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Ten Box Trial (Posted on 2022-04-29) Difficulty: 2 of 5
  • There are ten boxes containing 4, 8, 27, 31, 34, 35, 44, 45, 51, and 59 balls of either red, or blue, or green in color.
  • Some boxes contain only red balls, some boxes contain only blue balls, and remaining boxes contain only green balls.
  • None of the boxes contain balls having more than one color.
One salesman sold only one box out of them and he declared, "I have the same number of red, blue, and green balls left over."

Which box is sold out?

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
soln | Comment 1 of 10
logic: 
Sum of all 10 is 338
Sum of 9 out of 10 must be multiple of three
That gives four possibilities: remove 4, 35, 44, 59
Number of balls left is 330, 303, 294, 297 respectively
Then the colors each number 110, 101, 98, or 93 
Try each in turn:

Try 110 (don't use 4)
Make 110 from 59 and an addition of baskets to (51): none such

Try 101 (don't use 35)
      Make 101 from 59 and an addition (without 35, 59) of baskets to (42): 8 and 34
Also, make 101 from 51 and an addition (without  8, 34, 35, 51, 59) to (50): none such

Try 98 (don't use 44) 
      Make  98 from 59 and an addition (without 44, 59) of baskets to (39): 8 and 31
Also, make  98 from 51 and an addition (without 8, 31, 44, 51, 59) baskets to (47): none such

Try 93 (don't use 59)
      Make 93 from 51 and an addition (without 51, 59) to (42): 8 and 34
Also, make 93 from 45 and an addition (without 8, 34, 45, 59) to (48): 4 and 44

So, leaving out 59, we have (8, 34, 51), (4, 44, 45), and (27, 31, 35) each adding to 93 

  Posted by Steven Lord on 2022-04-29 07:40:03
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