Question

What is University ?!?!?!?!

Thnq ❤️​

Answer 1

Answer:

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Such problems can be considered using min-max arguments since its a finite type of question.

The summands that can be used is 8, 88, 888. (8888 onwards are clearly not eligible)

If the max summand is 888 all answers have shown that 888+88+8+8+8 is a way. We cant have 888 and two 88 as it will exceed 1000, we can consider the sum as 888+k*8 where k is a natural number and show that k=14. So 888+ 14 copies of 8 is also another way.

If the max is 8 a sum of 125 copies of 8 is also a way.

Lastly consider the case where max summand is 88. Consider the diophantine equation where 1000=88k+8m where k, m are integers. Then we have 125=11k+m. By Chinese Remainder theorem(?) or any other facts, this eqn is solvable, (one solution is (11,4)) and any other solutions is in the form (11-n,4+11n) for all n integers. Lastly we need k>0 and m positive (our premise is we have 88 as a summand so k cannot be 0) (m has to be positive as the qn does not allow subtraction) that gives 10 distinc solutions. So 10 ways here.

Hence 13 ways if order of summands does not matter.

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