Check Your Thinking .!.!.!

Let the no. be ...abc7

Since it is given that

 5( ...abc7 ) = 7...abc

We find that c = 5. Putting this value of c back in equation we have 5(...ab57) = 7...ab5 we gives b= 8. Continuing this way till we get 7 for the first time, we find required number is 142857

For variety's sake, here's a number theory approach to the problem ....

Letting $n$ be the number we are after, we subtract $7$ from $n$ , divide that result by $10$ to 'shift' the remaining digits to the right, leaving room to put the $7$ at the front of the newly formed number, which we are told is equal to $5n$ . This process is summarized by the equation

$\dfrac{n - 7}{10} + 7*10^{a} = 5n$ ,

where $a$ is the order of magnitude of $n$ . Simplifying this equation gives us that

$n - 7 + 7*10^{a+1} = 50n \Longrightarrow 49n = 7*(10^{a+1} - 1) \Longrightarrow 7n = 10^{a+1} - 1$ .

This is equivalent to stating that $10^{a+1} \equiv 1 \mod{7}$ . Now by Fermat's Little Theorem we know that $10^{6} \equiv 1 \mod{7}$ , so equating exponents we see that $a + 1 = 6 \Longrightarrow a = 5$ .

Thus $7n = 10^{6} - 1 = 999999 \Longrightarrow n = \boxed{142857}$ .

$let\quad number\quad =\quad .....7\\ ....7\quad *\quad 5\quad =\quad 7......\\ this\quad is\quad what\quad the\quad question\quad says\\ now\quad notice\quad when\quad we\quad multiply\\ 7\quad with\quad 5\quad unit\quad digit\quad of\quad new\quad no\\ would\quad be\quad 5\quad and\quad hence\quad it\quad would\quad \\ be\quad the\quad 2nd\quad last\quad digit\quad of\quad original\quad \\ number.\quad number\quad =\quad ......57\\ now\quad supposing\quad number\quad to\quad be\quad of\quad \\ different\quad digits.\\ case\quad 1\\ a\quad 3\quad digit\quad number\quad \\ a57\quad *\quad 5\quad =\quad 7a5\\ in\quad expanded\quad form\quad we\quad find\quad the\\ above\quad equation\quad but\quad get\quad a\quad decimal\quad \\ value\quad of\quad a.\quad which\quad is\quad not\quad possible\\ we\quad continue\quad the\quad same\quad process\quad for\quad a\quad 6\\ digit\quad number\quad as\quad before\quad it\quad in\quad all\\ case\quad you\quad will\quad get\quad decimal\quad value\quad of\quad ab\quad or\quad abc\\ case\quad 2\\ a\quad 6\quad digit\quad number\quad \\ abcd57\quad *\quad 5\quad =\quad 7abcd5\\ (100000a\quad +\quad 10000b\quad +\quad 1000c\quad +\quad 100d\quad +\quad 57)5\quad =\quad \\ 5(140000\quad +\quad 2000oa\quad +\quad 2000b\quad +\quad 200c\quad +\quad 2d\quad +\quad 1)\\ solving\quad we\quad get\\ 98(1000a\quad +\quad 100b\quad +\quad 10c\quad +\quad d)\quad =\quad 139944\\ abcd\quad =\quad 1428\quad \\ note\quad :\quad abcd\quad represents\quad 1000a\quad +\quad 100b\quad +\quad 10c\quad +\quad d\\ therefore\quad least\quad such\quad number\\ =\quad abcd57\quad =\quad 142857$ .

we have a number with n digits, were the last is 7:

d1 d2 d3 ... d(n-1) dn = d1 d2 d3 ... d(n-1) 7

When the last digit is carried of to the beginning, we have

7 d1 d2 d3 ... d(n-1)

But, when it holds the number becomes 5 times larger. So, we have:

7 d1 d2 d3 ... d(n-1) = 5 x (d1 d2 d3 ... d(n-1) 7)

Well, note first that 5 x 7 = 35

So, we must have 5 x (d1 d2 ... d(n-1) 7 ) ending in 5, which means that d(n-1) = 5 and:

7 d1 d2 d3 ... d(n-2) 5 = 5 x (d1 d2 d3 ... d(n-2) 57)

Note now that 5 x 57 = 285

So, 5 x (d1 d2 d3 ... d(n-2) 57) ends in 85, which means that d(n-2) = 8 and:

7 d1 d2 d3 ... d(n-3) 85 = 5 x (d1 d2 d3 ... d(n-3) 857)

Now, 5 x 857 = 4285, which means that d(n-3) = 2

Continuing on this way, we eventually obtain the number 142857 which, multiplied by 5 becomes 714285.