Problem:
There exists a unique positive integer a for which the sum
U=n=1∑2023⌊5n2−na⌋
is an integer strictly between −1000 and 1000. For that unique a, find a+U. (Note that ⌊x⌋ denotes the greatest integer that is less than or equal to x.)
Solution:
Answer (944):
First, compute
V=n=1∑20235n2−an=302023⋅2024⋅4047−a⋅102023⋅2024=102023⋅2024⋅(1349−a)
For a positive integer k, let r(k) be the remainder when k is divided by 5 . Because 0≤r(k)≤4 for all positive integers k, it follows that
V−54⋅2023≤U=V−n=1∑20235r(n2−an)≤V
If −1000<U<1000, then −1000<V<1000+54⋅2023 and thus
1349−2023⋅202410000−20248<a<1349+2023⋅202410000
Therefore a=1349 and V=0. Note that
r(n2−1349n)=⎩⎪⎪⎨⎪⎪⎧012 if n≡0 or 4(mod5) if n≡2(mod5) if n≡1 or 3(mod5)
Hence
U=−n=1∑20235r(n2−1349n)=−405⋅(52+51+52)=−405.
The requested sum is 1349+(−405)=944.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions