Q: find the no. of ways of painting a decagon ( 3d) can be painted with 7 different colors?
@techgeek2050 said:bhai can u plz explain the part in bold.. ?
by 2(5k+3) i meant that when you add 2 of the smallest 5k+3 form ke numbers you get (5k+3) + (5k+3) = 10k+6 = 10k+5+1 or 5k+1 form
and when u add another 5k+3 form ka number to it you get (5k+1) + (5k+3) = 5k+4 form ka number
so basically to find the smallest such number multiply the smallest 5k+3 form ka number with 3 ie 13*3 = 39..
but we can also get 5k+4 by adding two 5k+2 form ke numbers -> 17+17 = 34
so the smallest 5k+4 form ka number that we have is 34, hence the largest 5k+4 form ka number we cannot have is 34-5 = 29
hope it is clear now
@sumit99 said:A cloth merchant says due to slump in the market, he sells the cloth at 5% loss, but he uses a fake metre scale and actually gains 5%. What is the length of the scale?A- 92.75 cmB-90.25 cmC-1900/21D-90.8
1900/21
cost of cloth that he sell 0.95y
its real value xy/1000
0.95y - xy/1000/xy/1000 = 5/20
x = 1900/21
cost of cloth that he sell 0.95y
its real value xy/1000
0.95y - xy/1000/xy/1000 = 5/20
x = 1900/21
@Logrhythm said:I did it more intuitively rather mathematically n(n+1) + 1n*(n+1) are 2 consecutive terms and co-prime they would have different factor(s) and +1 won't make it so that it gets to a perfect square...or you can do it this way...n^2+n+1 = k^24k^2 - (2n+1)^2 = 3(2k+2n+1)(2k-2n11) = 3 = 3*12k+2n+1 = 3 and 2k-2n-1 = 1 (k=1 and n=0)2k+2n+1 = 1 and 2k-2n-1 = 3 (k=1 and n=-1)hence, only 2 such integer values exist
question says n has to be natural number to 0 hona chahiye na
Let k be the number of five-element subsets that can be chosen from the set of the first 14 natural numbers so that at least two of the five numbers are consecutive. Find the remainder when k is divided by 1000.?
@abhishek.2011 said:question says n has to be natural number to 0 hona chahiye na
0 is not taken as a natural number i guess
@abhishek.2011 said:question says n has to be natural number to 0 hona chahiye na
0 whole number me ata hai na :P
@albiesriram said:Let k be the number of five-element subsets that can be chosen from the set of the first 14 natural numbers so that at least two of the five numbers are consecutive. Find the remainder when k is divided by 1000.?
14C4 mod 1000=1??
@ScareCrow28 said:0 whole number me ata hai na
arey bhai n^2 + n + 1 has to be a perfect square for a natural n
u r putting n as 0 and -1 , which u cant do as n is natural here ???????
m i wrong????
@abhishek.2011 said:arey bhai n^2 + n + 1 has to be a perfect square for a natural nu r putting n as 0 and -1 , which u cant do as n is natural here ???????m i wrong????
Achaaa... If you say, n can't be
You are right..
@abhishek.2011 said:arey bhai n^2 + n + 1 has to be a perfect square for a natural nu r putting n as 0 and -1 , which u cant do as n is natural here ???????m i wrong????
@ScareCrow28 said:Achaaa... If you say, n can't be You are right..
mujhe aap loagon ki discussion samajh nahi aayi coz mein neend mein hun.....but i was sure (when i solved) that for natural numbers answer wld be 0 and for integers answer would be 2
@albiesriram said:no..
choose two consecutive nos=13 ways
chose any 3 from the rest 12=12c3
12c3*13 mod 1000=860??
@albiesriram said:Let k be the number of five-element subsets that can be chosen from the set of the first 14 natural numbers so that at least two of the five numbers are consecutive. Find the remainder when k is divided by 1000.?
(14c5 - 10c5) % 1000 = 750 ??
select 5 elements in 14c5 ways and then remove where 5 elements are not together....
for the second part i assumed 5 x's and 9 y's.....so these 5 x's can be placed between the 9 y's (10 gaps) in 10c5 ways
so 14c5 - 10c5 = 1750 % 1000 = 750
is this correct?? i am off to sleep...gn
@Logrhythm said:mujhe aap loagon ki discussion samajh nahi aayi coz mein neend mein hun.....but i was sure (when i solved) that for natural numbers answer wld be 0 and for integers answer would be 2
yup thats what m saying , i posted answer as 0 to this question just because n has to be natural number :)
@albiesriram said:Let k be the number of five-element subsets that can be chosen from the set of the first 14 natural numbers so that at least two of the five numbers are consecutive. Find the remainder when k is divided by 1000.?
@Logrhythm said:(14c5 - 10c5) % 1000 = 750 ??select 5 elements in 14c5 ways and then remove where 5 elements are not together....for the second part i assumed 5 x's and 9 y's.....so these 5 x's can be placed between the 9 y's (10 gaps) in 10c5 waysso 14c5 - 10c5 = 1750 % 1000 = 750is this correct?? i am off to sleep...gn
750 is correct
14C5-10C5=1750/1000=> 750 ansBTW Gd mrg :)
14C5-10C5=1750/1000=> 750 ans
BTW Gd mrg :)
When playing the lottery there are 6 different balls ranging between 0 and 60 that can be chosen from. How many different possibilities are there when picking lottery numbers?
@iLoveTorres said:When playing the lottery there are 6 different balls ranging between 0 and 60 that can be chosen from. How many different possibilities are there when picking lottery numbers?
i think that is 6 and not 60............
if it is 6 then......... 2^6 -1=63..
if it is 6 then......... 2^6 -1=63..
@Dexian said:i think that is 6 and not 60............if it is 6 then......... 2^6 -1=63..
it is 60 only.