Jerry wrote down the ones digit of each of the first 2025 positive squares: 1,4,9,6,5,6,…. What is the sum of all the numbers Jerry wrote down?
Answer Choices:
A. 9025
B. 9070
C. 9090
D. 9115
E. 9160
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Jerry wrote down the ones digit of each of the first 2025 positive squares:
12,22,32,…,20252.
The ones digit of n2 depends only on the ones digit of n. For n=0,1,…,9 we have
ones digit of nones digit of n200112439465566798491
So in each block of 10 consecutive integers, the ones digits of the squares are
1,4,9,6,5,6,9,4,1,0,
whose sum is
1+4+9+6+5+6+9+4+1+0=45.
We are considering the first 2025 squares, i.e. 12 through 20252. Since
2025=202⋅10+5,
there are 202 full blocks of 10 squares, plus 5 extra squares. The 202 full blocks contribute
202⋅45=9090
to the sum of the ones digits. The remaining 5 squares are 20212,20222,20232,20242,20252, which have the same ones digits as 1,4,9,6,5. Their sum is
1+4+9+6+5=25.
Therefore, the total sum of the ones digits Jerry wrote down is
9090+25=(D) 9115.
The problems on this page are the property of the MAA's American Mathematics Competitions