anonymous
  • anonymous
Challenge question from the 2015 Euclid competition. Two lists each consist of 6 consecutive positive integers. The smallest integer in the first list is a, the smallest integer in the second list is b, and a
Mathematics
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SOLVED
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chestercat
  • chestercat
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terenzreignz
  • terenzreignz
I haven't a clue, but if you guys don't mind, I'm going to "think loudly" on this thread. Apologies in advance if I appear to state the obvious XD
anonymous
  • anonymous
Hint: Consider the clues in the order presented, develop lists of (a,b) pairs and eliminate possibilities as you work through the clues. Clue 1: There are only two pairs of positive integers that multiply to give 49.
terenzreignz
  • terenzreignz
I got those. Hang on ^^

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terenzreignz
  • terenzreignz
a must be an integer from 1-6, it seems
anonymous
  • anonymous
There will be more than one pair of numbers that meet all the conditions, but you are correct. Keep going.
terenzreignz
  • terenzreignz
Well, if a is 1, b has to be 49. Anything else would put a multiple of 64 in the third set
anonymous
  • anonymous
Well done. That's one possibility, but there are others.
terenzreignz
  • terenzreignz
Thanks. I'm not normally this slow, but OS is dreadfully laggy on my end T.T Please have patience :>
terenzreignz
  • terenzreignz
If a = 2, there will be no 1 in the first set and so the only way for there to be a 49 in the third is if there was a 7 in the second set. If b = 7, then all conditions are met (maybe barely) with 84 being the element of the third set which surpasses 75 so a = 2 and b = 7 is possible. If b were any less than 7, then there will be no element greater than 75 in the third set as they will be bounded by 72.
anonymous
  • anonymous
Fantastic, @terenzreignz . You got another one! Not finished yet, however.
terenzreignz
  • terenzreignz
Without a 1 in the first set (let's just call it A), there must be a 7 in both sets (let's call the other one B) If 2 < a < 7, then 8 will be in A. This means there must not be an 8 in B or else there will be a 64 in the third set (let's call it C) The only way for there to be no 8 in B is if b = 2. But this will mean a > b which is not allowed. So I think I'm done?
anonymous
  • anonymous
The first bullet tells us that 49 is the product of an integer in the first list and an integer in the second list. Since 49 = 72 and 7 is prime, then these integers are either 1 and 49 or 7 and 7. If 1 is in one of the lists, then either a = 1 or b = 1. Since 1 ≤ a < b, then it must be that a = 1. If 49 is in the second list, then one of b, b + 1, b + 2, b + 3, b + 4, b + 5 equals 49, and so 44 ≤ b ≤ 49. Therefore, for 1 and 49 to appear in the two lists, then (a, b) must be one of (1, 49),(1, 48),(1, 47),(1, 46),(1, 45),(1, 44) . If 7 appears in the first list, then one of a, a + 1, a + 2, a + 3, a + 4, a + 5 equals 7, so 2 ≤ a ≤ 7. Similarly, if 7 appears in the second list, then 2 ≤ b ≤ 7. Therefore, for 7 to appear in both lists, then, knowing that a < b, then (a, b) must be one of (2, 3),(2, 4),(2, 5),(2, 6),(2, 7),(3, 4),(3, 5),(3, 6),(3, 7),(4, 5),(4, 6),(4, 7),(5, 6),(5, 7),(6, 7) .
terenzreignz
  • terenzreignz
Oops ^^
terenzreignz
  • terenzreignz
I missed (2,6) ?
anonymous
  • anonymous
The second bullet tells us that no pair of numbers in the first and second lists have a product that is a multiple of 64. Given that the possible values of a and b are 1, 2, 3, 4, 5, 6, 7, 44, 45, 46, 47, 48, 49, then the possible integers in the two lists are those integers from 1 to 12, inclusive, and from 44 to 54, inclusive. (For example, if the first number in one list is 7, then the remaining numbers in this list are 8, 9, 10, 11, 12.) There is no multiple of 32 or 64 in these lists. Thus, for a pair of integers from these lists to have a product that is a multiple of 64, one is a multiple of 4 and the other is a multiple of 16, or both are multiples of 8. If (a, b) = (1, 48),(1, 47),(1, 46),(1, 45),(1, 44), then 4 appears in the first list and 48 appears in the second list; these have a product of 192, which is 3 · 64. If (a, b) = (1, 49), there is a multiple of 4 but not of 8 in the first list, and a multiple of 4 but not of 8 in the second list, so there is no multiple of 64 in the third list. If (a, b) = (3, 4),(3, 5),(3, 6),(3, 7),(4, 5),(4, 6),(4, 7),(5, 6),(5, 7),(6, 7), then 8 appears in both lists, so 64 appears in the third list. If (a, b) = (2, 3),(2, 4),(2, 5),(2, 6),(2, 7), then there is no multiple of 8 or 16 in the first list and no multiple of 16 in the second list, so there is no multiple of 64 in the third list. Therefore, after considering the first two bullets, the possible pairs (a, b) are (1, 49),(2, 3), (2, 4),(2, 5),(2, 6),(2, 7).
anonymous
  • anonymous
The third bullet tells us that there is at least one number in the third list that is larger than 75. Given the possible pairs (a, b) are (1, 49),(2, 3),(2, 4),(2, 5),(2, 6),(2, 7), the corresponding pairs of largest integers in the lists are (6, 54),(7, 8),(7, 9),(7, 10),(7, 11),(7, 12). The corresponding largest integers in the third list are the products of the largest integers in the two lists; these products are 324, 56, 63, 70, 77, 84, respectively. Therefore, the remaining pairs (a, b) are (1, 49),(2, 6),(2, 7) Having considered the three conditions, the possible pairs (a, b) are (1, 49),(2, 6),(2, 7).
terenzreignz
  • terenzreignz
I fail T.T But that's how we learn ^_^
anonymous
  • anonymous
Congrats, @terenzreignz ! You got all three possibilities. Well done!

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