Find the last digit is 7 and which becomes 5 times larger when this lastmdigit is carried to the beginning of this number.
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Unit digit is 7. After operation,{ that is multiplying by 5 and shifting 7 to fist position,} 7 * 5=35, so unit digit must be 5. So the last two digits would be 57. Go to logical solution. ~ 57 * 5=285. So let us take the number as 2857. 7285/5=1457 and not 2857. So next step. 2857 * 5=14285 . So let us take the number as 142857. 142857 * 5=714285 so this is the number.
I do not know how this works. But this is a special number!!! 1 4 2 8 5 7 ∗ 1 , 1 4 2 8 5 7 ∗ 2 , 1 4 2 8 5 7 ∗ 3 , 1 4 2 8 5 7 ∗ 4 1 4 2 8 5 7 ∗ 5 1 4 2 8 5 7 ∗ 6 all have the same digits shifted!!! But 142857 * 7=999999!!! Logical solution. Well the logical way is as follows. x, y, z are unknown digits, xyz means 100+10y+z. After the operation, the first digit is 7. So before the operation, first two digits must be some thing more that 14. 14x * 5=714. So x is only 2 so that we can have 7 1 4 . But 4 would come if next digit is 8, (8*5=40). So the number likely is 142857. Check:- 142857*5 is equal to 714285.