1840 United States presidential election in Kentucky
The 1840 United States presidential election in Kentucky took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.
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Elections in Kentucky | ||||||
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Kentucky voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Kentucky by a margin of 28.4%.
With 64.20% of the popular vote, Kentucky would prove to be Harrison's strongest state in the 1840 election.[1]
Results
United States presidential election in Kentucky, 1840[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 58,488 | 64.20% | 15 | 100.00% | ||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 32,616 | 35.80% | 0 | 0.00% | ||
Total | 91,104 | 100.00% | 15 | 100.00% |
gollark: Equality Concord.
gollark: That could never go wrong!
gollark: What if you give everyone brain implants which tie their concept of happiness to serving The State™?
gollark: With guns! That's how it always worked.
gollark: "I could harvest 50% more crops on this communal farm if I worked harder, but I get food anyway."
References
- "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1840 Presidential General Election Results - Kentucky". U.S. Election Atlas. Retrieved 23 December 2013.
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