Lingyis said:
the 332-232 problem:
really, really, trivial. just a repeated application of (a2-b2) = (a-b)(a+b)
from that procedure, however, you get only 3 prime numbers under 100: 5, 13 and 97. now there's the less trivial part of finding a 4th prime. unless you count 1 as a prime, which you really shouldn't do.
if i were in a high school math competition, due to time pressure, i would take advantage of the question saying 4 prime numbers under 100. the easiest thing would be to calculate the next factor, which is 38+28. this is reasonably fast to calculate even without a calculator, and you get 6561+256=6817. this number happens to be easily factorizable, by inspection--clearly, 6817=17*401. (both 68 and 17 are multiples of 17). and i would move on to the next question, having gotten what the question asks for: 5, 13, 17 and 97.
there might be more sub-100 factors left in the 216+316 part, but in a competition you would just move on. i verified that this number has factors 3041 and 14177, i.e. no sub-100 primes. there might be an elegant way of showing that 216+316 does not have a sub-100 prime.
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Spot on! This is exactly my same procedure! Congratulations on finding the answer!
Basically you have 332-232, which means (316+216)(316-216). Then you grab (316-216) and factor it until you get the following:
(316+216)(38+28)(34+24)(34-24).
(34+24) is 97 which is a prime number below 100.
(34-24) is 65, which has two factors, 5 and 13, both prime numbers below 100.
Now, (318+218) is equal to 6817, which has two factors, 17 and 401. 17 is a prime number below 100.
And then, we have our answers.
Great work Lingyis! That was the procedure I was looking for.
During the test I was only able to obtain the 5 and the 13 because I was doing it through a completely alternate method. I arrived at the (right) conclusion that when we have 32x-22x, x being any natural number, 5 is a multiple the answer. When we have 34x-24x, 13 is a multiple of the answer. I arrived at the same conclusion for 97 when you have 38x-28x, but had contradictory answers so I left it with the first two.
Thinking back though, this problem is not hard at all and I could have been able to do it in under 20 minutes. That's more or less what it took me to do it on paper a few moments ago.
I'll have to leave soon. Here I'm going to post a much harder problem. It's from the Fourth International Olympiads. I take pride in being able to do it without help and under an hour. They normally give 4:30 hours to complete 3 problems from this Olympiads. Here goes:
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In an international olympiads awards ceremony there were m number of medals to be given in n days, n>1. The first day, 1 medal was given plus 1/7 of the resting medals. The second day, 2 medals were given plus 1/7 of the resting medals. And so on consecutively. The n-day, the n number of medals that were left were given. Find the number of medals and the number of days.
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Good luck!
