Q: Which is greater a^b or b^a ?
*************
ans-
if ab^a
else
vice versa
Q: Which is greater a^b or b^a ?
*************
ans-
if ab^a
else
vice versa
you are little mistaken
take examples
1^3and
2^4=4^2
can anyone plz help me to find the answer to the following to questions:
1.Arun Bikas and Chetakar have a total of 80 coins among them. Arun triples the number of coins with the others by giving them some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the number of coins he had at the beginning?
a. 11
b. 10
c. 9
d. 12
2.What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(3!)?
a..1
b.7!
c.8!
d.9!
3.Find the last two digits of 65*29*37*63*71*87
can anyone plz help me to find the answer to the following to questions:
1.Arun Bikas and Chetakar have a total of 80 coins among them. Arun triples the number of coins with the others by giving them some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the number of coins he had at the beginning?
a. 11
b. 10
c. 9
d. 12
2.What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(3!)?
a..1
b.7!
c.8!
d.9!
3.Find the last two digits of 65*29*37*63*71*87
For the 1 question , i think the bikas should have 20 coins at the beginning ..
Let A, B, C have x, y and z coins respectively in the beginning ....
Then x + y + z = 80
Initially ...
A x
B y
C z
After first step ....
A x - 2y - 2z
B 3y
C 3z
After Second Step ....
A 3(x - 2y - 2z)
B 3y - 2(x - 2y - 2z) - 6z = -2x + 7y - 2z
C 9z
Now No. of coins with B at the end : -2x + 7y -2z = 20 -------------(1)
Also x + y + z = 80.
this means -2x -2y - 2z = -160 --------------(2)
Subtracting (2) from (1) we get ,
9y = 180
y = 20...
Please tell me if i am wrong at any step ...
For 3. Question : The last two digits are 35 ....
5*9 = 45.
neglect 4..
5*7 = 35...
35*3 = 105
neglect 1...
05*1 = 05.
05*7 = 35..
so the last two digits of 65*29*37*63*71*87 = 35 ...
Please verify ...
3.Find the last two digits of 65*29*37*63*71*87
is it 95 ?
---Quote (Originally by krishnaghosh)---
can anyone plz help me to find the answer to the following to questions:
1.Arun Bikas and Chetakar have a total of 80 coins among them. Arun triples the number of coins with the others by giving them some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the number of coins he had at the beginning?
a. 11
b. 10
c. 9
d. 12
2.What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(3!)?
a..1
b.7!
c.8!
d.9!
3.Find the last two digits of 65*29*37*63*71*87
---End Quote---
3.
65*29*37*63*71*87/100=13*29*37*63*71*87/20
Now dividing 29by 20 u get 9 13/20=-7 and so on
this becomes -7*9*17*3*11*7/20 = -7*63*-3*33/20=21*3*13/20=19/20=-1
20-1=19
19*5=95 (5 was crossed out)
ans=95
2 .2(8!)-21(6!)=16(7!)-3(7!)=13(7!)
14(13!)=14!
14(7!)/13(7!) rem=1*7!=7!
14!/13(7!) = no rem
ans=7!
1. i not able to solve bt 20 is not even an option in sharma nd ans given is 10
My answer-"a"
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
My answer-"a"
Q.1 I have a drawing room of 12*10. I want to place tiles on the floor and I do not have monetary problem so tiles can be maximum. So how many tiles can be place.
Q.2 13^17+14^17-15^17/ 17 what is the remainder.
Q.3 if N^3/7, N^2/5 and N^5/11 so what are possible remainders of all cases.
Q.4 (N^p - N)/ P so the number can be divided by instead of P where P is a prime number.
"No options are given bcoz after options it's like no questions".........
My answer-"a"
Explanation Please
dinesh1787 SaysExplanation Please
this is fermat theorem
n^(P-1)k / P remainder is always 1. where is P is prime number.
Q: What is the largest number that can be formed using four 3's ?
Most probably these would have been discussed in some quant thread, but i couldn't find them in my 10 min search of the forum, hence asking the pro's for help
Great to see Number System thread is alive again. I thought PG official quant thread has eclipsed all
Q: What is the largest number that can be formed using four 3's ?
There was a question in AIMCAT 1120 of this sort and the answer should be 3^3^33
Q.1 I have a drawing room of 12*10. I want to place tiles on the floor and I do not have monetary problem so tiles can be maximum. So how many tiles can be place.
Q.2 13^17+14^17-15^17/ 17 what is the remainder.
Q.3 if N^3/7, N^2/5 and N^5/11 so what are possible remainders of all cases.
Q.4 (N^p - N)/ P so the number can be divided by instead of P where P is a prime number.
"No options are given bcoz after options it's like no questions".........
Q.1 I have a drawing room of 12*10. I want to place tiles on the floor and I do not have monetary problem so tiles can be maximum. So how many tiles can be place.
There should be some more constraints here as i can place a single tile of 12*10 and I can also place infinite tiles.
If the tiles are square (as a constraint), then the minimum number of square tile is of the size 2*2 (HCF of 12 and 10). The number of tiles = 12*10/2*2 = 30 (Thats the minimum).
Q.2 13^17+14^17-15^17/ 17 what is the remainder.
The remainder should be 13 + 14 - 15 = 12
Q.3 if N^3/7, N^2/5 and N^5/11 so what are possible remainders of all cases.
Need options.
Q.4 (N^p - N)/ P so the number can be divided by instead of P where P is a prime number.
Didn't understand the question at all.:-(
The question i had was which is greater 5^7 or 7^5, the calculations look humongous, is there a trick to it ?
For the 2nd Q, the answer is as you have written, whats the logic behind it ?
Kindly explain as and when you get time, Thanks :)
Hi
Basically the rule says that for any number greater than 3,
(smaller number)^(larger number) is greater than (larger number)^(smaller number)
i.e
if agreater than 3
Great to see Number System thread is alive again. I thought PG official quant thread has eclipsed all
Q: What is the largest number that can be formed using four 3's ?
There was a question in AIMCAT 1120 of this sort and the answer should be 3^33^3
i think it should be
3^3^33.
Correct me,if wrong
25! has how many trailing 0's..?
i think it should be
3^3^33.
Correct me,if wrong
Its absolutely correct, could you plz. share how got to the answer ?
What if the question was largest number with four 6's or with five 7's ?
topper@IITK Says25! has how many trailing 0's..?
Ans. 6 (right ?)
internet.fanboy SaysAns. 6 (right ?)
6 zeroes
25/5 = 5
5/5=1
total=5+1=6
Its absolutely correct, could you plz. share how got to the answer ?
What if the question was largest number with four 6's or with five 7's ?
look,you just have to do some arrangements.
The numbers formed will be
1.3^3^3^3
2.3^3^33
3.3^333
4.33^3^3
5.33^33
6.333^3
now clearly 3,4,5,6 ruled out..(hope u get it)
and among 1,2 see
1.3^3^3^3=3^3^27
which is clearly smaller than 2.
So,answer is 3^3^33