can anyone help me solve this problem
(3^555)+(5^333) is divisible by
a)2
b)3
c)37
d)11
e)all
can anyone help me solve this problem
(3^555)+(5^333) is divisible by
a)2
b)3
c)37
d)11
e)all
Answer should be 2. I checked by options and 2 divides it whereas 3 doesn't. So the answer can't be "all"
Hi all,
Can u plz tell me how to solve the following sum. This is 4m time material.
X is a set of all 5 digit nos in base 3. A set Y is formed by taking decimal equivalent of each element of set X as its elements. How many elements of Y are divisible by 3 but not divisible by 9.
Thanking you in advance.
Hi all,
Can u plz tell me how to solve the following sum. This is 4m time material.
X is a set of all 5 digit nos in base 3. A set Y is formed by taking decimal equivalent of each element of set X as its elements. How many elements of Y are divisible by 3 but not divisible by 9.
Thanking you in advance.
let the 5 degit no in base 3 be abcde
decimal equivalent of each degit will be
e=e*3^0
d=d*3^1
c=c*3^2
b=b*3^3
a=a*3^4
degits in base 3 can be 0,1,2 hence a,b,c,d,e can be 0,1,2
so only d will not be devisible by 9
A very simple question though .. I am missing some basic Funda !!
Can somebody please explain what I am missing ?
Its based on probability
Probem :
A question paper consists of two sections A and B having respectively 3 and 4 questions. Four questions are to be solved to qualify in that paper.One question from section A and 2 questions from section B are compulsory/In how many ways can a candidate select the questions to qualify in that paper?
Solution :
The solution obtained by ,
3C2 * 4C2 +3C1 * 4C3 = 30 is correct .
But say If I use the following method , what am I doing wrong .
Select the compulsory questions to be solved first and then select the remaining questions to be solved ,
1 out 3 compulsory ( section A ) == 3C1
2 out of 4 compulsory ( Section B ) == 4C2
Now have to select 1 out of 2 from section A or 1 out 2 from section B
(2C1 + 2C1 )
So solution is
( 3C1 * 4C2 ) * (2C1 +2C1 ) == 18 * 4 = 72
Or
select 4 out of the remaining 4 after selecting the compulsory ones
( 3C1 * 4C2 ) *4C1 = = 18 * 4 = 72
Something is getting counted more than once ? Can some one explain me the exact problem . Also whether approach followed by me is correct ? If so when to use which one ?
Select the compulsory questions to be solved first and then select the remaining questions to be solved ,
1 out 3 compulsory ( section A ) == 3C1
2 out of 4 compulsory ( Section B ) == 4C2
Now have to select 1 out of 2 from section A or 1 out 2 from section B
(2C1 + 2C1 )
So solution is
( 3C1 * 4C2 ) * (2C1 +2C1 ) == 18 * 4 = 72
Or
select 4 out of the remaining 4 after selecting the compulsory ones
( 3C1 * 4C2 ) *4C1 = = 18 * 4 = 72
Something is getting counted more than once ? Can some one explain me the exact problem . Also whether approach followed by me is correct ? If so when to use which one ?
Let the questions be numbered 1 to 7 (Section A - 1 to 3 Section B - 4 to 7).
According to your solution:
> Selecting Q.1 (compulsory from section A), Q.4 and Q.5 from section B (compulsory)
> Selecing Q.7 (4C1)
is equivalent to
> Selecting Q.1 (compulsory from section A), Q, 4 and Q.7 from section B (compulsory
> Selection Q.5 (4C1)
.. and so on.
---
Disclaimer: I haven't counted exactly how many such extra cases are included.
can ypu please explain as to how u solved it
how to solve
what is the remainder when (1!)^3+(2!)^3+.....+(1152!)3 is divided by 1152?
a)125
b)225
c)325
d)205
how to solve
find the 28383rd term of the series 12345678910111213......
a)3
b)4
c)9
d)7
how to solve
what is the remainder when (1!)^3+(2!)^3+.....+(1152!)3 is divided by 1152?
a)125
b)225
c)325
d)205
In such questions the first thing to be done is the prime factorisation of the divisor. Here 1152=2^7 * 3^2
Now if we start checking (1!)^3 will not have all the factors and hence will be part of remainder
Similarly,
(2!)^3 = 2^3 also does not have allfactors
(3!)^3 = 3^3 * 2^3-The power of 3 is fine but the power of 2 is still less
(4!)^3=2^9 *3^3 - From here on all the numbers will be perfectly divisible as they will have all factors of 1152.
So effectively our remainder is (1!)^3 +(2!)^3 + (3!)^3=225
Hope that answers ur query!!!
how to solve
find the 28383rd term of the series 12345678910111213......
a)3
b)4
c)9
d)7
Now we need to go digit wise for such a question
number of digits from 1-9=9
from 10-99=90*2=180
from 100-999=900*3=2700
from 1000-9999=9000*4=36000
hence 28383 will lie in between 1000-9999
Now remaining digits until 1000 will be =28383-(9+180+2700)
=25494
Dividing it by 4 to get the nth term will be
25494/4
=6373 term+ 2nd term of 6374th one
Means from 1000 the 6374th term will be
7374
hence the last digit will be 3
how to solve
what is the remainder when (1!)^3+(2!)^3+.....+(1152!)3 is divided by 1152?
a)125
b)225
c)325
d)205
Terms From (4!^3)...Upto (1152!^3)..when divided by 1152 = 24^2 * 2....wpould give remainder Zero...
So Our remainder = (1!)^3+(2!)^3+(3!)^3 = 216 + 8 + 1 = 225
how many integral values of x and yare there such that 4x+3y=3,while xa)144
b)141
c)143
d)142.
Please some one explain me to solve it
how to solve -
what is the highest power of 3 in the expression 58! - 38!?
a)17
b)18
c)19
d)none
please explain.
please explain how to solve it
what is the remainder when 2(8!) - 21(6!) divides 14(7!) - 14(13!)?
a)1
b)7!
c)8!
d)9!
The answer is 7!.
how to solve -
what is the highest power of 3 in the expression 58! - 38!?
a)17
b)18
c)19
d)none
please explain.
The answer should be 17 as the highest power of 3 in 38! is 17.
avinav2712 SaysThe answer should be 17 as the highest power of 3 in 38! is 17.
Hi Avinav,
Why did u consider only 38!??
Hi Avinav,
Why did u consider only 38!??
It's because you can take 38! common in the expression. Is the answer correct?
how to solve -
what is the highest power of 3 in the expression 58! - 38!?
a)17
b)18
c)19
d)none
please explain.
the answer will be option a) 17..
we can get it by checking the highest power in 38!...
please explain how to solve it
what is the remainder when 2(8!) - 21(6!) divides 14(7!)- 14(13!)?
a)1
b)7!
c)8!
d)9!
The answer is 7!.
7!(14 -14*13*12*11*10*9*
/7!(2*8 - 3)7!(14 -14*13*12*11*10*9*
/7!*13as (14 -14*13*12*11*10*9*
/13, leaves remainder 1so the answer is 7!