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Your approach is good. 1,1,1,1,1 2,4,8,6,2 3,9,7,1,3 4,6,4,6,4 5,5,5,5,5 6,6,6,6,6 7,9,3,1,7 8,4,2,6,8 9,1,9,1,9 0,0,0,0,0

You can also use that: 1,5,6,0 is repeated with every 1st power (You can ignore power) 4,9 is repeated with every 2nd power (odd|even power = 1st|2nd power) _________________

As I understand it, whenever there is a product of multiple of 5's and multiple of 2's , we will always end up with a last digit of zero's. Example: 5 ^1* 2^1 => last digit is 0 5^2 * 2^2 => last 2 digits are 0 so 5^19 * 2^17 will have 17 0's on the right hand side. Someone please correct me if I'm wrong...

Hi all, i was just wondering if it is always possible to use the approach of "intern" with those kind of questions?

It seems the most easiest way to me.

Many thanks

Check this: math-number-theory-88376.html. Chapters: LAST DIGIT OF A PRODUCT and LAST DIGIT OF A POWER are dedicated to this subject. _________________