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Mathematics 19 Online
OpenStudy (anonymous):

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?

OpenStudy (anonymous):

Did it all out and got 21 mod 105 . so i think its wrong !! not sure where i went wrong though

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

its ment to be done using CRT

OpenStudy (anonymous):

@Lcoohill what did you do so far?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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).

OpenStudy (anonymous):

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 !

OpenStudy (anonymous):

Good luck:-)

OpenStudy (anonymous):

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 ??

OpenStudy (anonymous):

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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