@scrabbler said:Shayad 70 aana chahiye, trying to solve orally Someone please do the hard work and confirm... @ChirpiBirdregardsscrabbler
Sir, aap bhi aise bologe!!!! ๐
sry for spamming
Sir, aap bhi aise bologe!!!! ๐
sry for spamming
@jain4444 said:How many 4-term geometric sequences have all four terms positive integers less than or equal to 100?
Considering increasing and decreasing GP's same, we can write the four terms as
a, ar, ar^2, ar^3
Clearly 'a' has to be a positive integer, but key fact is that 'r' need not to be. :)
Let r = p/q where p and q are co-prime positive integers and also p > q as I am considering increasing GPs only.
So the four terms of GP become: a, a(p/q), a(p/q)^2, a(p/q)^3
It can be easily inferred, for each term of GP to be a positive integer, that a is of the form k(q)^3.
Now checking cases:
(Ai) q = 1, p = 2 and a varies from 1 to 12 i.e. 12 sequences
{1, 2, 4, 8}, {2, 4, 8, 16}, {3, 6, 12, 24}, {4, 8, 16, 32}, {5, 10, 20, 40}, {6, 12, 24, 48}, {7, 14, 28, 56}, {8, 16, 32, 64}, {9, 18, 36, 72}, {10, 20, 40, 80}, {11, 22, 44, 88}, {12, 24, 48, 96}
(Aii) q = 1, p = 3 and a varies from 1 to 3 i.e. 3 sequences
{1, 3, 9, 27}, {2, 6, 18, 54}, {3, 9, 27, 81}
(Aiii) q = 1, p = 4 and a can be 1 only i.e. 1 sequence
{1, 4, 16, 64}
(B) q = 2, p = 3 and a = k(2)^3 = 8k where k varies from 1 to 3 i.e. 3 sequences
{8, 12, 18, 27}, {16, 24, 36, 54}, {24, 36, 54, 81}
(C) q = 3, p = 4 and a = k(3)^3 = 27k where k can be i only i.e. 1 sequence
{27, 36, 48, 64}
So total GPs possible are: 12 + 3 + 1 + 3 + 1 = 20 :)
I hope there is no missing case.
Team BV - Kamal Lohia
a, ar, ar^2, ar^3
Clearly 'a' has to be a positive integer, but key fact is that 'r' need not to be. :)
Let r = p/q where p and q are co-prime positive integers and also p > q as I am considering increasing GPs only.
So the four terms of GP become: a, a(p/q), a(p/q)^2, a(p/q)^3
It can be easily inferred, for each term of GP to be a positive integer, that a is of the form k(q)^3.
Now checking cases:
(Ai) q = 1, p = 2 and a varies from 1 to 12 i.e. 12 sequences
{1, 2, 4, 8}, {2, 4, 8, 16}, {3, 6, 12, 24}, {4, 8, 16, 32}, {5, 10, 20, 40}, {6, 12, 24, 48}, {7, 14, 28, 56}, {8, 16, 32, 64}, {9, 18, 36, 72}, {10, 20, 40, 80}, {11, 22, 44, 88}, {12, 24, 48, 96}
(Aii) q = 1, p = 3 and a varies from 1 to 3 i.e. 3 sequences
{1, 3, 9, 27}, {2, 6, 18, 54}, {3, 9, 27, 81}
(Aiii) q = 1, p = 4 and a can be 1 only i.e. 1 sequence
{1, 4, 16, 64}
(B) q = 2, p = 3 and a = k(2)^3 = 8k where k varies from 1 to 3 i.e. 3 sequences
{8, 12, 18, 27}, {16, 24, 36, 54}, {24, 36, 54, 81}
(C) q = 3, p = 4 and a = k(3)^3 = 27k where k can be i only i.e. 1 sequence
{27, 36, 48, 64}
So total GPs possible are: 12 + 3 + 1 + 3 + 1 = 20 :)
I hope there is no missing case.
Team BV - Kamal Lohia
@vishcat said:guys pls help me with this---->what is the coefficient of x^2*y^3*z^5 in (x+y+z)^10?
A general rule for questions like co-efficient of
(p^a)(q^b)(r^c)(s^d)............... in (p+q+r+s...........)^n is given by
C(n,a) * C(n-a,b) * C(n-a-b,c) * C(n-a-b-c,d) *................
there is a very nice pattern if you observe hence not very difficult to remember. For the given question co-efficient is C(10,2)*C(10-2,3)*C(10-2-3,5)
=C(10,2)*C(8,3)*(5,5)
=45*56*1 = 2520.
