A class of 40 students has unique weights between 30 to 69 kg. What is the difference between the average of the weight and the average of the weight taken by interchanging digits of each weight?
OPTIONS 1)0 2)2 3)5 4)10 5)7
i am also getting 0....
i am also getting 0....
the answer shud be d
take n=1
7^6-6^6 = (7^3)^2 - (6^3)2 so use a^2-b^2 and one factor wud be 127 and one wud be 13 (7^3+6^3 = a ^3 + b^3)
yes the answer is right and perfectly explained.
sirjee apply Euler's technique...
As 75 & 2 are coprime.
75=(5^2)(3)
then x = 75(1-1/5)(1-1/3)= 40
& 2^40/75 will give 1 as a remainder.
so ((2^40)^5)/75 will give you 1 as a remainder.
I THINK THIS COULD BE SOLVED MORE EASILY BY DOING:
(1+1)^200=(1K+1^200)/75 WILL GIVE U REMAINDER 1.WHERE K IS MULTIPLE OF 75.
CONCEPT :(a+b)^n=ak+b^n,by binomial.
ex. 64^12564/9 =(63+1)^12564/9
(63k +1^12564)/9
FIRST PART OF THIS EXPRESSION IS DIVISIBLE BY 9 OTHER WILL GIVE U THE REMAINDER
IT HAS A VERY IMPORTANT IMPLEMENTATION:thumbsup:
Q. The integers m and p are such that 2 1 ?
1. the greatest common factor of m and p is 2
2. the least common multiple of m and p is 30
Answer: it can be answered either with 1 alone or 2 alone.
If gcd(m,p)=2, it means they are even number. so, r will always be >1,irrespective of the values of m and p.
if lcm(m,p)=30, the values of m and p can be 5&6 or 3&10 . no other values can satisfy. in both the cases, r=1.
Puys Could anyone clearly explain me the for solving questions like this??? which is given in Arun Sharma!!!!!!!
find the highest power of 175 which divides 344!!!!!!!!
Arun sharma has made a note to be careful. and in this case, we need to check the num,ber of 5^2 and number of 7's.. but i am not able to understand it clearly....
Thanking in advance..........
Puys Could anyone clearly explain me the for solving questions like this??? which is given in Arun Sharma!!!!!!!
find the highest power of 175 which divides 344!!!!!!!!
Arun sharma has made a note to be careful. and in this case, we need to check the num,ber of 5^2 and number of 7's.. but i am not able to understand it clearly....
Thanking in advance..........
we can write 175 = 5^2 x 7
now we'll find the number of 5's and 7's in 344!
the number of 5's will be 83
and the number of 7's will be 57.
now to get get 175 we require 5^2 (i.e. two 5 is required) and one 7 is required.
In 344! we have 83 5's...we can make atmost 41 pairs of 5^2
(83/2= 41.5)
and we have 57 7's...
but since number of 5's is less...
therefore the highest power will be 41 { i.e. (5^2x7)^41 }
hope its clear now..why we have to check for 5^2 and 7....
How many 4-digit numbers divisible by 7 have at least two zeroes?
OPTIONS 1)34 2)32 3)30 4)28 5)37
Hi PUYS,
I found a website where the number of factors of a number which are perfect squares and perfect cubes is given in a very lucid manner and also a formula is given. The address is
VA-QA-AR-NAS-PF-Application 6 Vedic Aptitude
VA-QA-AR-NAS-PF-Application 7 Vedic Aptitude
How many 4-digit numbers divisible by 7 have at least two zeroes?
OPTIONS 1)34 2)32 3)30 4)28 5)37
i am getting total numbers as 31 (am i missing some numbers...)
the numbers are 1400,2100,2800............9800-(13 numbers)
1001,2002,3003............9009(9 numbers)..
1050,2030,3010,4060,5040,6090,8050,9030,3080(9 numbers)...
i am getting total numbers as 31 (am i missing some numbers...)
the numbers are 1400,2100,2800............9800-(13 numbers)
1001,2002,3003............9009(9 numbers)..
1050,2030,3010,4060,5040,6090,8050,9030,3080(9 numbers)...
When 2 zeros, three possibilities:-
i) a00b
b - a should be divisible by 7
when b = a, 9 cases
b - a = 7, 2 cases
a - b = 7, 2 cases
13 cases
ii) ab00
ab should be a multiple of 7
=> 13 such numbers
iii) a0b0
=> 3b - a should be be divisible by 7 (a should not be 0)
=> 3b = 7k + a
=> (a, b) can be (1, 5), (2, 3), (3, 1), (3,
, (4, 6), (5, 4), (6, 2), (6, 9), (7 7), (8, 5)=> 10 such numbers
When 3 zeros, one possibility
a000
=> one such number
=> Total = 37 such numbers
Hi friends!!!!!!
can some one explain Euler's technique......
hey puys,
pls help in solving this!!
Find 'n' in (2/3)^10 + (3/2)^10 = (2/5)^(2n-2)..
Q)find the sum of all the co-primes to 2016 which are less than 2016
can some1 help me with the solution of numbers question 65 LOD-II
Q.If you form a subset of integers chosen from 1-3000 such that no 2 integers add upto a multiple of 9.what can be the max number of elements in this subset.
a.1688 b.1332 c.1333 d.1334
abhi108890 SaysQ)find the sum of all the co-primes to 2016 which are less than 2016
Hi.. there is a formula for ques lik these...
