Solution to 'The Amusing Number'
In the amusing number, the last digit at the right is transferred to the left of the first digit. Hence the total number of digits remains the same. This suggests that the new number cannot be 10 or more times the original number since then the number of digits will change.
Let the original number be multiplied by 'a'. Now the number starts with 10112… During the course of multiplication, i.e.,
When we come to the first digit from the left there cannot be any carry over from the previous step of multiplication, since the second digit is 0 and since a is <10. Hence, the first digit of the new number (also last digit of the old number) must be a. Thus we know that new number starts with a (i.e.,a10110…) and the old number is 10112…..a
By giving different values (9,8,7 etc.) to a and by actual division we find that only when a=9 the quotient starts with 10112 & in all other cases this is not obtained as seen below:
when a = 8
(810110)/8 = 101264 (remainder 0)
when a = 7
(710112)/7 = 101444 (remainder 5)
Therefore, the quotient increases as 'a' decreases. Therefore a must be 9.
(910112...)/9 = 101123...
The process of division as shown above leaves a remainder of 5. Now 3 follows 2 in the quotient. But the intermediate digits are in the same sequence in both the numbers (Only one digit is transferred and we get the next digit.).
(9101123)/9 = 1011235 (remainder 8)
Since 5 follows 3 in the quotient, it should do so in the other side also. By placing the last digit obtained by division to the right of the upper number, the entire original number can be obtained as :
No wonder the earthling forgot this number. Transfer 9 to the left and see whether the new number is 9 times the original number.
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