1840 United States presidential election in Maine
Appearance
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Elections in Maine |
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The 1840 United States presidential election in Maine took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose ten representatives, or electors to the Electoral College, who voted for President and Vice President.
Maine voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Maine by a margin of 0.46%.
Results
United States presidential election in Maine, 1840[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 46,612 | 50.23% | 10 | 100.00% | ||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 46,190 | 49.77% | 0 | 0.00% | ||
Total | 92,802 | 100.00% | 10 | 100.00% |
References
- ^ "1840 Presidential General Election Results - Maine". U.S. Election Atlas. Retrieved 23 December 2013.