A housewife is travelling to market with all her eggs in one basket. She has between 100 and 200 eggs in the basket. Counting in threes there are 2 eggs leftover, counting in fives there is 1 egg leftover and counting in sevens there are 2 eggs leftover. How many eggs are in the basket?
Did it all out and got 21 mod 105 . so i think its wrong !! not sure where i went wrong though
This isn't a methodical solution but an inspection of the tables of the three numbers suggests that 191 is a solution. Since 191/7 gives a remainder 2. 191/5, remainder =1, 191/3, remainder = 2. One can also do this by looking at multiples of 21 = 3*7 between 100 and 200, adding 2 to each of them ( since we get a remainder 2 in both cases ) and dividing the resulting numbers by 5 to see which ones give a remainder of 1.
its ment to be done using CRT
@Lcoohill what did you do so far?
i found out the m1, m2, m3 , y1,y2,y3 added them all togther with a1,a2,a3 etc but i didnt get the right answer
x mod 3 and 35 -> -23*3 + 2*35 = 1 x mod 5 and 21 -> -4*5 +1*21 =1 x mod 7 and 15 -> -2*7 +1*15 =1 so 2*70 + 1*21 + 2*15 = 191
well i got the minimum on to be 86 mod 105 now i am stuck as to what to do next. i know 191 is answer its just figuring out the method so can do it all out in an exam
No idea, you have evidently been taught some specific method for solving the congruences and I don't know what it is. The method i am using above is one version of the extended Euclidean algorithm but there are others (which give the same answers).
ok ya i realize there is different methods thanks for your help although i did get 86 mod 105 . and when you add 105+86 you get 191 !! so maybe i will try another one and see does it work out the same !
Good luck:-)
if had had same example between 100 & 200 eggs x=2 mod 3 x=1 mod 5 x=4 mod 7 what answer would you get solving your way estudier ??
ok i got 191 mod 105 first . then to get minimum i got 191 mod 105 which gave me 86 mod 105 . so i went a step too far !! cheers
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