This seems easy

Number Theory Level 4

gcd ⁡ ( 2 3 0 10 − 2 , 2 3 0 45 − 2 ) = 2 x − 2 \gcd( 2^{30^{10}}-2,2^{30^{45}}-2)=2^{x}-2

If x = a n b n c n x=a^n b^n c^n where a , b a,b and c c are prime integers, find a + b + c a+b+c .

Source: PUMAC


The answer is 10.

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Akshat Sharda
Mar 15, 2016

Theorem: For natural a , m , n a,m,n ,

gcd ⁡ ( a m − 1 , a n − 1 ) = a gcd ⁡ ( m , n ) − 1 \gcd(a^m-1,a^n-1)=a^{\gcd(m,n)}-1

gcd ⁡ ( 2 3 0 10 − 2 , 2 3 0 45 − 2 ) 2 gcd ⁡ ( 2 3 0 10 − 1 − 1 , 2 3 0 45 − 1 − 1 ) 2 [ 2 gcd ⁡ ( 3 0 10 − 1 , 3 0 45 − 1 ) − 1 ] 2 [ 2 3 0 gcd ⁡ ( 10 , 45 ) − 1 − 1 ] 2 [ 2 3 0 5 − 1 − 1 ] = 2 3 0 5 − 2 x = 3 0 5 = 2 5 3 5 5 5 ⇒ 2 + 3 + 5 = 10 \gcd( 2^{30^{10}}-2,2^{30^{45}}-2) \\ 2\gcd( 2^{30^{10}-1} -1, 2^{30^{45}-1}-1) \\ 2\left[ 2^{\gcd( 30^{10}-1,30^{45}-1 )}-1 \right] \\ 2\left[ 2^{30^{\gcd(10,45)}-1}-1 \right] \\ 2\left[ 2^{30^{5}-1}-1 \right]=2^{30^{5}}-2 \\ x=30^{5}=2^53^55^5\Rightarrow 2+3+5=\boxed{10}

Bonus: Prove the theorem.

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