ATDH.
@iLoveTorres said:85515?
Nopes...I just calculated, it's not matching...Pls check again
Team BV - Kamal Lohia
Team BV - Kamal Lohia
@bodhi_vriksha said:Nopes...I just calculated, it's not matching...Pls check againTeam BV - Kamal Lohia
there was a small calculation mistake.. is it 85510?
@iLoveTorres said:there was a small calculation mistake.. is it 85510?
It's still not matching. It'd be better if you post your complete solution.
Team BV - Kamal Lohia
Team BV - Kamal Lohia
@bodhi_vriksha said:Considering increasing and decreasing GP's same, we can write the four terms as a, ar, ar^2, ar^3Clearly 'a' has to be a positive integer, but key fact is that 'r' need not to be. Let r = p/q where p and q are co-prime positive integers and also p > q as I am considering increasing GPs only.So the four terms of GP become: a, a(p/q), a(p/q)^2, a(p/q)^3It can be easily inferred, for each term of GP to be a positive integer, that a is of the form k(q)^3.Now checking cases:(Ai) q = 1, p = 2 and a varies from 1 to 12 i.e. 12 sequences {1, 2, 4, 8}, {2, 4, 8, 16}, {3, 6, 12, 24}, {4, 8, 16, 32}, {5, 10, 20, 40}, {6, 12, 24, 48}, {7, 14, 28, 56}, {8, 16, 32, 64}, {9, 18, 36, 72}, {10, 20, 40, 80}, {11, 22, 44, 88}, {12, 24, 48, 96}(Aii) q = 1, p = 3 and a varies from 1 to 3 i.e. 3 sequences{1, 3, 9, 27}, {2, 6, 18, 54}, {3, 9, 27, 81}(Aiii) q = 1, p = 4 and a can be 1 only i.e. 1 sequence{1, 4, 16, 64}(B) q = 2, p = 3 and a = k(2)^3 = 8k where k varies from 1 to 3 i.e. 3 sequences{8, 12, 18, 27}, {16, 24, 36, 54}, {24, 36, 54, 81}(C) q = 3, p = 4 and a = k(3)^3 = 27k where k can be i only i.e. 1 sequence{27, 36, 48, 64}So total GPs possible are: 12 + 3 + 1 + 3 + 1 = 20 I hope there is no missing case.Team BV - Kamal Lohia
sir they are considering decreasing GP too and GP's whose common difference is 1
so , final answer is 20*2 + 100 = 140
source :- https://brilliant.org/i/qocXKC/
@bodhi_vriksha said:It's still not matching. It'd be better if you post your complete solution.Team BV - Kamal Lohia
a+b+c = 2014
so the longest side can be 2014/2-1=1006
Now a'+b'+c' = 3*1006=3018
3018 is 3018-2014=1004 greater
so a'+b'+c'=1004
This will have (1004+2)C2 solutions possible = 505515
Now finding the ordered pairs
case 1) of the form aab = 3 times
possibilities (0,0,1004)(1,1,1002).....(502,502,0) so totally 503 possibilities
case 2) of the form abc = 3! = 6 times
505515=6x+(503)*3
6x=504006
x=84001
the required number of triangles = 84001+503=84504?
so the longest side can be 2014/2-1=1006
Now a'+b'+c' = 3*1006=3018
3018 is 3018-2014=1004 greater
so a'+b'+c'=1004
This will have (1004+2)C2 solutions possible = 505515
Now finding the ordered pairs
case 1) of the form aab = 3 times
possibilities (0,0,1004)(1,1,1002).....(502,502,0) so totally 503 possibilities
case 2) of the form abc = 3! = 6 times
505515=6x+(503)*3
6x=504006
x=84001
the required number of triangles = 84001+503=84504?
@jain4444 said:sir they are considering decreasing GP too and GP's whose common difference is 1 so , final answer is 20*2 + 100 = 140 source :- https://brilliant.org/i/qocXKC/
Increasing and Decreasing GPs can be considered different, no issues at all. :)
Regarding GP's with common difference 1, you asked about GP's with 'distinct' terms but 'Brilliant' is not saying anything about that. So they will certainly consider it but in your question, it shouldn't be.
