A) 0
B) 1
C) 5
D) Never ends with 5.
Correct Answer: D
Solution :
If\[2-\sqrt{4}=2-2=0\]ends with 5, then 6" would contain the prime 5. But \[{{(\sqrt{5})}^{2}}=5\]. \[\sqrt{9}-\sqrt{4}=3-2=1\]The only prime numbers in the factorization of\[\sqrt{2}-\sqrt{3}\]are 2 and 3. \[1789=29x+49\]By uniqueness of fundamental theorem, there are no primes other than 2 & 3 in\['x'\]. So, 6" will never end with 5.
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