1) the sum of the three smallest primes and one other prime
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1) the sum of the three smallest primes and one other prime is 77.what is the product of these four primes?
2)what are both integers n for which (2^n^4)(2^n^3)(2^n^2)(2^N)=1?(里面的2^n^4代表2的n次方又n的4次方)..
2)what are both integers n for which (2^n^4)(2^n^3)(2^n^2)(2^N)=1?(里面的2^n^4代表2的n次方又n的4次方)..
1) 最小的3个质数和一天其他质数的和是77,求这4个质数的积.
The three smallest primes are 2,3,and 5.So the 4th prime is 77-2-3-5=67.Their product is 2*3*5*67=2010.
2) 满足等式(2^n^4)(2^n^3)(2^n^2)(2^N)=1的两个整数是什么?
(2^n^4)(2^n^3)(2^n^2)(2^N)=2^(n^4+n^3+n^2+n),
Because 2^0=1,n^4+n^3+n^2+n must be 0.
n^4+n^3+n^2+n=n(n^3+n^2+n+1)=n*(n+1)*(n^2+1) must be 0.
It's easy to see either n=0 or n+1=0.
So the answer is n=0 or n=-1.
The three smallest primes are 2,3,and 5.So the 4th prime is 77-2-3-5=67.Their product is 2*3*5*67=2010.
2) 满足等式(2^n^4)(2^n^3)(2^n^2)(2^N)=1的两个整数是什么?
(2^n^4)(2^n^3)(2^n^2)(2^N)=2^(n^4+n^3+n^2+n),
Because 2^0=1,n^4+n^3+n^2+n must be 0.
n^4+n^3+n^2+n=n(n^3+n^2+n+1)=n*(n+1)*(n^2+1) must be 0.
It's easy to see either n=0 or n+1=0.
So the answer is n=0 or n=-1.
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