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A nine digit number in which every digit occurs except zero and which ends in $5$ cannot be a square. I was reading the solution from Arthur Engel's book Problem Solving Strategies . I didn't get the idea of assuming $A=10a+5$. Also, by assuming $A$ and then finding about A, how is that enough in solving the problem since $D=A^2$ .The solution is as follows:

Suppose there is such a nine digit number $D$ so that $D=A^2.A=10a+5$ which means $A^2=100a^2+100a+25 =100a(a+1)+25$. Consequences: The next to last digit is 2 , The third digit from the right in $D$ is one ,which can be the final digit in a(a+1) that is 0,2,or 6 .See the table below: $a=0$

$a(a+1)(mod10)=0$

$a=1$

$a(a+1)(mod10)=0$

$a=2$

$a(a+1)(mod10)=2$

$a=3$

$a(a+1)(mod10)6$

$a=4$

$a(a+1)(mod10)=2$

$a=5$

$a(a+1)(mod10)=0$

$a=6$

$a(a+1)(mod10)=0$

$a=7$

$a(a+1)(mod10)=2$

$a=8$

$a(a+1)(mod10)=6$

$a=9$

$a(a+1)(mod10)=2$

But $0$ cannot occur ,and $5$ has already occured.

  • saying $A$ ends in $5$ is equivalent to saying that it is of the form $10a+1$. And tehy are supposing that it is a perfect square in the hopes of deriving a contradiction. – lulu Nov 16 '21 at 17:29
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    @lulu You surely mean "... is of the form $10a+5$ ..." – Peter Nov 16 '21 at 17:39
  • You misquoted the value for $A^2$ (which I have corrected). – Bill Dubuque Nov 16 '21 at 18:52
  • "$D = A^2$ ends in $5$" means $A^2 = 10n+5,,$ so $,5\mid A^2\overset{Euclid}\Longrightarrow 5\mid A,,$ so $A$ ends in $0$ or $5$; but only the latter has square ending in $5\ $ (or use digit $0$ excluded by hypothesis). More algorithmically you could solve $,x^2\equiv 5\pmod{!10},$ as here. Of course you could also brute force check all $10$ possible digits, but that's less instructive. – Bill Dubuque Nov 16 '21 at 19:24
  • @Peter Absolutely, thanks. – lulu Nov 16 '21 at 19:41
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    Please change the tag to 'elementary-number-theory' – Bill Dubuque Nov 16 '21 at 22:27

1 Answers1

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The proof is by contradiction; it starts by assuming that such a number does exist, and then shows that this leads to a contradiction.

Let $D$ be a nine digit number in which every digit occurs except zero, and which ends in $5$, and which is a perfect square. Then $D=A^2$ for some integer $A$, and $A$ ends in $5$ because $D$ ends in $5$. An integer ends in $5$ precisely if it is of the form $10a+5$ for some integer $a$. This is why $A=10a+5$. Then by basic algebra $$A^2=(10a+5)^2=100a^2+100a+25=100a(a+1)+25.$$ This shows that the last two digits of $A$ are $25$. Then by crudely checking all $10$ cases, the author verifies that the last digit of $a(a+1)$ is always either $0$, $2$ or $6$. As $A^2$ does not contain a digit $0$, and already contains a digit $2$ in the second place, the third digit must be $6$, so $A^2$ ends in $625$. Correspondingly $a\equiv2,7\pmod{10}$ and so $A$ ends in either $25$ or $75$.

Servaes
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  • Actually We are considering $D$ to be the nine digit number as well as a square so,$D=A^2$ ,$D$ ends in $5$ so as $A$ ....but we dont know the number of digits of $A$ ,although we know $D$ to be a nine digit ...let's say $A$ is an n digit number $n\leq 9$ so $A=10^{n-1}a_{n-1}+....+5$ so,$A=10(10^{n-2}a_{n-1}+....+a_1)+5$...so $a=(10^{n-2}a_{n-1}+....+a_1)$...$A=10a+5$...if$ A^2=100a(a+1)+25 $...Now if $a(a+1)=6 $ so,$100.6+25=A$...but this just does'nt make any sense.....is $a=(10^{n-2}a_{n-1}+....+a_1)$?.... –  Nov 17 '21 at 03:18
  • Are you familiar with modular arithmetic? – Servaes Nov 17 '21 at 05:36
  • Yup...sure..... –  Nov 17 '21 at 07:37
  • But I am just not getting the idea behind the solution ..... –  Nov 17 '21 at 16:13
  • The number of digits of $A$ is either $4$ or $5$ because $10^8<A^2<10^9$. The argument above shows that $A^2$ ends in $625$, and that $a$ ends in either $25$ or $75$. – Servaes Nov 17 '21 at 16:49
  • Where have u shown a contradiction? Or are we supposed to keep on continuing the process? – Alan Abraham Nov 17 '21 at 16:56
  • @AlanAbraham The text quoted by OP is clearly incomplete and full of (copying) errors, I just attempted to clarify that piece of text. – Servaes Nov 17 '21 at 17:00
  • @Servaes Well,I was not asking about the number of digits of $A$ and "clearly" I have not intended to reflect upon your text or anything like that...it's just that I am just confused by the idea stated by me above which is not anything " quoted " "clearly" and it seems that you haven't quite got what I am trying to convey through the "piece of text".....it's just that I want to know what is the value of $a$ ? Is it the one stated by me? If that's it, how can $6$ be the third to last digit...cause its gotta be the nth digit if $a$ is an $n$ digit number.... –  Nov 17 '21 at 17:52
  • @Franklin Okay. – Servaes Nov 17 '21 at 17:53