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1828 United States presidential election in Maine

From Wikipedia, the free encyclopedia

1828 United States presidential election in Maine
Flag of the United States (1822-1836).svg

← 1824 October 31 – December 2, 1828 1832 →
 
JohnQAdams.jpg
Andrew Jackson.jpg
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%

President before election

John Quincy Adams
Democratic-Republican

Elected President

Andrew Jackson
Democratic

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 President and Vice President.

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 8 electoral votes and Jackson received 1.

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 4 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 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

References

  1. ^ "1828 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1828 Presidential General Election Results - Maine". U.S. Election Atlas. Retrieved 28 February 2013.
This page was last edited on 18 August 2019, at 17:20
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