A problem of Secondary Regional 2020.

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sakib17442
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A problem of Secondary Regional 2020.

Unread post by sakib17442 » Wed Apr 21, 2021 3:39 am

What is the greatest 5-digit palindrome n such that 7n is a 6-digit palindrome??

This problem appeared as a problem in Regional Exam of Secondary Category 2020. I am in junior category. But, this seems as an easy problem, that's why I tried to solve this problem and I got n=88088 as my answer. But I don't have any explanation. I don't even know whether it is correct or not. Please let me know about the proper explanation. I am new to this forum, Please forgive me if I did any mistake.

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sakib17442
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Re: A problem of Secondary Regional 2020.

Unread post by sakib17442 » Tue May 18, 2021 5:31 pm

Well, after some researches and efforts I got the answer and I am really happy that my answer was correct. I got some explanations on internet. Gathering them all I am providing the full solution
Let,
$ 7 . (abcba) _{10} = (ABCCBA)_{10}$ ... ... ... $(1)$

Here, $$a,b,c,A,B,C \ \epsilon {0,1,2,...,9}, a \neq 0, A \neq 0$$

Since, $a * 10^4 \ < (abcba)_{10} < (a+1) . 10^4 $ \

and, $ A . 10^5 < (ABCCBA)_{10} < (A+1) . 10^5$

eqn $(1)$ gives the two inequalities

$ 7a < 10(A+1)$ and $ 10A < 7(a+1) ... ... ... (2)$

Since, $A \equiv 7a (Mod 10)$,and $ a \neq 0$, we have the following nine possibilities for (a,A):

$(1,7), (2,4), (3,1), (4,8), (5,5), (6,2), (7,9), (8,6), (9,3)$

The ordered pairs $(3,1), (6,2), (9,3)$ do not meet the condition $7a<10(A+1)$, and the ordered pairs $(1,7), (2,4), (4,8), (5,5), (7,9)$ do not meet

the condition $10A < 7(a+1)$. Thus there remains to check only the pair $(a,A)=(8,6)$

Note that,'
$(ABCCBA)_{10} = (10^5+1)A + (10^4+10)B + (10^3+10^2)C$

$ = ((10^2 . 10^3) + 1)A + 10(10^3+1)B + (10^2.11)C$

$ = -A+C (Mod 7)$ Since, $10^3+1 = 7 . 11 . 13$

Now, from eqn(1), we get $A \equiv C (Mod 7) $, so that $A = C=6$

Next, note that,
$(abcba)_{10} = (10^4+1)a + (10^3+10)b + 10^2c$

$= ((10^2)^2 + 1)a + 10(10^2+1)b + 10^2c$

$\equiv 2a - c (mod 101)$, since $10^2 \equiv -1 (mod 101) $
and

$(ABCCBA)_{10} = ((10 . (10^2)^2) + 1)A + 10 . ((10 . 10^2 ) + 1) + (10^2.11)C$

$ \equiv 11(A+B-C) (Mod 101) $

From, eqn(1), we get, $7 (2a - c) = 11(A + B - C)$

Hence, $7c = 101k - 11(B-1) $ for some $ k \epsilon \mathbb{Z}$

Since, $7c+11(B-1 \le (7 . 9) + (11 . 8) < 2 . 101 $ $ k \epsilon {0,1}$. Some trials give $c = 0$ $B = 1$ if $k=0$,

$c=5, B = 7$ if $ k = 1$

Thus the two possibilities for the number in the RHS of eqn (1) are $616616$ and $676676$

The corresponding numbers for the LHS are $\frac{1}{7} . 616 . 1001 = 616 . 143 = 88088$ and

$\frac{1}{7} . 676 . 1001 = 676 . 143 = 96668$; only the first one is palindrome.

The only solution to eqn (1) is

$7 . 88088 = 616616$

Thus, the solution and the final answer of this question is \[\boxed{88088}\]


The solution is quite complicated especially for me. Sources: https://quora.com
Games You can't win because you'll play against yourself.
---Dr.Seuss

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