1840 United States presidential election in Maine

Main article: 1840 United States presidential election

1840 United States presidential election in Maine

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Nominee William Henry Harrison Martin Van Buren Party Whig Democratic Home state Ohio New York Running mate John Tyler none Electoral vote 10 0 Popular vote 46,612 46,190 Percentage 50.23% 49.77%

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

President before election Martin Van Buren Democratic Elected President William Henry Harrison Whig

A presidential election was held in Maine on November 2, 1840 as part of the 1840 United States presidential election. ^[1] Voters chose ten representatives, or electors to the Electoral College, who voted for president and vice president.

Maine voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won the state by a very narrow margin of 0.46%.

Maine was typically a Democratic state during the Second Party System, however, with Harrison narrowly winning the state, this would be the only time that a Whig presidential candidate would win Maine.

Results

1840 United States presidential election in Maine [2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Whig	William Henry Harrison of Ohio	John Tyler of Virginia	46,612	50.23%	10	100.00%
	Democratic	Martin Van Buren of New York	Richard M. Johnson of Kentucky	46,190	49.77%	0	0.00%
Total				92,802	100.00%	10	100.00%

See also

• United States presidential elections in Maine

References

1. Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
2. "1840 Presidential General Election Results - Maine". U.S. Election Atlas. Retrieved December 23, 2013.