1> The arithmatic mean of two numbers is smaller by 24 than the larger of the two numbers and the GM of the same number exceeds by 12 the smaller of the numbers find the numbers?
2> fine the numbers between 200 and 300, both included,whih are not divisible by 2,3,4,and 5
3> Given x an n are integers,(15n^3+6n^2+5n+x)/n is not an integer for what condition ?
a> n is positive
b> x is divisible by n
c> x is not divisible by n
d (a) and (c)
e> none of these
4> The unit digit in the expression 36^234*33^512*39^180 - 54^29*25^123*31^512 will be
(a) 8 (b)0 (c)6 (d)5(e)4
5> The difference of 10^25-7 and 10^24+x is divisible by 3 for x=?
(a)3 (b) 2 (c)4(d)6 (e)7
6> If a number is multipiled by 22 and the same number is added to it .then we get a number that it half the square of that number.find the number
7> 12^55/3^11 + 8^48/16^18 will give the digit at units place as
(a)4 (b)6(c)8(d)0 (e)5
8> the mean of 1,2,2^2....2^31 lies in between
(a) 2^24 to 2^25 (b) 2^25 to 2^26 (c) 2^26 to 2^27 (d) 2^23 to 2^24
hey guys pls solve i for me..
thanks
sumit
hii......i want 2 kno abt arun sharma quants....hw is it????????is it of cat standard??????will it help 1 for preparing cat & other allied mba xams?????
- N=202*20002*200000002*20002(15 zeroes)*20002(31 zeroes). The sum of digits of N is
- The remainder when 22 + 222 + 2222+ 22222+..2222(49 twos)2 is divided by 9 is:
Im getting the ans as 3 which is not an option only
- n ias a number such that 2n has 28 factors and 3n has 30 factors. 6n has?There must be a set pattern to solve this question. Kindly tell me.
hi guys, i have got on dumb doubt
what is this euler's theorem anyway? is it in arun sharma?
i searched about in on wiki but it is uncomprehendingly wierd.
please post any text u have about this stuff. :laugh:
hi guys, i have got on dumb doubt
what is this euler's theorem anyway? is it in arun sharma?
i searched about in on wiki but it is uncomprehendingly wierd.
please post any text u have about this stuff. :laugh:
wat to laugh on this???anywyas dont mind...but if u dont know abt anything...try to ask ur frndz...or in realted threads....may b it could b very helpful for u.....may b u couldnt find it on wiki...coz probably ur serach wasn't sufficient...google it once...u will get euler's theorem there for sure...one more thing...this is really helpful........u can't find it in arun sharma :)..........i will ask my frnd to post eluer therom here 😃
nidhisoni24 Sayswat to laugh on this???anywyas dont mind...but if u dont know abt anything...try to ask ur frndz...or in realted threads....may b it could b very helpful for u.....may b u couldnt find it on wiki...coz probably ur serach wasn't sufficient...google it once...u will get euler's theorem there for sure...one more thing...this is really helpful........u can't find it in arun sharma :)..........i will ask my frnd to post eluer therom here :)
thanks buddy. if u find a link anywhere here or on net, do post it. it wud be of gr8 help. thanks again.
Hi,
I have one doubt in Probability, My answer doesn't match with the answer given in Arun Sharma. Please help. I am new in this thread, I even tried to search whether anyone has posted the question earlier but all in vain. Here's the question:
Q. Three students appear in an examination of Mathematics. The probability of their success are 1/3, 1/4,1/5 respctively. Find the probability of success of at least two
a) 1/6 b) 2/5 c) 3/4 d) 3/5
Please post the method toooo. :boat:
Is there an index of all the problems posted in this tread?
Since the problems already discussed need not posted once again
vivs_d Saysthanks buddy. if u find a link anywhere here or on net, do post it. it wud be of gr8 help. thanks again.
Euler's theorem - Wikipedia, the free encyclopedia
:) hope it will help u
Is there an index of all the problems posted in this tread?
