21Easy one..
hint:
number is of the form 2k+1, 3a+2,4b+3, 5c+4+ 6c+5
which can be written as 2q-1,3w-1,4e-1,5t-1,6e-1 and also a divisible by 7
(From Brahmagupta, 7th century A.D.)
A girl was carrying a basket of eggs, and a man driving a horse hit the basket and broke all the eggs. Wishing to pay for the damage, he asked the girl how many eggs there were. The girl said she did not know, but she remembered that when she counted them by twos, there was one left over; when she counted them by threes, there were two left over; when she counted them by fours, there were three left over; when she counted them by fives, there were four left; and when she counted them by sixes, there were five left over. Finally, when she counted them by sevens, there were none left over. `Well,' said the man, `I can tell you how many you had.' What was his answer?
using chinese remainder theorem it cant be solved..........................
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13 Answers
211using division algorithm no. is of form 2k+1,3k+2,4k+3,5k+4,6k+5,7k
now by trying a few values(based on 2k+1,5k+4 and 7k form) we shortlist our possible cases.
and then its easy to check that answer is 119.
1mine solution was based on trial and error....so too time taking..generalising by trial and error would take even longer....how did u generalise shubhodip?
49oh subhodip had already posted! :P
dekho what he said is :
the number is 2k-1 or 3l-1 or 4m-1 or 5n-1 or 6p-1
thus the number next to the reqd no is divisible by 2,3,4,5,6
L.C.M = 60
so the reqd no is 60k-1
119 is the only number of form 60k-1 that is divisible by 7! :D
1that's fine....that much i also got..but i did not get how he came up with the underlined part(see below)..
numbers of the form 60(4k+2) - 1 are solutions. k = 0,1,2,....
49My answer "119 is the only number of form 60k-1 that is divisible by 7! :D" is a bit wrong
but subhodip ka ye 60(4k+2) - 1 wala baat to mujhe sahi nahi lag rha..! :s
I think answer is
60(7k+2) - 1 :D
1brothers and sister[3]
i don't want GENERALISED ANSWERS.i want how did u generalise?
21@ subhomoy plz show your working
im getting a big but correct ans.... i.e. 14219
21@ pirtish plz look your tagline says "I am an donkey" how can you be "an" donkey, you can be "a" donkey.........................but not an......... ok? according to the rules of eng. gram...........plz try to rectify it very soon.................
and look at this link given below:
http://captnemo.in/projects/iitjee/2011.html
and see if your name is "Thotakuri Sree Meghana" because its the name given for AIR 52 IITJEE'11
106in response to #8 ... showing only the generalization (think its a bit lengthy though)
required no is of the form 60k-1 and 7m ... k,m are whole nos.
60k = 7m+1
=> k = (7m+1)/60
k is an integer => 7m+1 ends with 0
=> 7m ends with 9
=> m is of the form 10n + 7 ... n is a whole no.
=> k = (70n + 50)/60
=> k = (7n+5)/6
7n is odd => n = odd = 2p + 1 ... p is a whole no.
=> k = (14p + 12)/6
=> k = 7p/3 + 2
7p/3 is an integer => p = 3t ... t is a whole no.
=> k = 7t + 2
=> the set of solutions is the set 60(7t+2)-1
=> Solution set 420t + 119 where t = 0,1,2,.....