1900 United States presidential election in North Dakota

1900 United States presidential election in North Dakota

← 1896 November 6, 1900 1904 →
 
Nominee William McKinley William Jennings Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Theodore Roosevelt Adlai Stevenson I
Electoral vote 3 0
Popular vote 35,898 20,531
Percentage 62.12% 35.53%

County Results
McKinley
  40-50%
  50-60%
  60-70%
  70-80%
  80-90%


President before election

William McKinley
Republican

Elected President

William McKinley
Republican

The 1900 United States presidential election in North Dakota took place on November 6, 1900. All contemporary 45 states were part of the 1900 United States presidential election. Voters chose three electors to the Electoral College, which selected the president and vice president.

North Dakota was won by the Republican nominees, incumbent President William McKinley of Ohio and his running mate Theodore Roosevelt of New York. They defeated the Democratic nominees, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan and his running mate, former Vice President Adlai Stevenson I. McKinley won the state by a margin of 26.59% in this rematch of the 1896 presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley to score a decisive victory.

With 62.12% of the popular vote, North Dakota would be McKinley's second strongest victory in terms of percentage in the popular vote after Vermont.[1] McKinley had previously won the state four years earlier, but more than doubled his margin from 1896. Bryan would later lose North Dakota again to William Howard Taft in 1908.

  1. ^ "1900 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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