Abraham Jacob Hollandersky - Height Birthday Zodiac Biography

Abraham Jacob Hollandersky’s height is 5ft 4in (1.63 m)

Abe “The Newsboy” Hollandersky (December 3, 1887 – November 1, 1966) became the first American boxer to win the Panamanian Heavyweight Title when he defeated Jack Ortega in nine rounds in Panama City on May 30, 1913. American congressmen, Naval personnel, and canal workers were among the crowd of nearly two thousand who watched Hollandersky gain victory over an opponent who outweighed him by over thirty pounds.

Born: 3 December, 1887
Died: 1 November, 1966
Height: 5ft 4in (1.63 m)
Astrological Sign: Sagittarius

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