Since the problems already discussed need not posted once again
hmm well here is no index of all prbs..but u can visit realted threads...like number system thread for number system prbs 😃
Euler's theorem - Wikipedia, the free encyclopedia
:) hope it will help u
This is the first time I came across this theorum and seriously for a neo like me its hard to understand:baaa:. Is there not any simpler version of this?? If yes, then please post it here or PM me if possible.
Thanks.
- xy is a number that is divided by ab where xy
(a) 11 (b) 33 (c)99 (d) 66 (e) 88
2. A number xy is multiplied by another number ab and the result comes as pqr,where r=2y, q=2(x+y) and p= 2x where x,y (a)11 (b)13 (c)31(d) 22 (e) none of these
3. denotes the greatest integer value just below x and
{x} its fractional value. The sum of ^3 and {x}^3 is -7.91 find x.
(a) -2.03 (b) -1.97 (c) -2.97 (d) -1.7 (e) none of these
4. 16^15+2^15 is divisible by
(a) 31 (b) 13 (c) 27 (d) 33 (e) 11
5. IF AB+XY=1XP,where A not equal to 0 and all the letters signify different digits from 0 to 9 ,then the value of A is :
(a) 6 (b) 7 (c) 9 (d) 8 (e) Any value above 6
6. x-3 + 2|x+1 =4 find the possible integral values of x
(a) 1 (b) -1 (c) 3 (d) 2 (e) There are many solutions
7.If 4 ^n+1 + x and 4^2n -x are divisible by 5,n being an even integer,fnd the least value of x.
(a) 1 (b) 2 (c) 3 (d) 0 (e) none of these
8. X^2 + |x-2=1 find the possible integral values of x.
(a) 1 (b) -1 (c) 0 (d) 1 or 0 (e) -2
hey guys pls solve all these questions ..
Thank you!!!
- xy is a number that is divided by ab where xy
(a) 11 (b) 33 (c)99 (d) 66 (e) 88
2. A number xy is multiplied by another number ab and the result comes as pqr,where r=2y, q=2(x+y) and p= 2x where x,y (a)11 (b)13 (c)31(d) 22 (e) none of these
3. denotes the greatest integer value just below x and
{x} its fractional value. The sum of ^3 and {x}^3 is -7.91 find x.
(a) -2.03 (b) -1.97 (c) -2.97 (d) -1.7 (e) none of these
4. 16^15+2^15 is divisible by
(a) 31 (b) 13 (c) 27 (d) 33 (e) 11
5. IF AB+XY=1XP,where A not equal to 0 and all the letters signify different digits from 0 to 9 ,then the value of A is :
(a) 6 (b) 7 (c) 9 (d) 8 (e) Any value above 6
6. x-3 + 2|x+1 =4 find the possible integral values of x
(a) 1 (b) -1 (c) 3 (d) 2 (e) There are many solutions
7.If 4 ^n+1 + x and 4^2n -x are divisible by 5,n being an even integer,fnd the least value of x.
(a) 1 (b) 2 (c) 3 (d) 0 (e) none of these
8. X^2 + x-2=1 find the possible integral values of x.
(a) 1 (b) -1 (c) 0 (d) 1 or 0 (e) -2
hey guys pls solve all these questions ..
Thank you!!!
Hi..will try to answer some..
1) c explanation: xy/100 wld be .xy
.xyxyxy... is close to (~>).xy so it should be divided by a number just less than 100 close to it.
or you can prove by example. just try the simplest 2 digit number which satisfies xy ie 10 and divide by 99.
2) d: simple substitution for all values.
note that xy means 10 x+y and similarly for ab and pqr after that simple subst u'll get 10a+b=22 ie the number ab is 22.
3) Not sure about the notation u used? was it the floor operation |_ x_ and not []? can't say. I don't have the book so I can't look it up either.
4) aiee..not sure how to do these types
5) c i think. same as 2..sub and you'll find 10 a+b+y=100+p sub in a=6 => b+y-p=40..not poss for any b,y,p cos they are single digits.
intuitively you should know that even 30,20 won't be possible so you are left with a=9, which will give you b+y-p=10 a lot of of combinations of b,y,p can give u that so a=9 is right.
