Find digits x and y such that the number $\overline{25xy5}$ is divisible by 3 and 7.

Question

I have some problems with the following task:
Find digits x and y such that the number $\overline{25xy5}$ is divisible by 3 and 7.

So far, I have come up with this solution:

$$ \begin{cases} 2+5+x+y+5=3p,\ p \in\mathbb{Z+} \\ 25xy-10=7q, \ q \in\mathbb{Z+} \end{cases} $$

$$ \begin{cases} 2+5+x+y+5=3p,\ p \in\mathbb{Z+} \\ 2500+10x+y-10=7q, \ q \in\mathbb{Z+} \end{cases} $$

$$ \begin{cases} x+y=3p-12,\ p \in\mathbb{Z+} \\ 10x+y=7q-2490, \ q \in\mathbb{Z+} \end{cases} $$

And the problem is that I don't know if I am on the right way to the solution and what to do with the parameters p and q.

Since $2+5+5\equiv 0\pmod 3$, you see that $x+y\equiv 0\pmod 3$ and you can get a similar relation $\pmod 7$. Ultimately, not a lot of pairs to check. — lulu, Mar 03 '24 at 22:09
Welcome to Mathematics Stack Exchange. $7|abctuv\iff 7|tuv-abd$, so you're looking for $7\mid xy-2$ — J. W. Tanner, Mar 03 '24 at 22:16
@J.W.Tanner: are you really giving people hints about the arcane secret that $1001 = 7 \cdot 11 \cdot 13$??? Oops! I gave it away $\ddot{\smile}$. — Rob Arthan, Mar 03 '24 at 22:20
Is it standard to use the vinculum $\overline{abc}$ to indicate $100a+10b+c$? — J. W. Tanner, Mar 04 '24 at 00:56
By congruence laws the divisibility holds $!\iff! \overline{xy}\equiv_7 2,\ \overline{xy}\equiv_3 0!\iff! \overline{xy}\equiv_{21} 9,$ by Easy CRT. $\ \ $ — Bill Dubuque, Mar 04 '24 at 01:15
Unlike the proposed duplicate question, OP here did not write congruences ("what to do with p and q?") — J. W. Tanner, Mar 04 '24 at 01:46
Or w/o CRT: $\bmod 21!:\ \color{#c00}{10\equiv\frac{!-!1}2},$ so $,\color{#0a0}{25}[\color{#c00}{10}^3]\equiv \color{#0a0}4\left[\color{#c00}{\frac{!!-1}2}!\right]^2!10\equiv \color{#0af}{10},,$ hence$,0\equiv \overline{\color{#0a0}{25}xy5}\equiv \color{#0af}{10}!+!\color{#c00}{10}\overline{xy}!+!5!\iff!$ $\color{#c00}{\frac{!-!1}2}\overline{xy}\equiv -15!\iff\overline{xy}\equiv 30\equiv 9\ \ $ — Bill Dubuque, Mar 04 '24 at 02:15
The $\rm\color{#c00}{prior}$ method is essentially the idea behind the reverse divisibility test for $21.\ \ $ — Bill Dubuque, Mar 04 '24 at 02:26

score 3 · Accepted Answer · answered Mar 03 '24 at 22:21

3

We have $$\overline{25xy5} =25005 + \overline{xy0}\underset7\equiv 1+10\cdot \overline{xy}\underset7\equiv 1+3\cdot \overline{xy}.$$

Hence $$1+3\cdot \overline{xy} \underset7\equiv 0,$$

or $$\overline{xy}\underset7\equiv 2.$$

Also we have that $$2+5+x+y+5 \underset3\equiv 0,$$

so $$x+y \underset3\equiv 0.$$

From here you just probably try all numbers of the form $7k+2$ and check if sum of two digits is divisible by $3$. From $02$, $09$, $16$, $23$, $30$, $37$, $44$, $51$, $58$, $65$, $72$, $79$, $86$, $93$ only $$09, 30, 51, 72, 93$$ are good.

answered Mar 03 '24 at 22:21

Aig

4,178

$\overline{xy}\equiv_7 2,\ \overline{xy}\equiv_3 0!\iff! \overline{xy}\equiv_{21} 9,$ by CRT. $\ \ $ – Bill Dubuque Mar 04 '24 at 01:07
Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Mar 04 '24 at 01:18

Find digits x and y such that the number $\overline{25xy5}$ is divisible by 3 and 7.

1 Answers1