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?
1well see that 6,5,4,3,2 all divide 120 (derived in my previous post) so the number has to be of the form
120k-1 (so as to give remainders 5,4,3,2,1 respectively )
further 120k-1 div. by 7 (given in problem)
thus we get
7|(120k-1)
7|119k+k-1
but 7|119k
so
7|(k-1)
7|k-1+7 => 7|(k+6)
so we get
k+6=7n
k=7n-6
1i gave the smallest possible number the general form would be as follows
120k-1 where k is of the form 7n-6 n= 1 to inf.
9also post a mathematical way of doing rather than trial and error method
this is a famous problem of 7thcentury AD
9can u find the set of all such nos possible ???
not very tough as uve come this far
9actually this has multiple ans !!!! there are many more ans
1u see the number should be of the form of x-1 where 6,5,4,3,2 all divide x and thus giuve remainders 1 less
so first we write x=2*
for 3 we write x=2*3*
for 4 we write x=2*3*2
for 5 we write x=2*3*2*5
for six observe the above is already divisible
thus
2*3*2*5-1
1I got it by the method thats it.u got the question back from an answer and i got the answer from the question.
1All the even moltiples of 7 gets cancelled.and even those ending with 5 and those multiple of three. and so by trial and error method for the rest u get the answer as 119.
1119/7=17...
119/6=19+5...
119/5=23+4....
119/4=29+3....
119/3=39+2....
119/2=59+1....
1hey see it is a multiple of seven. WHen she counted by the 5 there we 4 left. so the last digit of the no has to be 4 or 9.
When she counted by 2 there was 1 left over so it is an odd no.
Hence 4 cant be the lst digit .so 9 is the last digit
