Date: Wed, 30 Nov 1994 14:39:06 -0500 (EST) From: Anthony DAuria Subject: I have a problem for you to solve!!!! Dear Dr. Math or one of his henchmen, I have a good problem for you to try to solve. The problem requires an algebraic formula and it is really easy to solve. Be careful, this problem is very TRICKY!!!!! Here is the problem that I would like you to solve and write back with an answer. Thank you for your time. I hope that this will benefit your studies in any way. Thank you! Find the largest possible 2 integers such that the larger integer is more than 3 less than 3 times the smaller one. HINT: No guessing or trial or error is involved. Anthony D'Auria
Date: Wed, 30 Nov 1994 19:06:40 -0500 From: Dr. Math (Margaret Patterson) Subject: Anthony's question Hi Anthony - Thanks for writing to us. My name is Margaret and I am one of the Doctors Math. The other doctors are other women and men at Swarthmore College. I couldn't tell from your question whether or not you have already solved this problem or if it is just for our enrichment. We only answer questions that you need help with, not ones that you already understand. Well, this is going to be tricky, first notice that there are many possible pairs of numbers for which the larger is more than3 less than 3 times the smaller one. let x and y be the numbers you are looking for. Let x > y. Then we know that x > 3y-3 or x > 3(y-1) Some possibilities are x=10, y=2 or x=100, y=30, or even x = 1000 y=300 But, you have asked for the largest possible numbers for which this is true. I afraid there is no limit to how big the numbers can be. The only constraint is that y be about 1/3 of x. But x could be any number - so I don't think there is an answer to your question. Please write back if you have more questions about this or another problem. -Margaret Patterson, Math Doctor on call
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