6) b: for modulus, just chk 3 diff inequalities.
x>3=> x=5/3 not valid soln
-1x x=-1
so b.
7) sheitz got this after some thinking. i assume ur qn is ((4^(n+1))+x) & ((4^2n)-x) are divisible by 5.
since n is even n+1 is odd 4^any odd number will end in 4 so add 1 and its divisible by 5.
4^2n is 16^n which will always end in 6 so subtract 1 and it becomes divisible by 5.
so ans is a)1
8 ) modulus again 2 inequalities x2 none of which yield an integer value of x so 0 solutions. since the options are so small u might as well substitute the actual values and see!
:bigear:
no feedback huh..
3. denotes the greatest integer value just below x and
{x} its fractional value. The sum of ^3 and {x}^3 is -7.91 find x.
(a) -2.03 (b) -1.97 (c) -2.97 (d) -1.7 (e) none of these
anyway, this question's ans would be e i guess. you know xto make things easier make evthing +ve
^3 + {x}^3=7.91
which is less than 8=2^3
so cannot be 2, only option left for is 1.
=> {x}^3=67.91-1^3=6.91.
so x=cube root of 6.91
ie 1this is nearer to 8 so x is near 2.
1.9^3 is 6.859 very close to what we want 6.91
..so its smwhere arnd 1.91(approximating)
none of the options is 1.91 so (e)
and the 4th one i asked my ims maths teacher and he couldn't find any patterns. although if u try to do it by elimination you might be able to solve it. Answer is 27, but I don't know how to get there.
So the 4th question is a real humdigger still.
I am still waiting for the answer for my question pasted above.
a)112 b)160 c)144 d)cannot be determined
- N=202*20002*200000002*20002(15 zeroes)*20002(31 zeroes). The sum of digits of N is
a)2 b)5 c)6 d)7
- The remainder when 22 + 222 + 2222+ 22222+..2222(49 twos)2 is divided by 9 is:
1)
Take 2 out from each of the terms. You will be left with 32 *101*10001 ....
Now if u multiply the first 2 terms u will get 10101010. That is 4 ones seperated by zeroes. And if u multiply the first 3 terms u will get 8 ones seperated by zeroes. Same lines if u multiply all the terms you will get 32 ones seperated by zeroes.now when u mutilpy this with 32, then u get
323232...(32 times). So the sum will be 32*5=160
2) 22 can be written as 10*2+2. When u divide this by 9 ,for the first term remainder will be 1(since any power of 10 leaves a remainder of 1 when divided by 9).so for first term remainder is 4
222 can be written as 100*2+10*2+2. This will leave a remainder of 6 using the same login as above.
third term on similar lines will yield 8.
This is an AP 4+6+8....48 terms
Sum=24(4+4
=24*54this when divided by 9 yields a remainder of 6.
Am I right with the answers??
This is the first time I came across this theorum and seriously for a neo like me its hard to understand:baaa:. Is there not any simpler version of this?? If yes, then please post it here or PM me if possible.
Thanks.
Euler Theorem states that a^{f(n)} mod n=1
where a and are co prime and f(n)= no. of numbers lesser than n and co prime with it.
Ex:3^5 mod 8 =1
here n=8 ;f(n)=5 { numbers that are lesser than 8 and co prime with it are 1,3,5,6,7}
Going ahead further if n a prime number, then f(n)=n-1 (since all numbers lesser than it would be co prime with it)
So then a^(n-1) mod n =1....This is Fermat Theorem.
Hope this helps
hi guys,
what is the time limit of review questions after every block?
is it same for all? tell me bout first block.
hi manu..
ans to first is correct but for second.. remainder whn 22 is dividd by 9 is 4 and not 1..you can check manually..(18+4)
and besides the actualt question i wanted to ask was'
2^2 + 22^2+ 222^2+ 2222^2 + .+ 222(49 times)^2.. remainder from 9
(earlier there was a typing error)..plz let me know if u solve this..
is the remainder 6?
confirm please....i will post the method soon....m in a bit of hurry