To find out the sum of all the co-primes to N which are lesser than N goes like,
Sum = (N/2)*N*(1-(1/a)) (1-(1/b) and so on.. where N= a^m . b^n ....so on..where a,b,c etc are prime factors..
So here
2016= (2^5) * 7 * 9
Sum= 2016*1008* (1-(1/7)) * (1-(1/2))* (1-(1/9)) = 774144..
Correct me if i'm wrong!!
Number of co-primes to N that are lesser than N is N*(1-(1/a)) (1-(1/b) and so on..
can some1 help me with the solution of numbers question 65 LOD-II
Q.If you form a subset of integers chosen from 1-3000 such that no 2 integers add upto a multiple of 9.what can be the max number of elements in this subset.
a.1688 b.1332 c.1333 d.1334
This clearly implies that none of the multiples of 9 can be chosen for the subset..
Next, take 1.. now numbers lik 8, 17, etc cant come.. so any form 9n-1 shouldnt be chosen..
Similarly for 2, 9n-2 numbers cant be taken.. Similarly for 3 and 4, numbers 9n-3 and 9n-4 will not come..
So by observing, we conclude like numbers as (1,2,3,4) , (10,11,12,13) , (19,20,21,22), wil go on like this..
So 9n+1, 9n+2, 9n+3, 9n+4 terms till 3000 can be chosen..
So totally 1332 numbers + 4 = 1336... (1,2,3,4 nos.)
But the last number
So 1336 -1 = 1335..
And 1 multiple of 9 can be chosen as it wont affect de condition..
So 1336.. and excluding the values 1 and 3000
the answer is 1334...
can some1 help me with the solution of numbers question 65 LOD-II
Q.If you form a subset of integers chosen from 1-3000 such that no 2 integers add upto a multiple of 9.what can be the max number of elements in this subset.
a.1688 b.1332 c.1333 d.1334
This clearly implies that none of the multiples of 9 can be chosen for the subset..
Next, take 1.. now numbers lik 8, 17, etc cant come.. so any form 9n-1 shouldnt be chosen..
Similarly for 2, 9n-2 numbers cant be taken.. Similarly for 3 and 4, numbers 9n-3 and 9n-4 will not come..
So by observing, we conclude like numbers as (1,2,3,4) , (10,11,12,13) , (19,20,21,22), wil go on like this..
So 9n+1, 9n+2, 9n+3, 9n+4 terms till 3000 can be chosen..
So totally 1332 numbers + 4 = 1336... (1,2,3,4 nos.)
But the last number
So 1336 -1 = 1335..
but no such option.. pls help me wher i went wrong!!
I cant make out that particular step(the step in bold)...
my method:
we can take (1,2,3,4), (10,11,12,13),...
so 1,10,19...2989 this series has 332 terms
...& 4,13,22,...,2992 has also 332 terms
so total we get = 332 * 4 = 1328
in the next recursion, we only get 2998, 2999 & 3000
so 1328 + 3 = 1331
but to maximize, we can always take any 1 single multiple of 9 , which will suddenly go with the condition...
so max no will be = 1331 + 1 = 1332
correct me if I am wrong...
can some1 help me with the solution of numbers question 65 LOD-II
Q.If you form a subset of integers chosen from 1-3000 such that no 2 integers add upto a multiple of 9.what can be the max number of elements in this subset.
a.1688 b.1332 c.1333 d.1334
Are you sure about the options, as answer should be 1336
Since sum of no two numbers can be a multiple of 9, we can say that we can have all numbers of form 9n + 1, 9n + 2, 9n + 3, 9n + 4 and one number of form 9n
Numbers of form 9n + 1 = 334
Numbers of form 9n + 2 = 334
Numbers of form 9n + 3 = 334
Numbers of form 9n + 4 = 333
So, maximum number of elements in the set = 334*3 + 333 + 1 = 1336
Edited after seeing the next post
the options r
(1)1668 (2)1332 (3)1333 (4)1334
though I found a slight change in the language...the sum says "integers chosen from between 1-3000"
OA is given as 1333
Then answer should be 1334, as we have to exclude 1 and 3000
can some1 help me with the solution of numbers question 65 LOD-II
Q.If you form a subset of integers chosen from 1-3000 such that no 2 integers add upto a multiple of 9.what can be the max number of elements in this subset.
a.1688 b.1332 c.1333 d.1334
Are you sure about the options, as answer should be 1336
Since sum of no two numbers can be a multiple of 9, we can say that we can have all numbers of form 9n + 1, 9n + 2, 9n + 3, 9n + 4 and one number of form 9n
Numbers of form 9n + 1 = 334
Numbers of form 9n + 2 = 334
Numbers of form 9n + 3 = 334
Numbers of form 9n + 4 = 333
So, maximum number of elements in the set = 334*3 + 333 + 1 = 1336
the options r
(1)1668 (2)1332 (3)1333 (4)1334
though I found a slight change in the language...the sum says "integers chosen from between 1-3000"
OA is given as 1333
ya answer is 1333....