Number System - Questions & Discussions

guravus · August 22, 2011, 7:09pm

hii... plz tel me how to solve this...

what is the remainder when 128^(1000) is divided by 153? :banghead:

by eulers theorem since 128 and 153 are coprime
128^96 will give rem as 1
128^960 will give it as 1
now 128^40 will give it as 52

was a really long calculation for me....pls if some 1 has a short cut post it:cheers:

spunkyarora · August 22, 2011, 7:19pm

@om
128=2^7
thrfre we need to find the rem. of 2^7000 with 153
i'l b using chinese rem theorm
153=9x17
so we need to find the smallest value of x and y for the equation 9x + 17y=0
simply x=2, y= -1
rem of 2^7000 by 9 comes out to be 7
by 17 rem wud be one
so overall rem by 153 wud be 1x(9x2)+7(-1x17)= -135
so rem wud be 18

dext3r · August 22, 2011, 7:55pm

grm_bh Says
is d ans 384??

yes! please explain

dext3r · August 22, 2011, 7:58pm

@om
128=2^7
thrfre we need to find the rem. of 2^7000 with 153
i'l b using chinese rem theorm
153=9x17
so we need to find the smallest value of x and y for the equation 9x + 17y=0
simply x=2, y= -1
rem of 2^7000 by 9 comes out to be 7
by 17 rem wud be one
so overall rem by 153 wud be 1x(9x2)+7(-1x17)= -135
so rem wud be 18

hey I am unable to understand that how does this theorem works. could you explain please....

deepakkr97 · August 23, 2011, 1:03pm

Plz any one post the rules of "reverse eular theorem"..

deepakkr97 · August 23, 2011, 1:04pm

Plz any one post the rules of "reverse eular theorem"..

trojan12 · August 23, 2011, 7:21pm

i am stuck at bunch of question so guys please help me out .tell the concept also that you have used while solving :
1. find unit digit of N=17^27!^37!. options are 1,3,7,9.
.

grm_bh · August 23, 2011, 7:27pm

i am stuck at bunch of question so guys please help me out .tell the concept also that you have used while solving :
1. find unit digit of N=17^27!^37!. options are 1,3,7,9.
2.number of zeros at the end of 36!^36! options are :7x6!,8X6!,7X36!,8X36!.

27! and 37! are of the form 4n so following d cyclicity find last digit of 17^4=1 thus ans=1

36! is made up of 8 5's and no of zeros depend on no.of 5's
thus 8*36!

spunkyarora · August 24, 2011, 7:17am

rohit406 · August 24, 2011, 4:49pm

i am stuck at bunch of question so guys please help me out .tell the concept also that you have used while solving :
1. find unit digit of N=17^27!^37!. options are 1,3,7,9.
.

in such questions unit digit of 27!^37! will be 0 as it wil have certain nukmbers of 2s and 5s for sure so also it wil be divisible by 4 so 17 ^anything whose last digits are divisible by 4 has unit digit as 1 ...if second last digit is require then it will be 0

phoolan23 · August 24, 2011, 5:39pm

1.3 N is a number such that 200A. 12
B. 13
C. 14
D. 15

please explain the solution in detail!

grm_bh · August 24, 2011, 5:55pm

1.3 N is a number such that 200
A. 12
B. 13
C. 14
D. 15

please explain the solution in detail!

is the ans 13??? if yes will post d soln..

rohit406 · August 24, 2011, 5:56pm

1.3 N is a number such that 200A. 12
B. 13
C. 14
D. 15

please explain the solution in detail!

plz correct me if i am wrong :
since the number has exact 6 divisors so number of factors should be (2+1)*(1+1)=6 or (5+1)=6
so number should be if the form x^2*y or x^5 where x and y are prime numbers

by this i am only able to find out 3 numbers ....kindly help if i am missing sumthing

tamal220187 · August 24, 2011, 6:03pm

1.3 N is a number such that 200A. 12
B. 13
C. 14
D. 15

please explain the solution in detail!

am getting 13 numbers
6 divisors mean either they are x^5 form or x*y^2 form...
from that, we are getting 13 values
for x^5, only one value : 3^5 = 243
now for xy^2 form:
for y = 2, x = 53, 59, 61, 67, 71, 73
for y = 3, x = 23, 29, 31
for y = 5, x = 11
for y = 7, x = 5
for y = 11, x = 2
so total no of possible N = 1 + 6 + 3 + 1 + 1 + 1 = 13

grm_bh · August 24, 2011, 6:04pm

plz correct me if i am wrong :
since the number has exact 6 divisors so number of factors should be (2+1)*(1+1)=6 or (5+1)=6
so number should be if the form x^2*y or x^5 where x and y are prime numbers

by this i am only able to find out 3 numbers ....kindly help if i am missing sumthing

6 factors
p^5 and pq
p^5 - 1 case: 3^5

pq
For q = 2, p has 6 values 53, 59, 61, 67, 71, 73
For q = 3, p has 3 values 23, 29, 31
For q = 5, p has 1 value 11
For q = 7, p has 1 value 5
For q = 11, p has 1 value 2
thus 13

dext3r · August 24, 2011, 8:43pm

hey how do you people reach at the conclusion that the only possible combinations of six factors is x^5 and xy^2. is there any method or just by common sense??????:banghead:

spectramind07 · August 24, 2011, 8:50pm

Dext3r Says
hey how do you people reach at the conclusion that the only possible combinations of six factors is x^5 and xy^2. is there any method or just by common sense??????:banghead:

If N is a number that can be expressed as (a^p)(b^q)(c^r).....,then the number of factors of N = (p+1)(q+1)(r+1)......

In the above formula,a,b,c are all prime nos.

No of factors is given as 6.A number will have 6 factors only if it is of the form (x)^5 or xy^2(Use the above formula to verify).

dext3r · August 24, 2011, 10:35pm

If N is a number that can be expressed as (a^p)(b^q)(c^r).....,then the number of factors of N = (p+1)(q+1)(r+1)......

In the above formula,a,b,c are all prime nos.

No of factors is given as 6.A number will have 6 factors only if it is of the form (x)^5 or xy^2(Use the above formula to verify).

hey i know this formula but seriously ,i don't know, what had happened to me i forgot this at that time .....

ny ways thanx..... 😃

srihari123 · August 25, 2011, 1:54am

6545454545454.....(54 comes 100 times) / 5454545454545...(45 comes 100 times) == ?

Please help me in solving this question

testdjjaydev · August 25, 2011, 2:53am

How many sets of two numbers will have the LCM as A^m * B^n where A and B are prime numbers and m and n are natural numbers??