So final answer to question posed by you should be 2*20 = 40 only considering increasing and decreasing GPs distinct. :)
Team BV - Kamal Lohia
Regarding GP's with common difference 1, you asked about GP's with 'distinct' terms but 'Brilliant' is not saying anything about that. So they will certainly consider it but in your question, it shouldn't be.
So final answer to question posed by you should be 2*20 = 40 only considering increasing and decreasing GPs distinct. :)
Team BV - Kamal Lohia
@jain4444 said:The total number of Integer solutions of x ลy หz ยด=9720000
9720000 = 2^6 * 5^4 * 7^3
x^3 = 2^3a * 5^3b * 7^3c
y^2 = 2^2p * 5^2q * 7^2r
z^4 = 2^4x * 5^4y * 7^4z
3a + 2p + 4x = 6
=> 3 solutions
3b + 2q + 4y = 4
=> 2 solutions
3c + 2r + 4z = 3
=> 1 solution
and we could have 2 values for "y" and "z"
so , total solutions = 3*2*1*2*2 = 24
@iLoveTorres said:a+b+c = 2014so the longest side can be 2014/2-1=1006Now a'+b'+c' = 3*1006=30183018 is 3018-2014=1004 greaterso a'+b'+c'=1004This will have (1004+2)C2 solutions possible = 505515Now finding the ordered pairscase 1) of the form aab = 3 times possibilities (0,0,1004)(1,1,1002).....(502,502,0) so totally 503 possibilities case 2) of the form abc = 3! = 6 times505515=6x+(503)*36x=504006x=84001the required number of triangles = 84001+503=84504?
Now you got it finally....:)
But there was no need of such hectic calculations.
Here are my same steps with no (least) calculations
a' + b' + c' = 1004
C(1006, 2) = 3*503 + 6x = 1005*503
i.e. 6x = 1002*503
i.e. x = 167*503
Ans = x + 503 = 168*503 = 84504 :)
Team BV - Kamal Lohia
But there was no need of such hectic calculations.
Here are my same steps with no (least) calculations
a' + b' + c' = 1004
C(1006, 2) = 3*503 + 6x = 1005*503
i.e. 6x = 1002*503
i.e. x = 167*503
Ans = x + 503 = 168*503 = 84504 :)
Team BV - Kamal Lohia
@scrabbler said:No a+b + c = 15 then b + 5c = how many possible values? is what you must solve.regardsscrabbler
Yes..that's what Vineet meant. I just edited it above..
Team BV - Kamal Lohia
Team BV - Kamal Lohia
One question from my side before leaving - In a locality three newspapers are read; The Don, Jan Nayak, Lok Neta. Out of 120 families in the locality 93 has subscribed The Don, 79 has subscribed Jan Nayak and 85 has subscribed Lok Neta.
Find the maximum number of families that may have subscribed
(i) only Jan Nayak
(ii) only Lok Neta
(iii) Jan Nayak and Lok Neta but not The Don
Team BV - Kamal Lohia
PS: We are starting our CAT workshops from next weekend and first topic is SETs only, that's why the question. Whoever is interested may pm.
@nramachandran said:A person walking takes 26 steps to come down on a escalator and it takes 30 seconds for him for walking. The same person while running takes 18 second and 34 steps. How many steps are there ?
26 + 30e = 34 + 18e
e = 2/3
no. of steps = 26 + 30(2/3) = 46
@nramachandran said:What is the number of distinct triangles with integral valued sides and perimeter 14?
4 ?
@bodhi_vriksha said:One question from my side before leaving - In a locality three newspapers are read; The Don, Jan Nayak, Lok Neta. Out of 120 families in the locality 93 has subscribed The Don, 79 has subscribed Jan Nayak and 85 has subscribed Lok Neta.Find the maximum number of families that may have subscribed(i) only Jan Nayak(ii) only Lok Neta(iii) Jan Nayak and Lok Neta but not The DonTeam BV - Kamal LohiaPS: We are starting our CAT workshops from next weekend and first topic is SETs only, that's why the question. Whoever is interested may pm.
i)27
ii)27
iii)34
ii)27
iii)34
@nramachandran said:What is the number of distinct triangles with integral valued sides and perimeter 14?
4
@bodhi_vriksha said:Ext. Qs. What is the number of distinct triangles with integral valued sides and perimeter 2014?Team BV - Kamal Lohia
84504
N = 121212121212..................300 digits. find the remainder when N is divided by 99 ?
1.) 18 2.) 1 3.) 98 4.) none of these.
Regards,
pm89