United States presidential election in New York, 1836
Main article: United States presidential election, 1836
Elections in New Jersey | ||||||||
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The 1836 United States presidential election in New York took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose forty-two representatives, or electors to the Electoral College, who voted for President and Vice President.
New York voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won New York by a margin of 9.26%.
Results
United States presidential election in New York, 1836[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 166,795 | 54.63% | 42 | 100.00% | ||
Whig | William Henry Harrison of Ohio | Francis Granger of New York | 138,548 | 45.37% | 0 | 0.00% | ||
Total | 305,343 | 100.00% | 42 | 100.00% | ||||
References
- ↑ "1836 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 23 December 2013.
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