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7 solutions
Moderator note:
A nice standard approach.
Bonus question : Prove that 3 9 6 5 + 2 is a composite number. And prove that it has at least 2 distinct prime factors.
3 9 6 5 must end in a 3 , because 3 ( 4 n − 3 ) will always end in a 3 . Therefore, 3 9 6 5 + 2 must end in a 5 .
Numbers that end in 5 but are not equal in 5 will always be composite. Since 3 9 6 5 + 2 divided by 5 will not equal 5 or any power of 5 , it must have at least 2 distinct prime factors.
Great solution!!! Factoring out is always good when dealing with large exponents...
Dint get it! Explain in other easier way if possible! Please!
Let x = 3 3 9 5 , 9 x = 3 3 9 7
3 9 6 5 + 2 3 9 6 7 − 3 9 6 5 + 1 6 ⇒ x + 2 9 x − x + 1 6
x + 2 9 x − x + 1 6 = x + 2 8 x + 1 6 = x + 2 8 ( x + 2 )
x + 2 8 ( x + 2 ) = 8 ⋅ x + 2 x + 2 = 8 ⋅ 1
3 9 6 5 + 2 3 9 6 7 − 3 9 6 5 + 1 6 = 8
[3^965(9-1)+16]/3^965+2 =[3^965*8+16]/3^965+2 =8(3^965+2)/3^965+2 =8
Your L A T E X editing is not very good.
tx Anaghjyoti Ghosh
If you want to learn real L A T E X , read this
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Thanks for providing the LATEX site, expecting more related to above & other use-full math web sites
3 9 6 5 + 2 3 9 6 5 × ( 9 − 1 ) + 1 6 = 3 9 6 5 + 2 8 ( 3 9 6 5 + 2 ) = 8
really thanx the solution provided up is not much clear but for this no words but no.s
3^965 (9-1)+ 16÷3^965+2
3^965(8)+ (8)2÷3^965 3^965+2 cancel with each other and the answer is 8
excellent solution
3 9 6 5 + 2 3 9 6 7 − 3 9 6 5 + 1 6 = 3 9 6 5 + 2 9 ( 3 9 6 5 ) − 3 9 6 5 + 1 6 = 3 9 6 5 + 2 8 ( 3 9 6 5 + 2 )
= 8 .