1836 United States presidential election in New York

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1836 United States presidential election in New York
Flag of New York (1778-1901).svg
  1832 November 3 - December 7, 1836 1840  
Turnout70.5% [1] Decrease2.svg 13.7 pp
  Martin Van Buren circa 1837 crop.jpg William Henry Harrison by James Reid Lambdin, 1835 crop.jpg
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard Mentor Johnson Francis Granger
Electoral vote420
Popular vote166,795138,548
Percentage54.63%45.37%

New York Presidential Election Results 1836.svg
County Results

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

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.

Contents

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

1836 United States presidential election in New York [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 166,79554.63%42100.00%
Whig William Henry Harrison of Ohio Francis Granger of New York 138,54845.37%00.00%
Total305,343100.00%42100.00%

See also

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References

  1. Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
  2. "1836 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 23 December 2013.