John Brown
(1757-1837)

American lawyer and statesman, Virginia (1757-1837) – John Brown was born in Staunton (Independent city in Virginia, United States) on September 12th, 1757 and died in Lexington (City in and county seat of Fayette County, Kentucky, United States) on August 29th, 1837 at the age of 79. Today John Brown would be 266 years old.

Age

how old was John Brown when he died?

79

John Brown died in 1837 at the age of 79.


Biographical data

Birthday September 12th, 1757 (Monday)
Place of Birth Staunton
Independent city in Virginia, United States
Death Date August 29th, 1837 (Tuesday)
Death place Lexington
City in and county seat of Fayette County, Kentucky, United States
Birth sign (Zodiac) Virgo (The Maiden) ♍
Chinese Zodiac Ox 牛

Other personalities born on September 12

Also born in 1757

John Brown

Birthday

When was John Brown born?

John Brown was born on September 12th, 1757.

Death

When did John Brown die?

John Brown died on August 29th, 1837 at the age of 79 in Lexington (City in and county seat of Fayette County, Kentucky, United States). Today John Brown would be 266 years old.

Place of Birth

Where was John Brown born?

Staunton (Independent city in Virginia, United States).

Death place

Where did John Brown die?

Lexington (City in and county seat of Fayette County, Kentucky, United States).

Zodiac

What is the zodiac sign of John Brown?

John Brown was born in the zodiac sign Virgo (The Maiden).

Chinese Zodiac

Which Chinese zodiac sign does John Brown have?

John Brown was born in 1757 in the year of the Chinese zodiac Ox.

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Died on August 29th, 1837:
Anniversaries of death

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American lawyer and statesman, Virginia (1757-1837)