1824 United States presidential election in Maine
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File:Maine Presidential Election Results 1824.svg County Results
Adams 50–60% 70–80% 80–90% 90–100%
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| Elections in Maine |
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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. 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
| 1824 United States presidential election in Maine[1] | |||||
|---|---|---|---|---|---|
| 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
Notes
References
- ↑ "1824 Presidential General Election Results - Maine". U.S. Election Atlas. Retrieved February 27, 2013.
