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 42 representatives, or electors to the Electoral College, who voted for President and Vice President.
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| Turnout | 70.5%[1]  13.7 pp | |||||||||||||||||||||||||
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 County Results
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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%. Saratoga County would not vote Democratic again until 1964.
Results
edit| 1836 United States presidential election in New York[2] | ||||||||
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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% | ||||
See also
editReferences
edit- ^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
- ^ "1836 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved December 23, 2013.
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