Integer problem 5

Factorize 111111111111, which is a product of 7 primes.

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Solution 1

Factorizing the given integer in two ways

Let A=111111111111.
A=111111*1000001
   =111*1001*1000001
   =3*37*7*11*13*1000001 .....[1].
In another way,
A=1111*100010001
   =11*101*100010001 .....[2].
Since A is a product of 7 primes, from [1] and [2], A is written as:
A=3*7*11*13*37*101*p. (Note: 'p' is a prime.)
p=1000001/101=9901.
Hence, 111111111111=3*7*11*13*37*101*9901.

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Solution 2

Multiplying the given integer by 9 before prime factorizing

Let A=111111111111.
  9A
=999999999999
=10¹²-1
=(10⁶-1)(10⁶+1)
=(10³-1)(10³+1)(10²+1)(10⁴-10²+1)
=(10-1)(10²+10+1)(10+1)(10²-10+1)(10²+1)(10⁴-10²+1)
=9*111*11*91*101*9901.
A=3*37*11*7*13*101*9901.
Since A was written as a product of 7 integers, 9901 is a prime, too.
Hence, 111111111111=3*7*11*13*37*101*9901.

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Mathematical Problems

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© 2006 Takashi Shimazaki
Updated: April 14, 2013