Broken EggsDate: 12/12/96 at 18:46:01 From: Chris Walker Subject: Broken eggs I have had this problem for a month and I can't figure out what to do. Maybe you can help. A farmer is taking her eggs to the market in a cart, but she hits a pothole, which knocks over all the containers of eggs. Though she is unhurt, every egg is broken. So she goes to her insurance agent, who asks her how many eggs she had. She says she doesn't know, but she remembers somethings from various ways she tried packing the eggs. When she put the eggs in groups of two, three, four, five, and six there was one egg left over, but when she put them in groups of seven they ended up in complete groups with no eggs left over. What can the farmer figure from this information about the number of eggs she had? Is there more than one answer? I have already tried multiples of seven, but I can't figure the answer out. Could you please tell me how to get the answer and also tell me what it is? Thanks for your help. Date: 12/12/96 at 19:41:38 From: Doctor Wilkinson Subject: Re: Broken eggs This kind of problem is pretty tricky if you've never seen anything like it before. What you're looking for is a number that leaves a remainder of 1 when divided by 2, 3, 4, 5, or 6, but that is evenly divisible by 7. So as you already noticed, you only need to look at multiples of 7. Let's try 7. We get off to a good start. 7 leaves a remainder of 1 when divided by 2, 3, or 6. Unfortunately it doesn't work for 4. So let's see what we would have to do to get to a number which leaves a remainder of 1 when you divide it by 4, without losing what we've already got. Well, if you take a number that leaves a remainder of 1 when you divide by 2 and add 2 to that number, you get another number that leaves a remainder of 1. So if you add any multiple of 2, you still get a number that leaves a remainder of 1 when you divide by 2. Simliarly, if you take a number that leaves a remainder of 1 when you divide it by 3 and add any multiple of 3 to it, you get another number that leaves a remainder of 1 when you divide it by 3. Anything that leaves a remainder of 1 when divided by 2 or 3 automatically leaves a remainder of 1 when you divide by 6, so we get that for free. So if you take 7 and add anything which is a multiple of 2 and 3 and 7 to it, you get another multiple of 7 which leaves a remainder of 1 when you divide it by 2 or 3. So we don't lose anything if we add a multiple of 42 (that's 2 x 3 x 7) to 7. So lets' see if we can satisfy another of our conditions. 7 + 42 = 49. Hey, that's good! We've got a number that leaves a remainder of 1 when you divide by 4. It still doesn't work for 5, though. Can we use the same trick again? If we add a multiple of 4 we still get a number that leaves a remainder of 1 when you divide by 4. So if we add a multiple of 3 x 4 x 7, we'll get another number that satsifies all the conditions so far. There's only one to go! Okay, 49 + 84 = 133. No good. 133 + 84 = 217. No good. 217 + 84 = 301. Bingo!! So 301 is a possible solution. Are there any others? Well, yes, because you can add a multiple of 3 x 4 x 5 x 7 to get another one. But usually what you're looking for is the smallest solution, so 301 is probably the answer you want. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 08/05/97 at 18:03:26 From: Rachel Subject: Word problem - algebra? The problem: A woman with a basket of eggs finds that if she removes either 2, 3, 4, 5, or 6 at a time from the basket, there is always one egg left over. If she removes 7 eggs at a time from the basket, there are no eggs left over. If the basket holds up to 500 eggs, how many eggs does she have? I've solved the problem using trial and error. I found multiples of 7, and divided 2, 3, 4, 5, and 6 into all the multiples of 7. 301 seemed to be the answer because it is a multiple of 7, and when divided by 2, 3, 4, 5, 6, there was always a remainder of one. Is there a simpler and quicker way of solving this problem? Date: 08/12/97 at 09:09:34 From: Doctor Rob Subject: Re: Word problem - algebra? More systematic, perhaps, but simpler? You be the judge. Let N be the number. Since it leaves a remainder of 1 on division by 6, it must be of the form N = 6*M + 1, where M is an integer. (We willassume that all variables below are integers, too.) This also takes care of division by 2 and 3, since N = 6*M + 1 = 2*(3*M) + 1 = 3*(2*M) + 1. Since N leaves a remainder of 1 on division by 5, it must be of the form N = 5*L + 1 = 6*M + 1, so 5*L = 6*M, so L = 6*K and M = 5*K. Thus N = 30*K + 1. (This uses the fact that 5 and 6 have no common factor.) Furthermore, N leaves a remainder of 1 on division by 4, so N = 4*J + 1 = 30*K + 1, so 4*J = 30*K, or 2*J = 15*K. This means that J = 15*I and K = 2*I, so N = 60*I + 1. (This uses the fact that 2 and 15 have no common factor.) At this point, you could try values of I = 0, 1, ..., until you found a solution. Alternatively, you could continue in the following vein. Finally N leaves a remainder of 0 on division by 7, so N = 7*H = 60*I + 1 = (7*8 + 4)*I + 1, 7*(H - 8*I) = 4*I + 1, 7*(2*H - 16*I) = 8*I + 2, 7*(2*H - 17*I) - 2 = I, so put 2*H - 17*I = G, and I = 7*G - 2, so H = 60*G - 17. Then N = 420*G - 119. A number is of the above form if and only if it satisfies the divisibility conditions of the problem. The smallest positive answer occurs when G = 1, N = 301, which is the answer you found the long way. If you ever learn modular arithmetic, you will find that there is a systematic way of solving all such problems using the Chinese Remainder Theorem. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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