Where did John Quincy Adams win the majority of the votes in the 1824 election?

Answer 1

In the 1824 presidential election an individual needed to receive at least 131 electoral votes out of the 261 total electoral votes. Andrew Jackson received 99 electoral votes, John Quincy Adams received 84 electoral votes, William Crawford received 41 electoral votes, and Henry Clay received 37 electoral votes. The popular vote totals were Andrew Jackson 151,271 (42.9%), John Quincy Adams 113,122 (32.1%), Henry Clay 47,531 (13.5%), and William Crawford 40,856 (11.6%). Since no candidate received the required majority of 131 electoral votes, the president was elected by the U.S. House of Representatives in accordance with the Twelfth Amendment of the United States Constitution. John Quincy Adams won the election in the U.S. House of Representatives. If no candidate receives a majority of electoral votes, the Twelfth Amendment of the United States Constitution provides that the U.S. House of Representatives will select the president, with each state delegation casting one vote Adams was elected President on February 9, 1825, on the first ballot, with 13 states, followed by Jackson with 7, and Crawford with 4. Adams received votes from the New England states Connecticut, Maine, Massachusetts, New Hampshire, Rhode Island, Vermont, and other states Illinois, Louisiana, Kentucky, Maryland, Missouri, New York, and Ohio. Jackson received votes from Alabama, Indiana, Mississippi, New Jersey, Pennsylvania, South Carolina, and Tennessee. Crawford received votes from Delaware, Georgia, North Carolina, and Virginia.