1836 United States presidential election in Delaware
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Elections in Delaware |
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The 1836 United States presidential election in Delaware took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
Delaware voted for Whig candidate William Henry Harrison over the Democratic candidate, Martin Van Buren. Harrison won Delaware by a margin of 6.54%.
Results
United States presidential election in Delaware, 1836[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Whig | William Henry Harrison | 4,736 | 53.24% | 3 | |
Democratic | Martin Van Buren | 4,154 | 46.70% | 0 | |
N/A | Other | 5 | 0.06% | 0 | |
Totals | 8,895 | 100.00% | 3 |
References
- ^ "1836 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 4 August 2012.