Q> How many integers less than 300 are relatively prime to either 10 or 18?
1. 140
2. 141
3. 142
4. 139
5. 138
Please post the approach also
In the question is it integers or non-negative integers...
In the question is it integers or non-negative integers...
maclens SaysIn the question is it integers or non-negative integers...
It is not mentioned in the question but I ges it should be non negative integers.
Hi
I found this following problem....could some1 give me a detailed solution?
S={2^0,2^1,2^2,.......2^2009}
How many elements of set S start with digit 4?
Hi
I found this following problem....could some1 give me a detailed solution?
S={2^0,2^1,2^2,.......2^2009}
How many elements of set S start with digit 4?
Ans. 201
I have attempted this question purely on the basis of observation.
First term we come across is 4= 3rd term
Then, 4096 = 13th term
Then, 4194304 = 23rd term
So, observing the cyclicity, there shouls be 201 terms.
If someone has a better method , please share..
Ans. 201
I have attempted this question purely on the basis of observation.
First term we come across is 4= 3rd term
Then, 4096 = 13th term
Then, 4194304 = 23rd term
So, observing the cyclicity, there shouls be 201 terms.
If someone has a better method , please share..
Hi
I found this following problem....could some1 give me a detailed solution?
S={2^0,2^1,2^2,.......2^2009}
How many elements of set S start with digit 4?
Well,my method is very much like what Kashyap used...see if it can help...
2^2 = 4 ; 2^12 = 4096 (This we already know)
Just to confirm calculate 2^22...It will also be starting with 4
So we have an AP with a=2 and d=10(12-2)
Hence 2002=2+(n-1)10 (Took 2002 for so that it cud be divided by 10 after subtraction by 2)
This way we get n=201...:biggrin:
Hmmm...
method is kind of crude and not very elegant..
but i doubt that is the correct approach.
Your answer doesn't match anyways.
The answer given is 195
Hmmm...
method is kind of crude and not very elegant..
but i doubt that is the correct approach.
Your answer doesn't match anyways.
The answer given is 195
Its not crude...it took me just a minute to solve it...Just u have to be skillful with numbers....
Well,I am still adamant about 201...n i think 195 cant be answer...
I would wait till any other veteran tries it...
Its not crude...it took me just a minute to solve it...Just u have to be skillful with numbers....
Well,I am still adamant about 201...n i think 195 cant be answer...
I would wait till any other veteran tries it...
ur method will be useful
if options are 201 , 231 , 181 , 221
but having said that answer should be definitely less than 201..
see at one point , 2^10x + 2 will start with 5 ...
so we can only approximate it but not say accurate answer..
looking for elegant solution
ur method will be useful
if options are 201 , 231 , 181 , 221
but having said that answer should be definitely less than 201..
see at one point , 2^10x + 2 will start with 5 ...
so we can only approximate it but not say accurate answer..
looking for elegant solution
@bold part
Oh...will it?
If it will then yeah definately...this method cant be used...
Hi
I found this following problem....could some1 give me a detailed solution?
S={2^0,2^1,2^2,.......2^2009}
How many elements of set S start with digit 4?
2^0 -->1
2^1-->2
2^2-->4
2^3-->8
2^4-->1
2^5-->3
2^6-->6
2^7-->1
2^8-->2
2^9-->5
2^10-->1
2^11-->2
2^12-->4
hence cyclicity is 10
so in 2010/10 = 201 such nos. starting with 4
but for 2^2003--> the first digit is 9 instead of 8..from here the deviation starts..but i guess that wont effect the nos. starting with 4 in the set as there are no more nos. starting with 4...i go with 201...
2^0 -->1
2^1-->2
2^2-->4
2^3-->8
2^4-->1
2^5-->3
2^6-->6
2^7-->1
2^8-->2
2^9-->5
2^10-->1
2^11-->2
2^12-->4
hence cyclicity is 10
so in 2010/10 = 201 such nos. starting with 4
but for 2^2003--> the first digit is 9 instead of 8..from here the deviation starts..but i guess that wont effect the nos. starting with 4 in the set as there are no more nos. starting with 4...i go with 201...
the deviation will start from 2^100....naga is right...cyclicity cant be used here...
Abhinav90 Saysthe deviation will start from 2^100....naga is right...cyclicity cant be used here...
okk...considering the groan i got from the groanmaster:biggrin: "naga"...i think he is rt....i had checked for 2^101...but the deviation is frm 2^102:banghead:...so he is rtt...
it will be less than 201
HEY dude im in a dilema how to solve this q on no. systm or thez type to be specific i always get it wrong
(13^66-23)/183 watz the remainder ?? plz expln all the steps.....
thnx in advance
HEY dude im in a dilema how to solve this q on no. systm or thez type to be specific i always get it wrong
(13^66-23)/183 watz the remainder ?? plz expln all the steps.....
thnx in advance
(13^66-23)mod183
13^66mod183
183 = 3*61
13^66mod3 = 1
13^66mod61 = 13^6mod61 = 47^3mod61 = 47*13 = 1
hence 61k +1 = 3m + 1
for k = 3 we have 183k + 1
hence 1-23/183 = -22 = 161???
(13^66-23)mod183
13^66mod183
183 = 3*61
13^66mod3 = 1
13^66mod61 = 13^6mod61 = 47^3mod61 = 47*13 = 1
hence 61k +1 = 3m + 1
for k = 3 we have 183k + 1
hence 1-23/183 = -22 = 161???
hey grt man thts the answr jst clarify thez steps 13^6mod61 = 47^3mod61 = 47*13 = 1
(13^66-23)mod183
13^66mod183
183 = 3*61
13^66mod3 = 1
13^66mod61 = 13^6mod61 = 47^3mod61 = 47*13 = 1
hence 61k +1 = 3m + 1
for k = 3 we have 183k + 1
hence 1-23/183 = -22 = 161???
How did u get bold red part?
tac007 SaysHow did u get bold red part?
47^3 mod 61 = (-14)^3 mod 61
14^2 x (-14) mod 61 = 196 mod 61 x (-14) mod 61
Thus, 13 mod 61 x (-14) mod 61 (This is the step 47*13 mod 61)
Now, 13x(-14) mod 61 = -182 mod 61 = 1
Hope you got it now
guys
plz try this question
" If for N values of "p", where p1) N2) N=6
3) N=8
4) N=10
5)N>=12
guys
plz try this question
" If for N values of "p", where p1) N2) N=6
3) N=8
4) N=10
5)N>=12
option 5)
numbers of the form k*5 where k is prime and 77*5 = 35! will have 35^5
11*5 = 55! will have 55^5
it will make 12 values, only one option satisfy that.
plz point out any flaw
X=[579 * 580 * 581 * 582 * ......... * 635] + [ 52 * 57 ]
Find the remainder when X is divided by 57! .
a) 52
b) 0
c) 57! -1
d) None of these.