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Link to previous puzzle Simple Math Problem #1 (not linked to this one)

Here is another simple math problem. Can you solve it?

Examples:

  • $25 + 12 = 33$
  • $37 + 13 = 44$
  • $40 + 24 = 66$
  • $64 + 26 = 88$

Problem:

Find next statement

$\; ? \; + \; ? \; =\; ?$

HINT 1:

$\; ? \; + \; ? \; = 22$

HINT 2:

Mathematic calculations are used with a Twist so don't expect correct result always(mathematically):P

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  • $\begingroup$ Oh GOD, now there are three correct answers, not sure which one was first, anybody can help! $\endgroup$ – Preet Dec 19 '17 at 2:29
  • $\begingroup$ As, every answer has correct statement but with different reasoning(for partial or full), so will add my answer. Thanks everyone for trying this and providing different ideas! (+1 to all) $\endgroup$ – Preet Dec 19 '17 at 2:36
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It's:

80+48=22

Explanation:

The 'sums' are the first digits added, followed by the second digits subtracted, both done mod 10.

The tens digit of the first addend and the units digit of the last addend seem to be the digit present in the result of the previous sum (credits to @ibrahimmahrir). The units digit of the first addend is the sum of the units digits of the previous first addend mod 10 and the tens digit of the second addend is the difference between the digits of the previous second addend (units-tens).

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  • $\begingroup$ Sorry, this is not the answer i am expecting, there is a pattern as well and calculations are with a twist:P (not correct mathematics), will add hint soon $\endgroup$ – Preet Dec 19 '17 at 1:24
  • $\begingroup$ @Preet maybe it would be good to extend the pattern, then. 3 pieces of information aren't really enough to determine a pattern most of the time. $\endgroup$ – boboquack Dec 19 '17 at 1:25
  • $\begingroup$ @Preet is this correct now? $\endgroup$ – boboquack Dec 19 '17 at 1:29
  • $\begingroup$ Not now as well sorry, but can say you are close, just keep in mind that calculations are with a twist no need of Mod at all. $\endgroup$ – Preet Dec 19 '17 at 1:33
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    $\begingroup$ mod 10 and discarding carry are exactly the same; modulus is the formalised version $\endgroup$ – Wen1now Dec 19 '17 at 10:27
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Partial answer:

8? + ?8 = 22

As:

The left digit of the first number and the right digit of the second number are: $22$, $33$, $44$ and $66$ which is kind of the same sequence as in the results: $33$, $44$, $66$ and $88$ phased by one line. So the left and right numbers in the next equation are $88$.

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  • $\begingroup$ You don't need to put ?? as the result is known its 22, also I must say you are close to result even though reasoning is not correct(doesn't matter:P) $\endgroup$ – Preet Dec 19 '17 at 1:51
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I believe it's

80 + 48 = 22

Let us label the digits as AB + CD = EF

As ibrahim mentioned, A and D of each row seems to be based on the right-hand result of the 'previous' row. Therefore, we know that A and D are 8.

Additionally, in every previous row, A+C is always equal to E. But considering that 8 is larger than 2, my only idea is adding it by 4 (to 12) and assuming it'll always modulo the result by 10.

Likewise, in all previous cases F seems to be equal to (B-D mod 10). Since D is 8, I'm guessing B would be 0.

Therefore, the answer is 80 + 48 = 22

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  • $\begingroup$ Result is correct! In my approach I just discard carry, will add my answer too $\endgroup$ – Preet Dec 19 '17 at 2:33
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I think the answer is:

80 + 48 = 22

Second Addend

The second addend in each line has the following pattern:

25+12=33

37+13=44

13 = 1 & 3, where 1 is the difference between the digits of the previous line's second addend (2-1) and 3 their sum (1+2).

40+24=66

24 = 2 & 4 = (3-1) and (1+3)

64+26=88

26 = 2 & 6 = (4-2) and (2+4)

Extending this, we get the second addend of the sequence as:

(6-2) and (2+6) = 4 & 8 = 48

First Addend

For the first addend, a similar approach:

25

37 = 5-2 & 5+2

40 = 7-3 & 7+3, with a carry of 1

64 = 10-4 & 10+4, with a carry of 1 (the carry from above changes the 0 to a 10)

Taking this further, we get the first addend of the sequence as:

80 = 14-6 & 14+6, with a carry of 2 (the carry from the previous step changes the 4 to a 14)

The Sum

The sum in each line is calculated as:

  1. Take the sum of the digits of the second addend
  2. Keep only the unit digit
  3. Multiply by 11

    (1+2) x 11 = 33

    (1+3) x 11 = 44

    (2+4) x 11 = 66

    (2+6) x 11 = 88

And...

(4+8) = 12, keep 2 and multiply by 11 to get 22.

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  • $\begingroup$ +1 Perfect, You got the answer!! even though reasoning is different than me:( I made each operand in second statement from both operands in first statement; Can you also explain the result part(not necessary if cant) as it was also part of puzzle(easy for you now), will accept it soon! $\endgroup$ – Preet Dec 19 '17 at 2:23
  • $\begingroup$ @Preet - I added my interpretation of the resulting sum, although I'm not sure if it's how you framed it when creating the puzzle. $\endgroup$ – Phylyp Dec 19 '17 at 2:23
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    $\begingroup$ @Preet - it would be nice if you could also explain your approach, to see how we differed (particularly for the sum). $\endgroup$ – Phylyp Dec 19 '17 at 2:27
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    $\begingroup$ ok, will add another answer:P $\endgroup$ – Preet Dec 19 '17 at 2:31
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    $\begingroup$ This answer is much better than mine! Mine relies on the hint that the next row is = 22, but this answer is able to find a pattern and deduce the first and second addend without the hint. $\endgroup$ – votbear Dec 19 '17 at 2:42
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As, everyone got correct answers but with slightly different approaches so posting my answer with my approach:

  1. Every second statement is formed from its previous statement as:

enter image description here

  1. Thus, to find the last statement:

enter image description here

  1. Explanation for result part:

enter image description here

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