible by 15
C-The sum and no. of factors not divisible by 75
D-The sum and no. of factors not divisible by 24
Q2-Find the No. of divisors 544 which are greater than 3?
Q3-Find the No. of divisors 544 excluding 1 and 544?
Q4-Find the No. of divisors 544 which are perfect squares?
Q5-Find the number of zeroes in
100^1* 99^2* 98^3 * 97^4.....................* 1^100
Block 1 - Number system
Q1-For the no. 7200 find..
A-The sum and no. of factors divisible by 150
B-The sum and no. of factors divisible by 15
C-The sum and no. of factors not divisible by 75
D-The sum and no. of factors not divisible by 24
Q2-Find the No. of divisors 544 which are greater than 3?
Q3-Find the No. of divisors 544 excluding 1 and 544?
Q4-Find the No. of divisors 544 which are perfect squares?
Q5-Find the number of zeroes in
100^1* 99^2* 98^3 * 97^4.....................* 1^100
(1) 7200 = 2^5*3^2*5^2.
(A) 150 = 2*3*5^2.
So, take 150 common in 7200 -> 2*3*5^2 (2^4*3).
=> Sum of factors = {(2^5 - 1)/1}*{(3^2 - 1)/2} = 124.
(B) For 15, just take 3*5 common.
(C) For not divisible by 75, take sum of all factors and then subtract the sum of factors of those factors which are divisible by 75.
(D) Follow same procedure as above.
(2) 544 = 2^5*17.
Total factors = 6*2 = 12.
Factors So, factors > 3 = 12 - 3 = 9.
(3) Would be 10. (total factors - 2).
(4) Perfect squares = Power/2 + 1 = (5/2 + 1)*(1/2 + 1) = 3.
What is Reminder for 50^50^50^50^50/9
What is Reminder for 50^50^50^50^50/9??
Jaggu3107 SaysWhat is Reminder for 50^50^50^50^50/9??
Can anybody solve this through binomial theorem
2^99 divided by 99 (it can be solved easily by chinese remainder)
but in interview i was asked to solve it by binomial expansion
Hey Classpart can u tell me how to solve it by chinese remainder theorem???
Jaggu3107 Sayscan tell me how to solve it by chinese remainder theorem???
By cr method answer would be
6.4 1027
1- Find the number of divisors 544 which are greater than 3.
2- Find the number of divisors 544 excluding 1 and 544.
3- Find the number of divisors 544 which are perfect squares.
4- Find the number of zeros in 100^1*99^2*98^3*97^4*...................*1^100.
5- Find the max. value of n such that 477x42x37x57x30x90x70x2400x2402x243x343 is perfectly divisible by 21^n.
6- 100! + 200! =
7- 57*60*30*15625*4096*625*875*975=
8- 1!*2!*3!*4!*..................50! =
9- Find the remainder when 51^203 is divided by 7?
10- Find the remainder when 67^99 is divided by 7?
11- Find the remainder when 75^80 is divided by 7?
12- Find the remainder when 54^124 is divided by 17?
13- Find the remainder when 25^102 is divided by 17?
14- Find the units digit in each of the following case-
1^1*2^2*3^3........................*100^100
173^45x152^77x777^999
82^43*83^44*84^97*86^98*87^105*88^94
15- Find number of numbers between 300-400 both included that are not divisible by 2,3,4 and 5.
2- Find the number of divisors 544 excluding 1 and 544.
3- Find the number of divisors 544 which are perfect squares.
4- Find the number of zeros in 100^1*99^2*98^3*97^4*...................*1^100.
5- Find the max. value of n such that 477x42x37x57x30x90x70x2400x2402x243x343 is perfectly divisible by 21^n.
6- 100! + 200! =
7- 57*60*30*15625*4096*625*875*975=
8- 1!*2!*3!*4!*..................50! =
9- Find the remainder when 51^203 is divided by 7?
10- Find the remainder when 67^99 is divided by 7?
11- Find the remainder when 75^80 is divided by 7?
12- Find the remainder when 54^124 is divided by 17?
13- Find the remainder when 25^102 is divided by 17?
14- Find the units digit in each of the following case-
1^1*2^2*3^3........................*100^100
173^45x152^77x777^999
82^43*83^44*84^97*86^98*87^105*88^94
15- Find number of numbers between 300-400 both included that are not divisible by 2,3,4 and 5.
im getting ans for 3rd ques as 0..dnt knw...if wrong plz reply
can anyone teach me how to mod of a number????plz
what is the remainder when 44! is divided by 47
bhow bhow Sayswhat is the remainder when 44! is divided by 47
3...Though it will be negative
mbajamesbond Says3...Though it will be negative
please post how have you calculated this?
bhow bhow Sayswhat is the remainder when 44! is divided by 47
Abhinav3 Saysplease post how have you calculated this?
This is negative remainder ...Where the denominator is larger than the denominator we can use it....
If you divide 44/47 You get a remainder of -3(Because 47 is 3 lesser than 44)
If the numerator was 48 then then answer the would have been 1(48/47)
mbajamesbond Says3...Though it will be negative
dnt u think it should be 23?
Jaggu3107 SaysWhat is Reminder for 50^50^50^50^50/9??
reminder is 5.
Split this is as (5*10)^50^50^50^50
=> 10 divided by 9 gives reminder 1 and 5 divided by 9 gives reminder 5
so reminder is 5.
bhow bhow Sayswhat is the remainder when 44! is divided by 47
It should be 23
(n-2)! leaves remainder as 1 when divided by n if n is a prime number.
=> 45! leaves remainder by 47 as 1
=> 44! * 45 leaves remainder by 47 as 1
Let remainder of 44! be x and remainder of 45 = -2
=> -2*x leaves remainder of 1 by 47
=> x = -24 as -2*-24 = 48 => 1
=> Remainder of 44! by 47 = -24 = 23 .... 47-24=23