It is way big!

Algebra Level 3

If x = ( 123456789 ) ( 76543211 ) + ( 23456789 ) 2 x = (123456789)(76543211)+(23456789)^2 , then find the trailing number of zeros in x 1 / 4 x^{1/4} ?

1 3 4 5

Puneet Pinku by Puneet Pinku , 20, India

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1 solution

Hung Woei Neoh
May 14, 2016

Let n = 23456789 n=23456789

x = ( 100000000 + n ) ( 76543211 ) + 23456789 n = ( 100000000 ) ( 76543211 ) + 76543211 n + 23456789 n = ( 76543211 ) ( 1 × 1 0 8 ) + ( 76543211 + 23456789 ) n = ( 76543211 ) ( 1 × 1 0 8 ) + ( 100000000 ) n = ( 76543211 ) ( 1 × 1 0 8 ) + ( 23456789 ) ( 1 × 1 0 8 ) = ( 76543211 + 23456789 ) ( 1 × 1 0 8 ) = ( 100000000 ) ( 1 × 1 0 8 ) = ( 1 × 1 0 8 ) ( 1 × 1 0 8 ) = 1 × 1 0 16 x=(100000000+n)(76543211) + 23456789n\\ =(100000000)(76543211) + 76543211n + 23456789n\\ =(76543211)(1 \times 10^8) + (76543211+23456789)n\\ =(76543211)(1 \times 10^8) + (100000000)n\\ =(76543211)(1 \times 10^8) + (23456789)(1 \times 10^8)\\ =(76543211 + 23456789)(1 \times 10^8)\\ =(100000000)(1 \times 10^8)\\ =(1 \times 10^8)(1 \times 10^8)\\ =1 \times 10^{16}

x 1 / 4 = ( 1 × 1 0 16 ) 1 / 4 = 1 × 1 0 4 = 10000 x^{1/4} = (1 \times 10^{16})^{1/4} = 1 \times 10^4 = 10000

The number of trailing zeroes = 4 =\boxed{4}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Awesome answer.. Upvoted

Puneet Pinku - 5 years, 1 month ago

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