1836 United States presidential election in New York

1836 United States presidential election in New York
	November 7–9, 1836
Turnout	70.5% 13.7 pp
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

A presidential election was held in New York from November 7–9, 1836 as part of the 1836 United States presidential election.^[2] 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

1836 United States presidential election in New York^[3]
Party		Candidate	Running mate	Popular vote		Electoral vote
Party		Candidate	Running mate	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%

^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
^ "Presidential Elections". Weekly Messenger. November 12, 1836.
^ "1836 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved December 23, 2013.