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In a book (The nuts and bolts of Proofs) there is the exercise:

If $n$ is a three-digit number, whose unit digit is at most 4, whose hundred digit is between 1 and 5, and whose tens digit is the sum of the other two digits, then n is divisible by 11.

My attempt of proving this statement.

Since n is a three-digit number, $n= a_2a_1a_0 = a_2 \times 100+ a_1 \times 10+a_0$. By hypothesis, $0\leq a_0 \leq 4$, $1\leq a_2 \leq 5$, and $a_1 = a_2 + a_0$. Therefore

\begin{align*} n& = a_2 \times 100+ a_1 \times 10+ a_0 = a_2\times 100+(a_2+a_0)\times 10+a_0\\ &= a_2\times 100+a_2\times 10+a_0\times 10+a_0=a_2\times 110+a_0\times 11\\ &=11(10a_2+a_0) \end{align*}

I do not get where the part of the hypothesis $[0\leq a_0 \leq 4,~ 1\leq a_2 \leq 5]$ has been used. I apologize if I am missing something trivial. By the way, with the following code Python

lst = range(100,1000)
lst1 = []
lst2 = []
for i in lst:
    n = i
    digit_3 = n//100
    n = n % 100
    digit_2 = n // 10
    n = n % 10
    digit_1 = n
    if digit_2 == digit_1+digit_3:
        lst1.append(i)
print(lst1)
for i in lst1:
    if i%11 == 0:
        lst2.append(i)
print(lst2 == lst1)

I get the following set of 3-digits number whose tens digit is the sum of the other two digits, and they are divisible by 11: $[110, 121, 132, 143, 154, 165, 176, 187, 198, 220, 231, 242, 253, 264, 275, 286, 297, 330, 341, 352, 363, 374, 385, 396, 440, 451, 462, 473, 484, 495, 550, 561, 572, 583, 594, 660, 671, 682, 693, 770, 781, 792, 880, 891, 990]$.

Some (but not all) of them verify the $[0\leq a_0 \leq 4,~ 1\leq a_2 \leq 5]$ hypothesis part.

For the sake of clarity this the proof of the author. enter image description here

Dimitris
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    To emphasize, brute force is seen as inelegant for proofs of statements like these. You should be able to prove this without. Now... suppose the hundreds digit was $a$ and the units digit was $c$. Is it true that your number is then equal to $110a + 11c$? Is it true then that your number is equal to $11\cdot (10a + c)$ and is as such by definition a multiple of $11$? – JMoravitz Feb 08 '22 at 15:52
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    I guess that this condition is imposed to make sure that the sum $a_0 + a_2 \le 9$ is a single digit. – ViktorStein Feb 08 '22 at 15:53
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    As for "where was the hypothesis that $0\leq a\leq 4$ and $0\leq c\leq 5$" used, it wasn't really... apart from ensuring that the tens "digit" did not exceed nine. The statement you are asked to prove can be generalized where we don't care about them on an individual digit basis. More specifically, a number is divisible by $11$ iff the alternating sum and difference of digits from right to left (starting with summing the furthest right digit) is a multiple of $11$. Here for instance $143\mapsto 3 - 4 + 1 = 0=11\cdot 0$ is a multiple of $11$ so $143$ is as well. – JMoravitz Feb 08 '22 at 15:54
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    Similarly, $54939291\mapsto -5+1-9+2-9+3-9+4=-22$ is a multiple of eleven so $54939291$ is as well. – JMoravitz Feb 08 '22 at 15:58

1 Answers1

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Together, $0\leq a_0\leq 4$ and $1\leq a_2\leq 5$ give $a_1=a_2+a_0\in[1,9]$.

yurnero
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  • Thank you very much for your answer. But with this strict hypothesis, we lose the numbers [660,671,682,693,770,781,792,880,891,990] don't we? – Dimitris Feb 08 '22 at 15:58
  • So what? Those are multiples of eleven but the statement you are asked to prove neither asks you to show they are nor to show that they are not multiples of eleven. The statement was phrased as an if-then, not as an if-and-only-if. – JMoravitz Feb 08 '22 at 16:00
  • @JMoravitz Thanks a lot. I have updated my question adding the author's proof. – Dimitris Feb 08 '22 at 16:01