An old woman goes to market and a horse steps on her basket and crushes all the eggs. The rider offers to pay for the damages and asks her how many eggs she had brought. She does not remember the exact number, but when she had taken them out two at a time, there was one egg left. The same happened when she picked them out three, four, five, and six at a time, but when she took them seven at a time there were none left over. What is the smallest number of eggs she could have had?
Let \(x\) = number of eggs in her basket We need to find the minimum positive integer, \(x\), that satisfies : \[ x = 2a+1\\ x = 3b+1\\ x = 4c+1\\ x = 5d+1\\ x = 6e+1\\ x = 7f~~~~~~ \]
should the one for 7 also have a +1?
nope, because `but when she took them seven at a time there were none left over. `
oh okay! so when I'm doing the math to figure all of this out, am I doing 2*1+1 and so on to see which ones do not give me remainders??
You have the right idea which works but that would be very painful what if the oldie has million eggs in her basket ? would you be testing all million numbers ?
noooo lol
Look at this number : \[n = 2*3*4*5*6\] whats the remainder when \(n\) is divided by 2 ?
0!
and what about the remainders when \(n\) is divided by 3, 4, 5 and 6
0 for all of those too!
good, before it gets boring can you find a smaller positive number that is exactly divisible by 2, 3, 4, 5 and 6 ?
90??
let me rephrase the question : what is the "least" positive number that is divisible by 2, 3, 4, 5 and 6 ?
so you're telling me she had 90 eggs in that basket?
not at all
wait never mind 90 is divisible by all of them except 4
btw the least number is not 90 you can do better, try again
here is a hint : "lcm"
13?!
that fails when u divide by 5
wait no because there is R1
hm...
whats the lcm of 2,3,4,5,6 ?
2
Okay thats wrong Never heard of "lcm" and "gcd" before ?
would lcm be 720?
or 60?
60 is the lcm, yes im pretty sure you knew these lcm and gcd stuff
haha yeah, I'm just thinking way too into everything now sorry about that!
its fine so if the oldie has 60 eggs in her basket, then if she takes out 2 eggs at a time, 0 eggs will be left in the end. but we want 1 egg in the end, so let the number of eggs be 60+1 = 61 and see if it works
what are the remainders when you divide 61 by 2, 3, 4, 5, and 6 ?
i got R1 for all except 4, 4 was R5
are u really sure ?
hold on!
whah messed up big time there! theyre all R1
good so \(61\) satisfies all the equations except the last one \[x=7f\]
so then there are 61 eggs!
or do i need to adjust my answer since 61/7 isn't a perfect fit?
Exactly! we need to adjust it so that last equation is also satisfied
Notice that any number of form \(60k+1\) leaves the remainder \(1\) when divided by 2, 3, 4, 5, or 6
plugin k=1, 2, 3 till you get a number thats divisible by 7
k=2 : 60(2) + 1 = 121 not divisible by 7 k=3 : 60(3) + 1 = 181 not divisible by 7 k=4 : 60(4) + 1 : 241 not divisible by 7 k=5 : 60(5) + 1 : 301 divisible by 7 so the least possible eggs that the old woman could be having is 301
5!!
sorry, it took me a little while but i actually did get that answer!
thank you so much for your help and especially your patience!! i really appreciate it :)
np:)
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