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Given that the following are true:
44 and 182 make 101
118 and 197 make 76

Then, solve this:
143 and 72 make ____?

Hint 1:

The pattern is derived from the first two numbers in each line (e.g. 44 and 182). The solution should be used as a check to confirm you have the correct pattern (i.e. if you think you found the pattern, use the pattern on 44 and 182 to confirm you get 101).

Hint 2:

To produce a given solution from the two given numbers, the first of the two numbers will need to change in some way.

Hint 3:

The solution is logical.

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  • $\begingroup$ I can find a polynom that reproduces these results. The answer could be any number $\endgroup$ – fffred May 8 '18 at 12:59
  • $\begingroup$ @fffred I assure you once you find the pattern it will be quite obvious, there is only one possible solution to this. If there aren't anymore submissions in the near future I'll probably add a hint. $\endgroup$ – Colton May 8 '18 at 13:39
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Answer is

-9

The process seems to be as follows:

Sum the digits of the first number, add 1, square the result and substract that from the second number.

For 44 and 182:

4 + 4 = 8, 8 + 1 = 9, 9 squared is 81 and 182 - 81 = 101.

For 118 and 197:

1 + 1 + 8 = 10, 10 + 1 = 11, 11 squared is 121 and 197 - 121 = 76.

So for 143 and 72:

1 + 4 + 3 = 8, 8 + 1 = 9, 9 squared is 81 and 72 - 81 = -9.

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Answer is

79

As,

Of 44 and 182 , when you square individual digits of each of these numbers and add, you get 16+16+1+64+4 = 101

And,

118 and 197 give 77 instead of 76...

Therefore,

143 and 72 make 1+16+9+49+4 = 79

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  • $\begingroup$ Sorry, can you explain the second equation a little more please? When I use your formula from the first equation on the second I get 'Bar uhaqerq avargl frira', abg 'friragl fvk' be 'friragl frira'. $\endgroup$ – hagfy May 8 '18 at 17:50
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    $\begingroup$ All of the descriptions are correct as written, so operate under this assumption. The second series does NOT follow your pattern. $\endgroup$ – Colton May 8 '18 at 18:49
  • $\begingroup$ What I meant is, as an iteration, repeat the addition of sum of squares of individual digits, till you are left with a 2 digit number of each of the operands on the left hand side. In this case it is 66 and 131 after 1st iteration and 66 & 11 after second iteration thereby giving 66+11=77. $\endgroup$ – Mea Culpa Nay May 9 '18 at 0:43

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