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supposed to be solvable with no calculator

 Jan 22, 2022
 #1
avatar+117849 
+1

I did it partly by logic and partly by trial and error

 

999 base10  =  1 111 100 111 in base 2   there are 10 digits but not all can be a 1

so the maximum sum of the base 10 digits is 9

 

900 base 10  = 1 110 000 100    that is no good so it is less than 900

 

800 base 10 =  1 100 100 000   that is no good

the most I can add to it is 10 base10  so that is   1010 base two

which means that  1 100 will be the highest four base 2 digits which means that the max number of ones is 8 which is too small

so the number is less than 800

 

700, 701,702,710,711,720 are all no good  so it has to be less than 700

 

600,601,602,603,610, 611, 612, 620, 621, 630   all no good so it is less than 600

 

500 base 10 = 111 110 100 this doesn't work but it looks promising

there are 6 ones here  ... if I add 2 base 10 that will add 1 extra 1 ... sounds good

502 = 111 110 110    PERFECT

503 = 111 110 111    PERFECT

 

504,510,511,512,513,520,521,522,530,531,540  all no good

 

So the largest one is    503

 

Note: I cold easily have made a careless error.

 Jan 23, 2022
 #3
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+1

Great, this is the correct answer, thank you so much!

Guest Jan 23, 2022
 #2
avatar+117849 
0

A response from you (the asker) would be much apprecieated.

 Jan 23, 2022

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