1824 United States presidential election in Maine
(Redirected from United States presidential election in Maine, 1824)
This article relies largely or entirely on a single source. (June 2022) |
The 1824 United States presidential election in Maine took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for president and vice president.
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County Results
Adams 50–60% 70–80% 80–90% 90–100%
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During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the presidency. Maine voted for John Quincy Adams over William H. Crawford. Adams won Maine by a margin of 63.0%.
Results
edit1824 United States presidential election in Maine[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic-Republican | John Quincy Adams | 10,289 | 81.50% | 9 | |
Democratic-Republican | William H. Crawford | 2,336 | 18.50% | 0 | |
Totals | 12,625 | 100.0% | 9 |
See also
editNotes
editReferences
edit- ^ "1824 Presidential General Election Results - Maine". U.S. Election Atlas. Retrieved February 27, 2013.