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Magic

cinelli
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Magic

#77499

Postby cinelli » August 28th, 2017, 11:36 am

Astonish your family on this bank holiday Monday.

On a piece of paper write a 3-digit number (‘abc’) with all three digits different. Reverse the digits (‘cba’) and subtract the smaller from the larger to produce ‘def’. If it’s below 100 add zero at the beginning. Reverse the digits (‘fed’) and this time add to ‘def’. Turn your piece of paper upside down and read the sum. I predict that you will see

6801

I wonder why.

Cinelli

Gengulphus
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Re: Magic

#77510

Postby Gengulphus » August 28th, 2017, 12:39 pm

cinelli wrote:Astonish your family on this bank holiday Monday.

On a piece of paper write a 3-digit number (‘abc’) with all three digits different. Reverse the digits (‘cba’) and subtract the smaller from the larger to produce ‘def’. If it’s below 100 add zero at the beginning. Reverse the digits (‘fed’) and this time add to ‘def’. Turn your piece of paper upside down and read the sum. I predict that you will see

6801

I wonder why.

Spoiler:

Well, actually only a and c have to be different...

Suppose c > a. Then cba = 100c + 10b + a, abc = 100a + 10b + c and cba > abc, so the subtraction produces (100c-c) + (10b-10b) + (a-100a) = 99(c-a). So the result of the subtraction is a multiple of 99 from 1*99 to 9*99, which along the instructions to add a leading zero if below 100 mean that it is 099, 198, 297, 396, 495, 594, 693, 792 or 891. By inspection, d lies in the range 0 to 8, e=9 and f=9-d in all cases, and since def = 100d + 10e + f and fed = 100f + 10e + d, the addition produces (100d+d) + (10e+10e) + (f+100f) = 20e + 101(d+f) = 180 + 909 = 1089 in all cases. Then it's just a matter of rotating that figure by 180 degrees, producing 6801.

If instead a > c, the same argument applies, just with the roles of a and c swapped.

Gengulphus

modellingman
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Re: Magic

#77902

Postby modellingman » August 30th, 2017, 1:20 pm

Ignoring the turn upside down bit, the equivalent 2 digit puzzle always yields 99. A 4 digit puzzle yields either 10890 or 9999 (depending on whether the second digit of the larger number is greater than or smaller than the fourth digit). A 5 digit puzzle yields either 10890 or 99099 (again depending on the relation of second digit to its complement). A 6 digit puzzle yields one of 1098900, 1089990, 999999 or 991089.

cinelli
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Re: Magic

#79465

Postby cinelli » September 6th, 2017, 9:31 am

Well solved. Good little puzzle though, don't you think?

Cinelli


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