waf's challenge question (2 Viewers)

[withoutaface]

m and n are positive integers with m < n. The last three decimal digits of 1978^m are the same as the last three decimal digits of 1978ⁿ. Find m and n such that m+ n has the least possible value.

Not the most difficult question ever, but should get your lateral thinking skills going. Anyone to PM me with an answer gets rep points

 

[withoutaface]

I should have specified, I require working as well as a laterally thought out solution, as opposed to subbing in values.

 

[thejosiekiller]

I should have specified, I require working as well as a laterally thought out solution, as opposed to subbing in values.

damn im out

 

[breaking]

ahh yeah i know this, its easy

 

[withoutaface]

Oh and by the way, I actually don't remember how I did this in the first place, but I remember it being methodical, and I did it mid year 12 without using any 4u techniques, so yeah should be solvable for most of you.

 

[Slidey]

I guess programming algorithms is out?

 

[dawso]

try to just sub in values a<= 50, b<100, see what you come up withQUOTE]

in a pm to me about acmilan's question......lol, now thats wat i call a hypocrite

ill start on ur question now...

 

[withoutaface]

try to just sub in values a<= 50, b<100, see what you come up withQUOTE]

in a pm to me about acmilan's question......lol, now thats wat i call a hypocrite

ill start on ur question now...

dawso that method I suggested was actually more methodical than merely subbing values in, as you were decreasing the bounds constantly to accomodate each other. A similar approach would also be valid for this question.

 

[dawso]

ah ok, goodo...

*begins question*

 

[Slidey]

dawso that method I suggested was actually more methodical than merely subbing values in, as you were decreasing the bounds constantly to accomodate each other. A similar approach would also be valid for this question.

That's what I did for the problem initially. I observed that it was increasing by a small amount each time and then realised that once the amount built up enough it would reach one and you'd have a natural number - the answer. Then I used series.