1836 United States presidential election in New York
(Redirected from United States presidential election in New York, 1836)
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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