Triplets consisting of three different numbers are formed from number 1 to 10.1. How many of these triplets are such that sum of numbers is divisible by 3.37,42,36,382.How many of triplets formed are such that sum of numbers is divisible by 9 and do not have 9 in them? 7,6,9,10
Consider the set S = {1, 2 , 3 €Ś €Ś.. 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?(1) 3 (2) 4 (3) 6 (4) 7
Consider the set S = {1, 2 , 3 €Ś €Ś.. 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?(1) 3 (2) 4 (3) 6 (4) 7
x- no. of elements (x>=3) d=999/(x+1)
999=3^3*37 so 8 factors
x+1 can take 8 values out of which 999 doesn't hold
Triplets consisting of three different numbers are formed from number 1 to 10.1. How many of these triplets are such that sum of numbers is divisible by 3.37,42,36,382.How many of triplets formed are such that sum of numbers is divisible by 9 and do not have 9 in them? 7,6,9,10
My Take 42 and 9
Method
1) numbers are of the form 3K 3K+1 3K+2
3K = 3
3K+1 = 4
3K+2 = 3
SO total numers = 3*4*3 = 36
@pankaj1988 : Yes bhai. Got confused with three different numbers .. first i had done 42 but ye dekhe thoda confuse hua isliye 6 cases nikaldiya 😛 😛
all 3k = 1
all 3k+1 = 4C3 = 4
All 3K+2 = 1
Edited : total 42
2) In order to be divisible by 9
lets take 3K1 3K2+1 3K2+2
so 3(K1+k2+K3) + 3 = 9K
Maximum sum we get = 7+8+6 = 21 without 9
Since number has to be divisble by 9 maximum sum = 18
So k1+k2+k3 = 5
So number of non negative solutions = 4C2 = 6
@krum bhai this is wrong method?? if i take k1+k2+k3 = 2 then i am getting 2 cases since k1 cannot be 0 which is the one more cases i am missing??
What are the no. of factors of 1000000 which are less then 1000 but do not divide 1000 ?@pankaj1988@anytomdickandhary@krum@mbajamesbond - aajao sir letz start nw
x- no. of elements (x>=3)d=999/(x+1)999=3^3*37so 8 factorsx+1 can take 8 values out of which 1,3 doesn't holdso 8-2=6
S = {1, 2, 3, €Ś.. 1000} We need AP €™s in which the first term is 1 and last term is 1000. The common difference must be a factor of (1000 - 1) = 999 = 3^3*37
There are 8 factors of 999. viz 1, 3, 9, 27, 37, 111, 333 and 999. As we need at least 3 terms, we can take any of these values expect 999, i.e., 7 values.
Consider the set S = {1, 2 , 3 €Ś €Ś.. 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?(1) 3 (2) 4 (3) 6 (4) 7
Consider the set S = {1, 2 , 3 €Ś €Ś.. 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?(1) 3 (2) 4 (3) 6 (4) 7
My Take 36 and 6Method1) numbers are of the form 3K 3K+1 3K+23K = 33K+1 = 43K+2 = 3SO total numers = 3*4*3 = 362) In order to be divisible by 9lets take 3K1 3K2+1 3K2+2so 3(K1+k2+K3) + 3 = 9KMaximum sum we get = 7+8+6 = 21 without 9Since number has to be divisble by 9 maximum sum = 18So k1+k2+k3 = 5So number of non negative solutions = 4C2 = 6Please point out if any mistake
Bhai in 1) 6 cases are missing. OA is 42...Chk for krum bhai post no 323...
aur 2nd samajh me nhi aa rha hai... mere pass jo soln hai usme cases banae hai...OA is 9
(n-1)d=999 therefore n-1=999/d so n-1 should have whole number values
that is possible only if d is one of the factors of 999 but if you take 999 as well then n-1=1 and number of terms only 2 since we need minm 3 terms so number of factors of 999-1(999 itself)
that is 8-1=7
this is the generalized approach for this kind of problems
dude what you have done is totient fn this gives no. of coprimes of 1000 and less that it.here it should go like no. of factors of 1000000=2^6*5^6=49no. of factors of 1000000 which are less than 1000 since it is the square root of 1000 should be 48/2 +1(1000 itself)=25and now number of factors of 1000=2^3*5^3=16so here answer should be 25-16 =9 what do you think about this? i hope @gs4890 sir will make it clear
generalized result for the Q -
(total no. of factors of n^2 + 1) / 2 - total no. of factors of n
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I guess I need to say ki OA mat batana seedhe solve kar dena..\m/
and PG ke Thread should be renamed to XAT thread ..:D
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we all r almost of same age :)