How to find which of these two numbers is greater?
(22)^(1/2) or (33)^(1/3)
How to find which of these two numbers is greater?
(22)^(1/2) or (33)^(1/3)
make the powers of both the numbers same, and then compare bases
here, (22)^(1/2) =(22^3)^(1/6) and (33)^(1/3)= (33^2)^(1/6)
=> now 22^3 and 33^2 needs to be compared
raise both the numbers to power 6
then u would see 22^3 and 33^2 is what u need to compare. obviously the former is larger.
raise both the numbers to power 6
then u would see 22^3 and 33^2 is what u need to compare. obviously the former is larger.
yep.. u r correct, i was just telling the approach..
i didn't calculated just bcoz obviously the former is larger
raise both the numbers to power 6
then u would see 22^3 and 33^2 is what u need to compare. obviously the former is larger.
yep.. u r correct... i was just telling the approach
didn't calculated bcoz obviously the former is larger
make the powers of both the numbers same, and then compare bases
here, (22)^(1/2) =(22^3)^(1/6) and (33)^(1/3)= (33^2)^(1/6)
=> now 22^3 and 33^2 needs to be compared
Can i apply this process of making 'powers of both number same' in every such related questions?? i mean where comparison is asked??
chandanx22 SaysCan i apply this process of making 'powers of both number same' in every such related questions?? i mean where comparison is asked??
yes you can apply this anywhere when there is a possibility.
because we are not harming the actual numbersor the relation between them.
greater number remains greater when the two numbers are raised by same power
How to find which of these two numbers is greater?
(22)^(1/2) or (33)^(1/3)
(22)^(1/2) = y
If you observe carefully 4
(33)^(1/3) = z
If you observe carefully 3
So, y>z
This approach takes less time but is applicable when the powers are 1/2 or 1/3 for which you do should have a grasp of squares and cubes..
.. Which most PUYS have..
!!
puys,
im posting some very easy-basic questions on number systems.
lets c how fast we all can solve these:
kindly post solutions too:
1. If a number has 208 factors excluding 1 and itself,then the maximum no of factors that are prime is: a.2 b.4 c.3 d.5
2. An 89 digit no. is formed by writing first 49 natural nos. in front of each other as 1 2 3 4 5 6 .........48 49 .Find the remainder when this number is divided by 125: a.24 b.99 c.124 d.none
3. ABC denotes a 3 digit no.If A and B are interchanged,the value of the no. decreases by 90.How many possible values exist for A and B?
a.10 b.depends on value of B c.9 d.none
4.The ninth digit of ((707)power 3) is
a.9 b.4 c.3 d.none
5.The largest value of k so that 146! is divisible by 21k is
a.6 b.22 c.48 d.36
Gurdeesh
puys,
im posting some very easy-basic questions on number systems.
lets c how fast we all can solve these:
kindly post solutions too:
1. If a number has 208 factors excluding 1 and itself,then the maximum no of factors that are prime is: a.2 b.4 c.3 d.5
2. An 89 digit no. is formed by writing first 49 natural nos. in front of each other as 1 2 3 4 5 6 .........48 49 .Find the remainder when this number is divided by 125: a.24 b.99 c.124 d.none
3. ABC denotes a 3 digit no.If A and B are interchanged,the value of the no. decreases by 90.How many possible values exist for A and B?
a.10 b.depends on value of B c.9 d.none
4.The ninth digit of ((707)power 3) is
a.9 b.4 c.3 d.none
5.The largest value of k so that 146! is divisible by 21k is
a.6 b.22 c.48 d.36
Gurdeesh
1. (b) 4
2. (b) 99
3. (c) 9
4. (c) 3 - ->either way if you take from left or from right
5. (d) 36
Please confirm the answers, i'll post the approach once confirmed.!
hi
can someone pls solve this one for me:
find the product of (1+tan1)(1+tan2)(1+tan3)...............(1+tan45)
1. (b) 4
2. (b) 99
3. (c) 9
4. (c) 3 - ->either way if you take from left or from right
5. (d) 36
Please confirm the answers, i'll post the approach once confirmed.!
Wow dear..all are correct except the last one...the last one has option (b) correct...
plase post the solution methods dear
hi
can someone pls solve this one for me:
find the product of (1+tan1)(1+tan2)(1+tan3)...............(1+tan45)
2^23..!! can you please confirm..!!
Wow dear..all are correct except the last one...the last one has option (b) correct...
plase post the solution methods dear
1. If a number has 208 factors excluding 1 and itself,then the maximum no of factors that are prime is: a.2 b.4 c.3 d.5
208 factors excluding 1 and itself.. means i can take 210 factors in all.. on factorising 210 = 7*5*3*2
For any no. N = (x)^a * (y)^b * (z)^c
No of factors are (a+1)(b+1)(c+1)
So we can express the no. with 210 factors as (Prime1)^6*(Prime2)^2*(Prime3)^2*(Prime4)^1
Hence, a total of 4 prime nos..
2. An 89 digit no. is formed by writing first 49 natural nos. in front of each other as 1 2 3 4 5 6 .........48 49 .Find the remainder when this number is divided by 125: a.24 b.99 c.124 d.none
(12345....4)849 -- ()--> won't make any difference as the last 4000 is already a factor of 125.. So all we need to do is 849/125 which gives 99 as remainder..!!
