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More or less 6 days (Posted on 2019-01-24) Difficulty: 3 of 5
<begin>
A student has a book containing 411 pages.

She read a certain number of pages on the first day and created a rule to work out how many pages she had to read on each succeeding day.
She decided that the number of pages to be read on the next day should be equal to the square of the sum of the digits of the page she ended at.
She found that on the sixth day, the number of pages she had set herself to read took her exactly to the final page of the book.
How many pages did she read on the 1st day?
<end>
The above puzzle comes from a Cambridge University publication.

My question: list all the possibles scenarios , completing the book not necessarily in 6-day period, but still adherring to her rules.

See The Solution Submitted by Ady TZIDON    
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soln via computer - 6 or less | Comment 1 of 6
Reading to page 61 on day 1 does it but here are all the outcomes that get to 411

 13  29 150 186 411   0     day 5

 16  65 186 411   0   0     day 4

 29 150 186 411   0   0     day 4

 61 110 114 150 186 411     day 6

 65 186 411   0   0   0     day 3

110 114 150 186 411   0     day 5

114 150 186 411   0   0     day 4

150 186 411   0   0   0     day 3

186 411   0   0   0   0     day 2

190 290 411   0   0   0     day 3

241 290 411   0   0   0     day 3

290 411   0   0   0   0     day 2

        program pages

        implicit none

        integer i,j,k,l,sq,bookmark(8)

           do 2 i=1,410

                do l=1,8

                bookmark(l)=0

                enddo

           j=1

           bookmark(1)=i

1          k=bookmark(j)

           j=j+1

           bookmark(j)=bookmark(j-1)+sq(k)

            if(bookmark(j).eq.411)then

            print 10,(bookmark(l),l=1,6),' day ',j

10          format(6(i3,2x),a8,i1)

            go to 2

            elseif (bookmark(j).lt.411)then

            go to 1

            endif

2          continue

        end

        

        function sq(n)

        implicit none

        integer sq,n,i1,i2,i3

        i1=n/100

        n=n-i1*100

        i2=n/10

        n=n-i2*10

        i3=n

        sq=(i1+i2+i3)**2

        return

        end


Edited on January 25, 2019, 4:49 am
  Posted by Steven Lord on 2019-01-25 03:25:56

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