Primes!
The positive number k is the product of four different positive prime numbers. If the sum of these four prime numbers is a prime number greater than 20, and the least possible value of k is ((x^2)-10)) ,find:
x^5-(x-1)^5
The answer is 723901.
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1 solution
P l e a s e c o r r e c t a t y p o . i t s h o u l d b e . . . . ( ( ( X 2 ) − 1 0 ) o n e ) l e s s . Since the lest possible k is required, take first three smallest primes. Product of first three =2 * 3 * 5=30 and sum=2+3+5=10. I f P i s t h e 4 t h p r i m e , k + 1 0 = 3 0 ∗ P + 1 0 = X 2 . S o ( 3 P + 1 ) 1 0 = X 2 . Possible if 10|(3P+1), and 10|X. P must have 3 at unit place. . . ( ∗ ∗ ) First try, X=10, P=3. But 3 is the 2 n d p r i m e . N e x t X = 2 0 , 3 P = 3 9 , P = 1 3 , a p r i m e . So, k=2 * 3 * 5 * 13=390, k+10=400 = 2 0 2 ∴ X = 2 0 . X 5 − ( X − 1 ) 5 = 2 0 5 − 1 9 5 = 7 2 3 9 0 1 . A L T E R N A T E L Y : − F r o m ( ∗ ∗ ) After 5, next prime with 3 at unit place is 13. With P=13, X=20 turns out to be correct.