Between 1 to 3000 may mean that both exclusive . Then also we can take 9n+2 as must and any 3 among other 7 forms and one 9n So 334+333*3+1=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!!
ya both the nos. shoud be excluded.. and also i forgot to add one multiple of 9.. and 1334 is the answer..
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
kritikunal Says
ya answer is 1333....
How it's 1333..:lookround: pls explain!! if possible, post the answer/approach given in the book!!
Aset is formed by including some of the first one tousand natural numbers. S contains the maximum number of numbers such that they satisfy the following conditions: 1. No number of the set S is prime. 2.when the numbers of the set are selected 2 at a time, we always see co prime numbers. ANS=12
Q70 how many triangular numbers less than 1000 have the property that they are the difference of squares of two consecutive natural numbers.
Aset is formed by including some of the first one tousand natural numbers. S contains the maximum number of numbers such that they satisfy the following conditions: 1. No number of the set S is prime. 2.when the numbers of the set are selected 2 at a time, we always see co prime numbers. ANS=12
Q70 how many triangular numbers less than 1000 have the property that they are the difference of squares of two consecutive natural numbers.
ans=22
plzz explain....
thanks in advance:lookround:
Solution:
1. The solution should be the square of the prime numbers. The numbers are: 4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961
But i'm getting ly 11 numbers. wat i missed?
2. this question is interesting.
The triangular numbers: 1 = 1 1+2 = 3 1+2+3 = 6 1+2+3+4 = 10 so, generalising, the number can be obtained by n(n+1)/2. By the condition given, it should be
n(n+1)/2 by trail and error method, we get, n now, consider two consecutive no. a, a-1 (a)^2 - (a-1)^2 = 2a + 1 so, the no. should be of the form 2a-1
now, n= 2a-1 n 2a -1 a so, there are 22 possible values.
for the 1st question even i dont understand y the ans is diff..maybe the ans given in the book is wrong...if u get the correct ans do post it here..thanx..
1. The solution should be the square of the prime numbers. The numbers are: 4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961
But i'm getting ly 11 numbers. wat i missed?
2. this question is interesting.
The triangular numbers: 1 = 1 1+2 = 3 1+2+3 = 6 1+2+3+4 = 10 so, generalising, the number can be obtained by n(n+1)/2. By the condition given, it should be
n(n+1)/2 by trail and error method, we get, n now, consider two consecutive no. a, a-1 (a)^2 - (a-1)^2 = 2a + 1 so, the no. should be of the form 2a-1
now, n= 2a-1 n 2a -1 a so, there are 22 possible values.
for the first question:
squares of first 11 prime nos. and 1( 1 is not a prime no. but can be a co-prime no.) answer will be 1,4,9,25,49,121,169,289,361,529,841,961
you can always see that the power is always 2. So, for any (x+y)^n or (x+y+z)^n, on expressing, every individual ll have the power added to n. see, binomial theorem.
now, (a+b+c)^20 = a^x.b^y.c^z + .....+ 21 terms it ll take every possible value of x, y, z such tat it's sum is 20. only whole number. so, x+y+z=20. have to count how mny possibles are available.
x y z 0 0 20 0 1 19 . . 0 20 0 21 terms 1 0 19 1 1 18 . . 1 19 0 20 terms
it'll go like that, and the anwer = 21+20+19+....+1 = 231
using Euler's for remainder... 9^11^13 / 7 rem = ? or 2^11^13 / 7 rem = ? (9/7 rem = 2) now, 2 and 7 are coprimes. (we will use euler's) And, phi(7) = 6. So, (2^6k)/7 rem = 1. Now, we have to express the power of 2 (i.e.,11^13) in the form of 6k+p, so that we can have it in the form (2^6k)*(2^p)/7 and we can find the rem.
Again, (in order to express it as 6k+p) 11^13/6 rem = ? phi(6) = 2. And 11 and 6 are coprimes. So, 11^2k/6 rem = 1 => 11^13 = (11^12)*(11^1) => 11^13/6 rem = (11^12/6 rem) * (11/6 rem) => (Since 11^2k/6 rem = 1) 11^13/6 rem = 1*5 = 5 So, 11^13 can be written in the form of 6k+5
Now, 9^11^13/7 rem = 2^11^13/7 rem = 2^(6k+5)/7 rem = 1*(2^5)/7 rem = 32/7 rem = 4
