What is the ones digit of the sum
⌊ 1 k ⌋ + ⌊ 2 k ⌋ + ⌊ 3 k ⌋ + ⋯ + ⌊ 2025 k ⌋ ? ⌊ k 1 ⌋ + ⌊ k 2 ⌋ + ⌊ k 3 ⌋ + ⋯ + ⌊ k 2 0 2 5 ⌋ ?
(Recall that ⌊ x ⌋ ⌊ x ⌋ x x
Answer Choices:
A. 1 1 2 2 3 3 5 5 8 8
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We want the ones digit of
S = ⌊ 1 ⌋ + ⌊ 2 ⌋ + ⋯ + ⌊ 2025 ⌋ . S = ⌊ 1 ⌋ + ⌊ 2 ⌋ + ⋯ + ⌊ 2 0 2 5 ⌋ .
For each positive integer m m
m 2 ≤ n < ( m + 1 ) 2 ⟹ ⌊ n ⌋ = m . m 2 ≤ n < ( m + 1 ) 2 ⟹ ⌊ n ⌋ = m .
On this interval, there are ( m + 1 ) 2 − m 2 = 2 m + 1 ( m + 1 ) 2 − m 2 = 2 m + 1 n n m m ≤ 2025 ≤ 2 0 2 5
4 5 2 = 2025. 4 5 2 = 2 0 2 5 .
Thus, the sum is
S = 45 + ∑ m = 1 44 m ( 2 m + 1 ) = 45 + 2 ∑ m = 1 44 m 2 + ∑ m = 1 44 m = 45 + 2 ( 44 ) ( 44 + 1 ) ( 2 ⋅ 44 + 1 ) 6 + 44 ⋅ ( 44 + 1 ) 2 S = 4 5 + m = 1 ∑ 4 4 m ( 2 m + 1 ) = 4 5 + 2 m = 1 ∑ 4 4 m 2 + m = 1 ∑ 4 4 m = 4 5 + 2 6 ( 4 4 ) ( 4 4 + 1 ) ( 2 ⋅ 4 4 + 1 ) + 2 4 4 ⋅ ( 4 4 + 1 )
Since we only need the ones digit, we compute everything modulo 10 1 0
S ≡ 45 + 2 ( 44 ) ( 44 + 1 ) ( 2 ⋅ 44 + 1 ) 6 + 44 ⋅ ( 44 + 1 ) 2 10. S ≡ 4 5 + 2 6 ( 4 4 ) ( 4 4 + 1 ) ( 2 ⋅ 4 4 + 1 ) + 2 4 4 ⋅ ( 4 4 + 1 ) m o d 1 0 .
Therefore,
S ≡ 45 + ( 44 ) ( 15 ) ( 89 ) + 22 ⋅ 45 ≡ (D) 5 10. S ≡ 4 5 + ( 4 4 ) ( 1 5 ) ( 8 9 ) + 2 2 ⋅ 4 5 ≡ (D) 5 m o d 1 0 .
The problems on this page are the property of the MAA's American Mathematics Competitions