1836 United States presidential election in Kentucky

Main article: 1836 United States presidential election

1836 United States presidential election in Kentucky

←  1832 November 7–9, 1836 1840  →

←  MI NY  →

Nominee William Henry Harrison Martin Van Buren Party Whig Democratic Home state Ohio New York Running mate Francis Granger Richard Mentor Johnson Electoral vote 15 0 Popular vote 36,861 33,229 Percentage 52.59% 47.41%

County Results Harrison   50–60%   60–70%   70–80%   80–90% Van Buren   50–60%   60–70%   70–80%   80–90%

The 1836 United States presidential election in Kentucky was held between November 7 and 9, 1836 as part of the 1836 United States presidential election. ^[1] Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.

Kentucky voted for Whig candidate William Henry Harrison over the Democratic candidate, Martin Van Buren. Harrison won Kentucky by a margin of 5.18%.

Results

1836 United States presidential election in Kentucky [2]
Party		Candidate	Votes	Percentage	Electoral votes
	Whig	William Henry Harrison	36,861	52.59%	15
	Democratic	Martin Van Buren	33,229	47.41%	0
Totals			70,090	100.0%	15

References

1. "Presidential Elections". Weekly Messenger. November 12, 1836.
2. "1836 Presidential General Election Results - Kentucky". U.S. Election Atlas. Retrieved August 4, 2012.