8 website users clicked on a particular button a number of times, with each user using different devices, browsers and operating systems. At the end of the day, the total clicks amounted to 50. Below are the users’ respective devices and device software and the number of clicks each of the devices and software amounted to.

USER A uses Chrome Browser, Mobile and Android OS

USER B uses Brave Browser, Desktop and Windows

USER C uses Chrome Browser, Desktop and Linux

USER D uses Edge Browser, Desktop and Windows

USER E uses Opera Browser, Mobile and iOS

USER F uses Brave Browser, Mobile and Android OS

USER G uses Opera Browser, Desktop and Linux

USER H uses Edge Browser, Mobile and iOS

TOTAL CLICKS                    50

Chrome                                20

Brave                                    10

Edge                                      10

Opera                                   10

 

Mobile                                 29

Desktop                               21

 

Android                               10

Windows                            25

Linux                                     10

iOS                                         5

 

WITH THE INFORMATION ABOVE, HOW MANY CLICKS DID EACH USER MAKE?

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1 Answer

Let the user names represent their quantities of clicks. The table below lays out the given information:

The columns in the body of the table create equations:

Operating Systems: A+F=10, B+D=25, C+G=10, E+H=5

Browsers: B+F=10, A+C=20, D+H=10, E+G=10

Devices: B+C+D+G=21, A+E+F+H=29.

We appear to have inconsistencies. If B+D=25 and C+G=10 then B+C+D+G=25+10=35 (OS).

However, B+C+D+G=21 (devices).

Also: A+F=10 and E+H=5 so A+E+F+H=15 (OS), but A+E+F+H=29 (devices).

The discrepancies amount to 14 clicks in each case.

Because of these discrepancies I can only try to provide a solution which ignores the given device counts but permits the OS counts.

The device counts adjusted are 15 for mobiles and 35 for desktops. These add up to 50 clicks.

Please correct the inconsistencies.

The smallest total is E+H=5, so E=5-H (0≤E,H≤5);

E+G=10, so 5-H+G=10, G=H+5 (5≤G≤10);

C+G=10, so C+H+5=10, C=5-H (0≤C≤5);

A+C=20, so A+5-H=20, A=H+15 (15≤A≤20);

A+F=10, so H+15+F=10, F=-5-H (F<0). None of the click counts can be negative.

There can be no solution unless the information is incomplete or incorrect.

If we replace all the totals with the initial letters (lowercase) of the browsers and operating systems we get:

E=i-H, E+G=o, i-H+G=o, G=H+o-i;

C+G=l, C+H+o-i=l, C=l+i-o-H;

A+C=c, A+l+i-o-H=c; A=H+c+o-i-l;

A+F=a, H+c+o-i-l+F=a, F=a+i+l-c-o-H (a+i+l=25, c+o=30, resulting in F<0);

B+F=b, B+a+i+l-c-o-H=b, B=H+b+c+o-a-i-l;

B+D=w, H+b+c+o-a-i-l+D=w, D=w+a+i+l-b-c-o-H;

D+H=e, so w+a+i+l-b-c-o=e, or w+a+i+l=e+b+c+o=50 (as expected).

Also, A+E+F+H=a+i, B+C+D+G=w+l.

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