1836 United States presidential election in New York

Main article: 1836 United States presidential election

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.

1836 United States presidential election in New York

← 1832 November 3 – December 7, 1836 1840 →
Turnout	70.5%[1] 13.7 pp
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

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]
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

• United States presidential elections in New York

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 December 23, 2013.