Thrifty numbers Part 4

Logic Level 2

□ + □ = □ □ − □ = □ □ × □ = □ □ ÷ □ = □ \large{ \square \, + \, \square \, = \, \square \\ \square \, - \, \square \, = \, \square \\ \square \, \times \, \square \, = \, \square \\ \square \, \div \, \square \, = \, \square \\ }

Given that the integers 0 , 1 , 2 , … , 9 0,1,2,\ldots,9 are to be filled in all the twelve square boxes above such that all four equations above are satisfied.

What is the maximum possible distinct number of integers used?

Details and Assumptions

  • You can use the same integer more than once.

  • You are not required to use all nine integers.

  • Each square box can only fill in exactly one digit.

8 9 7 10

Chung Kevin by Chung Kevin , 45, Australia

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2 solutions

Denton Young
Dec 11, 2015

3 + 6 = 9

5 - 5 = 0

2 * 4 = 8

7 / 7 = 1

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Moderator note:

Ah, slightly amazed that we got all 10 digits. Though, it's not like there is any strong reason why we would be blocked.

Yes! Thank you for your solution. ;)

Chung Kevin - 5 years, 6 months ago

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You're welcome. Good problem!

Denton Young - 5 years, 6 months ago
A A
May 27, 2016

This types of problems can be solved by a little bit more abstract thinking as well , that is by finding abstractly the reasoning and grasp of the a priori form or conditions of possibility by which it can work or not anyways.

Suppose there is possible to use all 10 terms and now ask about what structural characteristics would make this possible and to guide the reasoning about it firstly observe the way 0 behaves in any of this operations. In any of them as a result of the a priori nature of it when it is used it will give a result which will be either 0 or the number with which it is used implying therefore that if it will be used in any of this operations there would be at most a number of 2 distinct terms that will appear this being valid for any possible configuration that uses 10 terms.

The remaining 8 terms that will be used should be in such a way used that they fit into the other remaining operations and because in 3 operations there are 9 squares to be filled in 2 operations there will be just different numbers while in the other operation remaining there will be possible to repeat one of them giving that from the entire configuration 2 numbers will repeat. This means that by considering the operations and their internal structure related to the numbers it is easy to understand all possible solutions as the thinking is done in a general note (from the abstract form considered of which I said earlier) and would be of particular interest to analyze of course the more restrictive operations first multiplication and division. However just by considering things in an abstract manner until this point it is easy to find some solutions for the problem. All that has to be done is to take 0 in conjunction with some other numbers such that the remaining numbers work for the other operations this propriety of 0 being enough to determine possible configurations already and by covering already pretty well the abstract understanding works pretty much synthetically also. This implies considering things for 2 cases , one in which 0 will be used for one of the restrictive operations and one in which it is used in subtraction or addition.

Considering the case of using 0 in the more restrictive operations (multiplication , division) observe that in the other restrictive operation should appear one of the number 1 , 2 or 3. If in the remaining restrictive operation a number will repeat then all the remaining numbers which are used in addition and subtraction should be different which implies that because this determines firstly if it is possible or not to be such a solution those triples of numbers used in addition and multiplication should firstly be chosen. For this from the numbers 1 to 9 eliminating at least one of the numbers 1 and 3 should be checked which 2 triples of numbers can be chosen such that they have completely distinct elements and when 2 of them are added will give the 3nd this giving all the possible solutions which should be possible. For the case in which in the remaining restrictive operation no number will repeat , thing possible just for 2 ,4 , 8 and 2 , 3 ,6 eliminate also the numbers used and see which 2 pairs of numbers are at a distance of one of the numbers remaining and then decide which other number will be used with 0 which is so because it allows the possibility of a nubmer repeating in addition and multiplication.

For the case in which 0 is taken in addition or subtraction then a pretty similar approach , which considers the way the numbers behave in multiplication and subtraction should be possible.

Note that by this simple abstract considerations it is anyways possible to guide a solution to this problem without just checking case as it provides insight into the a priori structure used. This problem would be a pretty good example as an introduction into elementary abstract thought I think.

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