1836 United States presidential election in New York
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| ← 1832 | November 3 – December 7, 1836 | 1840 → |
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| Turnout | 70.5%[1]  13.7 pp |
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| |  |  | | Nominee | Martin Van Buren | William Henry Harrison | | | Party | Democratic | Whig | | Home state | New York | Ohio | | Running mate | Richard Mentor Johnson | Francis Granger | | Electoral vote | 42 | 0 | | Popular vote | 166,795 | 138,548 | | Percentage | 54.63% | 45.37% | |
 County Results | Van Buren 50–60% 60–70% 70–80% | Harrison 50–60% 60–70% | |
President before election
Andrew Jackson
Democratic | Elected President
Martin Van Buren
Democratic | |
| Elections in New York State |
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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 42 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%. Saratoga County would not vote Democratic again until 1964.
Results
See also
References
- ^ 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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- See also
- Presidential elections
- Senate elections
- House elections
- Gubernatorial elections
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13.7 pp
