1828 United States presidential election in Maine

Summary

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%

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 edit

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 edit

References edit

  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.