Omar made a list of all the arithmetic progressions of positive integer numbers such that the difference is equal to $2$ and the sum of its terms is $200$. How many progressions does Omar's list have?
Problem
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Tags: Arithmetic Progression, algebra
Lamboreghini
16.09.2021 23:15
Let the progression have $n$ terms and first term $a.$ Then, the last term is $a+2(n-1)=a+2n-2.$ The average of the first and last terms is thus $\frac{a+a+2n-2}{2}=a+n-1.$ So the sum of the terms is $an+n^2-1=200.$
Notice that if $n$ is even, then the left side is odd while the right side is even. So the number of terms in all the arithmetic progressions in Omar's list is odd.
Also, if $n$ is a multiple of $5,$ then the left side is one less than a multiple of $5$ while the right side is. So $n$ cannot be a multiple of $5.$
Math_Is_Fun_101
17.09.2021 03:24
We claim the answer is $6$. Suppose the first term of the AP is $a$ and that the progression has $k$ terms. Note that $a,k\in\mathbb Z^+$. Then, the sum of the terms of the AP is \[ a+(a+2)+\ldots+(a+2k-2)=ak+2(1+\ldots+k-1)=ak+(k-1)k=k(a+k-1). \]Thus, we must have $k(a+k-1)=200$. Now, since $a\ge1$, we have $a+k-1\ge k$, so $k$ must be the smallest factor. Now, note that $200=2^3\cdot5^2$ has $12$ positive divisors, giving $6$ possible values for $k$. Each value of $k$ corresponds to a single value of $a$, so each of these $6$ values gives one of the progressions in Omar's list. Since no other progression can be on the list by the work above, the answer is $6$. $\blacksquare$ Extra: The 6 progressions are \begin{align*} 200\\ 99,101\\ 47,49,51,53\\ 36,38,40,42,44\\ 18,20,22,24,26,28,30,32\\ 11,13,15,17,19,21,23,25,27,29 \end{align*}