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Friday Puzzler 2 -- a+b-c=d

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Friday Puzzler 2 -- a+b-c=d

#1

Friday Puzzler 2 -- a+b-c=d

Alex Y

In case that other one is too easy, try this.

a+b-c=d

You need to find the values for a, b, c, and d so that the equation above is true in four different senses:

1) Taken as the normal arithmetical sense, e.g. 45+15-5=55

2) Taken as operating on the sets of digits in each number. e.g., 45+15-5 = 154, because {4,5} U {1,5) = {4,5,5,1}; take away {5}, and d can be any number made up of [all] the digits in {1,4,5}

3) Taken as operating in the same way on the sets of Roman Numerals in that representation of the equation,

4) Taken as operating on the sets of letters in the equation with each of a,b,c,and d spelled out in the English words for the numbers. E.g., forty-five + fifteen - five = whatever number you can spell using [all] the letters {e,e,f,f,f,i,n,o,r,t,t,y} (I'm pretty sure there is not one)

Re: Friday Puzzler 2 -- a+b-c=d

#2

Time for some hints ...

Larry Barrett

I don't know how to get started on this one.

Are a, b, c single digits?

Re: Friday Puzzler 2 -- a+b-c=d

#3

Re: Time for some hints ...

Alex Y

They are all less than three digits.

The Roman Numerals may be a good place to start. A little experimenting around will narrow your choices for B and C.

Re: Friday Puzzler 2 -- a+b-c=d

#4

Re: Time for some hints ...

Larry Barrett

How about 23 + 45 - 43 = 25

{23} + {45} = {2345}

{2345} - {43} = {25}

twentythree + fortyfive - fortythree = twentythreefortyfive - fortythree =

twentyfive

XXIII + XLV - XLIII = XXXLVIII - XLIII = XXV

If this is correct, there seem to be several other possibilities.

Re: Friday Puzzler 2 -- a+b-c=d

#5

Excellent!

Alex Y

(and I thought this was by far the harder of the ones I posted this time)

This was not the solution of which I was aware, but you are right that this opens up a whole family of solutions.

Another solution which does not depend on compound words for the names is

11+2-1=12.

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