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It has been swohn that to raed a txet the oedrr in which the lrtetes of each idniaduvl word aepapr is not ipmotanrt, so lnog as the fsrit and lsat ltetres are the correct oens. This is not the case with nmuebrs baecsue if one slcarbmes the ditgis of a nmbeur it is not psisolbe to wrok out what the ogirianl nemubr was.

Tehre are, hevoewr, cirtaen cesas in wchih tehre is sifuficnet inmoartfion to fnid out the onriiagl neumbrs if olny the interior diigts of each of them wree mixed wilhe the fsirt and lsat digtis wree lfet untaelred. Scuh is the case in the fonlwloig aditiodn:

Can you rseotre the oigrianl atdiiodn and its sum?

  34,614  
  52,876  
+ 72,548  
--------   
 187,308

Source: Beatriz Viterbo

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  • 2
    $\begingroup$ Ntoe: The aoiutpmssn in the fsrit ppargarah is dlbartsnomey uurtne, ellaicepsy wtih legnor wdros. (If you are having trouble reading this, reverse all the interior letters of the word (all but the first and last)). $\endgroup$ – GentlePurpleRain Dec 4 '18 at 17:26
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The solution is

$36.414 + 58.726 + 75.248 = 170.388$
There is a phenomenon in English called Typoglycemia, where, given some sentence, as long as the first and last letters of each word stay put, the middle letters can be scrambled beyond recognition and somehow the human brain is able to comprehend what the sentence means.

For example, although the paragraph in the question is scrambled, it is not hard to read it.

This puzzle applies the same idea, but to as a math problem. I assumed that the first and last digits of each number were correct, and that I had to scramble the other digits to get an equation that represented a "correct" addition problem.

Here's how I solved the puzzle.

First, I noticed that the tens digits in each of the summands add to $15$. Because we know that these digits are correct, the tens digit of the sum must be either $7$ or $8$, with a carry of $2$ or $3$ from the ones column. But, notice that it is not possible to make a carry of $3$ in the ones' column, and therefore the tens digit of the sum must be $7$.

Then, I set the digits in the ones' column to be the largest digit they could be. This yielded $6$ in the first summand, $8$ in the second summand, and $5$ in the final summand; for a total of $19$. To make the necessary carry of $2$ into the tens column, I assumed a carry of $1$ from the tenths' column. This made the sum of the ones' column $20$, placing a $0$ as the ones' digit of the sum and the necessary carry of $2$ into the tens' column.

At this point, I had
$36.\textrm{[14|41]}4 + 58.\textrm{[27|72]}6 + 75.\textrm{[24|42]}8 = 170.\textrm{[38|83]}8$
At which point, it is relatively easy to figure out the correct combination. I noticed that it is not possible to make a sum of $18$ in the tenths' column (remember, we need a carry of $1$ into the ones' column, for the previous paragraph to work). Therefore, we have a sum of $13$ in the tenths' column; the only way to do this is $4 + 7 + 2 = 13$.

This gives a final answer of
$36.414 + 58.726 + 75.248 = 170.388$

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Solution:

$36.414 +58.726 + 75.248 = 170.388$

Method:

First note that with the starting numbers staying in place the absolute min numbers (before the decimal) are 31, 52, 72, which add up to 155. The max (not including the decimals) adds to 169. With that in mind I set the value to be a 170 in the final answer and made the rest of the numbers be 36, 58, 75. Then since I knew the decimals of the added numbers has to add to over 1 (to turn the 169 to 170). With that observation in mind I found the final numbers placement.

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1
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If

the decimal point can be moved as well,

there are other solutions

which are trivial modifications of the one already found by others:
3.6414 + 5.8726 + 7.5248 = 17.0388
364.14 + 587.26 + 752.48 = 1703.88
3641.4 + 5872.6 + 7524.8 = 17038.8

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  • $\begingroup$ My original intention involved whole numbers, with dot separating thousands, as we use it in Spanish. I have corrected question. $\endgroup$ – Bernardo Recamán Santos Nov 29 '18 at 11:26
  • $\begingroup$ Oh, the dot separated thousands. I assumed that they were decimals in my answer, I'll correct for that. $\endgroup$ – Hugh Nov 29 '18 at 14:45

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