United States presidential election in Illinois, 1836
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Elections in Illinois | ||||||||
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The 1836 United States presidential election in Illinois took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.
Illinois voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Illinois by a margin of 9.38%.
Results
United States presidential election in Illinois, 1836[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic | Martin Van Buren | 18,369 | 54.69% | 5 | |
Whig | William Henry Harrison | 15,220 | 45.31% | 0 | |
Totals | 33,589 | 100.0% | 5 | ||
References
- ↑ "1836 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved 4 August 2012.
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