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1836 United States presidential election in New York

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Main article:1836 United States presidential election
1836 United States presidential election in New York

← 1832November 7–9, 18361840 →
← KY
LA →
Turnout70.5%[1]Decrease 13.7pp
 
NomineeMartin Van BurenWilliam Henry Harrison
PartyDemocraticWhig
Home stateNew YorkOhio
Running mateRichard M. JohnsonFrancis Granger
Electoral vote420
Popular vote166,795138,548
Percentage54.63%45.37%

County results

Van Buren

  50–60%
  60–70%
  70–80%

Harrison

  50–60%
  60–70%

Tie

  


President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic


Elections in New York
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Apresidential election was held inNew York from November 7–9, 1836 as part of the1836 United States presidential election.[2] Voters chose 42 representatives, or electors to theElectoral College, who voted forPresident andVice President.

New York voted for theDemocratic candidate,Martin Van Buren, overWhig candidateWilliam Henry Harrison. Van Buren won New York by a margin of 9.26%.Saratoga County would not vote Democratic again until 1964.

Results

[edit]
1836 United States presidential election in New York[3]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
DemocraticMartin Van Buren ofNew YorkRichard M. Johnson ofKentucky166,79554.63%42100.00%
WhigWilliam Henry Harrison ofOhioFrancis Granger ofNew York138,54845.37%00.00%
Total305,343100.00%42100.00%

See also

[edit]

References

[edit]
  1. ^Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
  2. ^"Presidential Elections".Weekly Messenger. November 12, 1836.
  3. ^"1836 Presidential General Election Results - New York". U.S. Election Atlas. RetrievedDecember 23, 2013.
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