1836 United States presidential election in New York
The 1836 United States presidential election in New York took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose 42 representatives, or electors to the Electoral College, who voted for President and Vice President.
.svg.png.webp){kind=small} | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| Turnout | 70.5%[1]  13.7 pp | |||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| Elections in New York State |
|---|
 |
New York voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won New York by a margin of 9.26%.
Results
| 1836 United States presidential election in New York[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 166,795 | 54.63% | 42 | 100.00% | ||
| Whig | William Henry Harrison of Ohio | Francis Granger of New York | 138,548 | 45.37% | 0 | 0.00% | ||
| Total | 305,343 | 100.00% | 42 | 100.00% | ||||
References
- Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
- "1836 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 23 December 2013.
State and district results of the 1836 United States presidential election | ||
|---|---|---|
 | ||
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.