3. ABC denotes a 3 digit no.If A and B are interchanged,the value of the no. decreases by 90.How many possible values exist for A and B?
a.10 b.depends on value of B c.9 d.none
100a+10b+c=100b+10a+c+90
90(a-b)=90 ==> a-b=1
Min. value of a is 1 as otherwise it'll be a 2 digit no. so ordered pair for (a,b) is (1,0);(2,1);(3,2);(4,3)....(9,
.. a total of 9 pairs..!!4.The ninth digit of ((707)power 3) is
a.9 b.4 c.3 d.none
(700+7)^3 ={ 700^3 + 3*700*7(700+7) }+ 7^3 ; {} - won't affect the answer as 343000000 ensures that the 9th digit is the units digit..
5.The largest value of k so that 146! is divisible by 21k is
a.6 b.22 c.48 d.36
2136 if taken as 4 digits of a number is a factor of 146! as 2136=89*3*8 whereas 2148=179*8 and therefore is not a factor of 146!
If it is 21*k, then the no. is definitely (c) 48.. but don't know how the answer is 22..!!
ekanshtiwari Says2^23..!! can you please confirm..!!
ya thats the ans. can u tell me the approach
Neha011 Saysya thats the ans. can u tell me the approach
(1+tan45) (1+tan44)... (1+tan1)
2*(1+tan44)(1+tan1)(1+tan43)(1+tan2)... (1+tan22)(1+tan23)
now, tan(A+B) = (tanA + tanB)/(1-tanA*tanB)
tanA*tanB = 1-{(tanA+tanB)/tan(A+B)}
take (1+tan44)(1+tan1), rest will be same
1+ tan44 + tan1 + tan44*tan1
1+ tan44 + tan 1 + 1 - {(tan1 + tan44)/tan (44+1)}
1+1+tan44+tan1-tan44-tan1
=2
Therefore, we'll get 2*2^22 ==> 2^23
Please tell me how to solve:
(3003^9000) / 9000
ans is 3003
as we know a^p-a/p is completely divided so
please confirm
ans is 3003
as we know a^p-a/p is completely divided so
please confirm
a^p-a/ p (remainder) is 0 when p is a prime number and "a" and "p" are relatively prime.
3003 = 3 * 7 * 11 * 13
(3003^9000) = 3 ^ 9000 * (7*11*13)^9000
Take out the common factors, the fraction reduces to / 1000
Euler's Number for 1000 = 1000 * (1/2) * (4/5) = 400
3^198 * 1 / 1000 (rem) = 3^198/ (2^3 * 5^3) .
3^198/(2^3) , Euler's Number for 2^3 = 8(1/2) = 4.
Remainder is 3^2/(2^3) = +1
3^198/(5^3) , Euler's Number is 5^3 = 125 (4/5) = 100
3^98/(5^3) (Rem) = 2^98/(5^3) = 3^14/(5^3) = +9 (Horrible Mistake)
it should be +14.
Hence using Chinese Remainder theorem
125A + 14 = 8B + 1 => 125A + 13 = 8B => 5A + 5= 8B , A = 7. Hence the first number that satisfies the equation is 875 + 14 = 889.
Hence the actual number should be 889 * 9 = 8001.
3003^9000 (mod 9000)
(1001)^2*(3003)^8998 (mod 1000)
= 1*(3003)^198 (mod 1000)
= 3^198 (mod 1000)
So, we need to find last three digits of 3^198
3^198 = 9^99 = (10 - 1)^99
Last three digits wil be given by:-
1000k - C(99, 97)*100 + 99*10 - 1
= 1000k - 100 + 990 - 1
= 889
=> Remainder will be 889*9 = 8001
a^p-a/ p (remainder) is 0 when p is a prime number and "a" and "p" are relatively prime.
3003 = 3 * 7 * 11 * 13
(3003^9000) = 3 ^ 9000 * (7*11*13)^9000
Take out the common factors, the fraction reduces to / 1000
Euler's Number for 1000 = 1000 * (1/2) * (4/5) = 400
3^198 * 1 / 1000 (rem) = 3^198/ (2^3 * 5^3) .
Now 3^198/ (2^3) Rem will always be +1. - Eq (1)
3^198/(5^3) Rem = 3^98/5^3(rem) = (-2)^98/(5^3)
Now 2^7 / 5^3 (rem) = +3
hence (-2)^98/(5^3) Rem = 3^14/(5^3) = 9 - Eq (2)
Combining eq 1 and 2 and using Chinese Remainder Theorem.
8A + 1 = 125B + 9 => 8A = 125B + 8 => 8A = 5B, B = 8
hence 2009 is the first number. Hence the actual Remainder should be 2009 * 9 = 18081
18081 / 3003 = 63.
Do confirm the answer and also I have skipped some steps. Let me know in case of any doubt.
3003^9000 (mod 9000)
(1001)^2*(3003)^8998 (mod 1000)
= 1*(3003)^198 (mod 1000)
= 3^198 (mod 1000)
So, we need to find last three digits of 3^198
3^198 = 9^99 = (10 - 1)^99
Last three digits wil be given by:-
1000k - C(99, 97)*100 + 99*10 - 1
= 1000k - 100 + 990 - 1
= 889
=> Remainder will be 889*9 = 8001
Easier still :
3^(899
* ( 7*11*13) ^ 2 mod 10003^8998 * 3^2 mod 1000 = 1
so
x*9 = 1000k+1
so X= 889
hence rem = 889*9 = 9= 8001
PS : i suck at anything Chinese