Number System - Questions & Discussions

help me in solving the following sum:

a natural number N when divided by 10 leaves a remainder of R and quotient of Q. the number N is equal to the sum of the product QR and the sum(Q+R). If N is a multiple of 9 and is less than 10000, ho0w many values can N assume?

a)110
b)112
c)999
d)1000


is it 112
Let N=10Q + R
N = QR + Q +R

Therefore

10Q+R = QR +Q+R
Therefore 9Q = QR
R = 9
Put
Q = 0,1,2......

N = 9,19,29,.......9999
= 1000 different value.

Is the answer Correct?

how can N be 19,29,.... it has to be a multiple of 9..........
is it 112

yup, u r correct once agn...do post the solution...
catwoman_09 Says
how can N be 19,29,.... it has to be a multiple of 9..........


Yes this is wrong I didnot consider the case that N is a multiple of 9
help me in solving the following sum:

a natural number N when divided by 10 leaves a remainder of R and quotient of Q. the number N is equal to the sum of the product QR and the sum(Q+R). If N is a multiple of 9 and is less than 10000, ho0w many values can N assume?

a)110
b)112
c)999
d)1000


my explanation
10Q+R = QR +Q+R
9Q = QR
R = 9
so d values of N can be of d form 9*(10k+1)
k can be 0,1,2,.....111
hence number of diff values of N=112.......

If the product of four positive integers is 10! What is the smallest possible value their sum can have?

a. 175
b. 176
c. 180
d. 181

I am getting 175 but it took too much time with hit and trial method. Any short method for this.

In a national hockey single elimination tournament, 303 teams are participating. How many games will be played before a team becomes the national champion?

a. 152
b. 302
c. 77
d. 303
In a national hockey single elimination tournament, 303 teams are participating. How many games will be played before a team becomes the national champion?

a. 152
b. 302
c. 77
d. 303



it should be b) 302

sample it for smaller value

u get if n team plays , n-1 matches are played.. so 302

the answer should be 302
to eliminate 302 team you need 302 matches

If the product of four positive integers is 10! What is the smallest possible value their sum can have?

a. 175
b. 176
c. 180
d. 181

I am getting 175 but it took too much time with hit and trial method. Any short method for this.

a+b+c+d >=45(approx)
numbers have to be around 44
40,42,45,48 = 175
used hit and trial only.
it should be b) 302

sample it for smaller value

u get if n team plays , n-1 matches are played.. so 302

dont think we need smaller samples...302 matches are required for 302 teams to loose ...:-P
If the product of four positive integers is 10! What is the smallest possible value their sum can have?

a. 175
b. 176
c. 180
d. 181

I am getting 175 but it took too much time with hit and trial method. Any short method for this.


45^4 = 2025^2 = 41xxxxxx > 10 !

so there is one possibility where in sum is less than 180

this rule out option 3 and 4

now write 10 ! = 6* 4 * 5 * 6 * 7 * 8 * 9 * 10

make this expression as product of four numbers in 40's

40 * 42 * 45 * 48


so sum = 175

i cant term this hit and trial .. dunno what method to term
In a national hockey single elimination tournament, 303 teams are participating. How many games will be played before a team becomes the national champion?

a. 152
b. 302
c. 77
d. 303


less than one to be a champion

b.302

Hello Aspirants ,,,

This is my 1st post as mee not comfortable with NumberSystem... Please help me out..... I have started my journey so help me outttt


1 : What is the remainder when 32^32^32 is divided by 9??

Hello Aspirants ,,,

This is my 1st post as mee not comfortable with NumberSystem... Please help me out..... I have started my journey so help me outttt


1 : What is the remainder when 32^32^32 is divided by 9??


answer is 4......
refer total gadha it helped me alot
Hello Aspirants ,,,

This is my 1st post as mee not comfortable with NumberSystem... Please help me out..... I have started my journey so help me outttt


1 : What is the remainder when 32^32^32 is divided by 9??

its 4
(32^32^32)/9
= ((125^10^32).(25^32))/9
= ((8^10^32).(7^32))/9
= ((64^5^32).(49^16))/9
= ((1^5^32).(4^16))/9
= ((1).(4^15)(4))/9
= ((64^5)(4))/9
4

2nd method is by Euler :32^32^32/9

sol- using Euler theorem:-

32 n 9 are co prime

Euler number of 9=6
6*5=30
32^30=1mod9

thus,
32^32^32
=(32^2)^32
=(5^2)^32 (using rem theorem)
=25^32
=7^2 (using rem theorem as well eulers theorem)
=49

49=4mod9

hence rem=4
In a national hockey single elimination tournament, 303 teams are participating. How many games will be played before a team becomes the national champion?

a. 152
b. 302
c. 77
d. 303


answer is 302
Hello Aspirants ,,,

This is my 1st post as mee not comfortable with NumberSystem... Please help me out..... I have started my journey so help me outttt


1 : What is the remainder when 32^32^32 is divided by 9??


Answer is 4....

99.......99 (2009 times) = n

whats the sum of digits of n^2

answer is 4.

i tried it by euler number

32^32^32/9

euter number for 9 is 6

32^32/6 will leave 4 as remainder

(32^6*x)*32^4/9
1*4=4

is it right