Jumping Seven

Let x be the number after removing 7 from the actual number. Let n be the number of digits of x.

Then, the actual number = $10x + 7$ and the number on keeping 7 at the front = $7 \times 10^{n} + x$ .

So, $5(10x + 7) = 7 \times 10^{n} + x$

On simplifying, $7x = 10^{n} - 5$

$10^{n} - 5$ will be off the form 999.....5 and must be a multiple of 7. I found it by using the inverse process of basic multiplication by keeping 5 at the unit's place and proceeding till the form 999.....5 is achieved.

Finally I got, $14285 \times 7 = 99995$

So, $x = 14285$ and $n = 5$ .

And $142857 \times 5 = 714285$

So, the answer is $142857$ .

My method is different not easy to explain first I wrote -------- 7 in unit place. multiply by 5 = 35 I wrote 5 in 10;s place and c/f 3 and continued the multiplication till I get 714285

Truston Sempf solution 49x + 7 = 7 × 10^n divide by 7

7x + 1 = 10^n so 7x = 10^n – 1

10^n – 1 is of the form ...99999 the property of 7, 11 and 13 six digit number 999999 is divisible by 7, 11 and 13

so 7x = 999999 and x = 142857

Let x=any positive integer

Let n= the number of digits in x

$5x=\left( \frac { x-7 }{ 10 } \right) +7\cdot { 10 }^{ n-1 }$

The left side of the equation is simple enough. The right side of the equation I have set up will result in the moving of the 7 from the end of the number, to the beginning of the number.

Now we get the all the x on one side of the equation.

$5x=\left( \frac { x-7 }{ 10 } \right) +7\cdot { 10 }^{ n-1 }\\ 5x-\left( \frac { x-7 }{ 10 } \right) =7\cdot { 10 }^{ n-1 }$

We did this by subtracting $\left( \frac { x-7 }{ 10 } \right)$ from both sides.

And now we need to combine like terms. First we need to multiply $5x$ by $\frac { 10 }{ 10 }$ to get a common denominator, leaving us with $\frac { 50x-x-7 }{ 10 }=7\cdot { 10 }^{ n-1 }$ . This simplifies to:

$\frac { 49x-7 }{ 10 } =7\cdot { 10 }^{ n-1 }$

Let's get rid of that pesky fraction though, no one likes fractions. We multiply both sides by 10 to give us:

$49x-7=7\cdot { 10 }^{ n-1 }\cdot 10$

Of course we know based on the properties of exponents that ${ 10 }^{ n-1 }\cdot 10=7\cdot { 10 }^{ n-1+1 }$ , which simplifies to $7\cdot { 10 }^{ n }$ .

Let's put it all together:

$49x-7=7\cdot { 10 }^{ n }$

Okay now that we have this equation in simpler terms, let's start playing with that "n" number. Remember this is the number of digits in x. And remember x must be a whole number by definition, anything that is not a whole number cannot be our solution.

If we substitute 1 for n:

$49x-7=7\cdot { 10 }^{ 1 }\\ 49x-7=70\\ 49x=70-7\\ 49x=63\\ x\approx 1.28571$

If we substitute 2 for n:

49x-7=7\cdot { 10 }^{ 2 }\\ 49x-7=700\\ 49x=693\\ x\approx 14.14286

If w substitute 3 for n:

49x-7=7\cdot { 10 }^{ 3 }\\ 49x-7=7000\\ 49x=6993\\ x\approx 142.714

If we substitute 4 for n:

$49x-7=7\cdot { 10 }^{ 4 }\\ 49x-7=70000\\ 49x=69993\\ x\approx 1428.428$

If we substitute 5 for n:

$49x-7=7\cdot { 10 }^{ 5 }\\ 49x-7=700000\\ 49x=699993\\ x\approx 14285.5714$

If we substitute 6 for n:

$49x-7=7\cdot { 10 }^{ 6 }\\ 49x-7=700000\\ 49x=6999993\\ x=142857$

And at last we have the correct answer! $x=142857$

I know this is a number theory problem, but please bear with my programming solution (Python solution). I felt there were too many cases to consider and I'm really looking forward to a person who can write a purely mathematical solution.

num = 0  
#this number represents the number with the property outlined in the problem statement  

counter = 0    
#this number controls the while loop, if it is 0, it increases the value of "num", if it is 1, it stops

reverse = 0    
#this is number "num", but the 7 is moved to beginning of "num"

while counter <> 1:
    num = num + 1 
    if (num % 10) == 7:    #test if the number ends in 7
        reverse = int("7" + str(num/10))    #move '7' to the beginning of the number
        if num * 5 == reverse:    #test the property outlined in the problem statement
            print num    
            counter = counter + 1    #if it finds the number, stop loop, output num

Upon running the program, it returns 142857 as the answer.

It is sure that the no. ends with 7 & the previous digit of 7 is 5[ as the no is divisible by 5 when 7 jumps to the 1st position]

Let consider any no which ends with 7 & the previous one is 5

Let the no is abc57{you can take any no as your preference}

According to the condition , the new no is 7abc5

Then,

      70000+1000a+100b+10c+5=5×[10000a+1000b+100c+50+7]

       490×[100a+10b+c]=70000-280=69720

        100a+10b+c=69720/490=142.285

As a,b,c are integers so 100a+10b+c can't be a fraction.

So to satisfy the condition ,

               [{7×10^(n-1)-280}÷490]be an integer

As the no is minimum n=6 that satisfy the condition

   Then ,
            100a+10b+c=1428[(700000-280)/490=1428]

    So, a=1, b=4 ,c=2, d=8

As the no is 142857.