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# Some Arithmetic Sequences

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The sum of four positive integers that form an arithmetic sequence is 46. Of all such possible sequences, what is the greatest possible third term?

There is a sequence of numbers such that every entry except for the first entry is the arithmetic mean of its two neighboring entries. The 27th entry is 94 and the 94th entry is 27. What's the first entry?

How many distinct, non-equilateral triangles with a perimeter of 60 units have integer side lengths a, b, and c such that a, b, c is an arithmetic sequence?

All positive integers whose digits add up to 11 are listed in increasing order: 29, 38, 47, . . . What is the 30th number in that list?

Jan 5, 2019

#1
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1. Let the first term be a, and let the common difference be d. Then the four positive integers are a, a+d, a+2d, and a+3d. The sum of these four positive integers is 4a+6d=46, so 2a+3d=23. Solving for d, we find that =(23-2a)/3. The third term is a+2d=a+2(23-2a)/3=(46-a)/3. Thus, to maximize this expression, we minimize a, so a must be 1. Substituting 1 in, we find that the largest possible third term is 15.

2. 94 =F + 26D,
27 =F + 93D, solve for F, D
F =120 - First term
D = -1 - Common difference

3.  As we begin, we note that no side of the triangle can exceed 29 units in length. Let’s start at the 20–20–20 triangle. We first count the number of triangles that have 2 sides of length less than 20. Let the third side be of length 20+k where 2≤k≤9. This we do by taking away k units in length from the first and the second sides, by subtracting at least one unit from each. For each k, We can do it in ⌊k/2⌋ ways. Thus, counting for each possible k, we have 1+1+2+2+3+3+4+4=201+1+2+2+3+3+4+4=20 cases.

Let’s now go back to the 20–20–20 triangle and count the number of cases where only one side has a length <20<20 units. We can take away 1 to 18 units from this one side and redistribute it to the other two such that no side exceeds 29 units in length. Let us take away k units from this one side, such that 1≤k≤9. In order for the triangles to be distinct, the redistribution can be done in ⌊k/2⌋+1⌊k/2⌋+1 ways. Then, for all possible values of k in this range, we have 1+2+2+3+3+4+4+5+5=29 cases. Now, let us consider the cases where 9
Finally, we add up all our cases:

1+20+29+25=751+20+29+25=75 distinct triangles. Also, subtract one because of the equilateral one.

4. I really can't think of another way other than listing.

Jan 5, 2019
#2
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Thanks asdf1243 for the first two, but the 3rd one doesn't seem to be right.

Could you check your answer and if you notice anything you could tell me?

Jan 5, 2019
#3
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Hmm.
I have a very similar question as a matter of fact.
The only difference is that I am trying to compute the remainder when a_2017 is divided by 45.
Any thoughts?
I couldn't find where the 44 was implemented in guest's answers, otherwise I would have attempted to solve this problem by plugging in 2017 in to his/her's method.
It is clearly not viable to type all the first 2017 numbers

You could sum up all the digits from 0 to 9 as follows:
(0, 1 608, 1 236, 1 806, 2 408, 3 010, 3 606, 4 207, 4 808, 5 409)
Total = 28 098. If you sum up this number, it =27 which is divisible by 9. And since it ends in 5(2015), it is also divisible by 5.
So now, you only have 2 numbers to worry about:
20162017 mod 45 =37

Jan 6, 2019
#4
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Bruh Sudhish, have you really stooped down to these levels to look for aops HW answers? Bruh

Jan 12, 2019