1828 United States presidential election in Maine

Main article: 1828 United States presidential election

The 1828 United States presidential election in Maine took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for the president and vice president.

1828 United States presidential election in Maine

← 1824 October 31 – December 2, 1828 1832 →

Nominee John Quincy Adams Andrew Jackson Party National Republican Democratic Home state Massachusetts Tennessee Running mate Richard Rush John C. Calhoun Electoral vote 8 1 Popular vote 20,773 13,927 Percentage 59.71% 40.03%

County Results Adams   50-60%   60-70%   70-80% Jackson   50-60% No Data/Vote

President before election John Quincy Adams Democratic-Republican Elected President Andrew Jackson Democratic

Maine voted for the National Republican candidate, John Quincy Adams, over the Democratic candidate, Andrew Jackson. Adams won Maine by a margin of 19.68%. Adams received eight electoral votes and Jackson received one.

With 59.71% of the popular vote, Maine would prove to be Adams' fifth strongest state in the 1828 election after Rhode Island, Massachusetts, Vermont and Connecticut.^[1]

This was the only time prior to the 2016 election (when Republican nominee Donald Trump received one of the state's four votes) that an electoral vote split occurred in Maine.

Results

1828 United States presidential election in Maine[2]
Party		Candidate	Votes	Percentage	Electoral votes
	National Republican	John Quincy Adams (incumbent)	20,773	59.71%	8
	Democratic	Andrew Jackson	13,927	40.03%	1
	N/A	Other	89	0.26%	0
Totals			34,789	100.0%	9

See also

• United States presidential elections in Maine

References

1. "1828 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
2. "1828 Presidential General Election Results - Maine". U.S. Election Atlas. Retrieved February 28, 2013.