numbers?????

Question

find a ten digit number containing each digit once, so the number formed by the first n digitd is divisible by n for each value of n between one and 10.

Answer 1

It is obvious that the last digit is 0, and the fifth digit is 5.

It is also obvious that the first, third, seventh, and ninth digits are odd, with the second, fourth, sixth, and eighth digits even, since divisibility by an even number requires the number to be even.

The first three digits, the first six digits, and the first nine digits must add to a multiple of three. This means the fourth, fifth, and sixth digits add to a multiple of three, and the seventh, eighth, and ninth digits add to a multiple of three.

For the middle three, this means that the numbers in the fourth and sixth places are either (2 and 8) and (4 and 6), with order to be determined.

Since 4 * 25 = 100, divisibility by 4 is determined solely by the last two digits of the number. Thus the third and fourth digits form a number divisible by 4. The only possibilities are 12, 16, 32, 36, 72, 76, 92, 96. Thus the fourth digit can only be 2 or 6. Thus the sixth digit can only (respectively) be 8 or 4.

Since 8 * 125 = 1000, divisibility by 8 is determined solely by the last three digits of the number. This the sixth, seventh, and eighth numbers form a multiple of eight. The sixth digit can only be 4 or 8. Since both 400 and 800 are divisible by eight, the seventh and eighth digits form a multiple of eight. The possibilities are 16, 32, 72, and 96. Thus the eighth digit can only be 2 or 6. This means that the second digit must be 4 or 8.

Here is what we have in terms of possibilities:

Digit one: 1, 3, 7 or 9
Digit two: 4 or 8
Digit three: 1, 3, 7 or 9
Digit four: 2 or 6
Digit five: 5
Digit six: 4 (with fourth digit 6) or 8 (with fourth digit 2)
Digit seven: 1, 3, 7 or 9
Digit eight: 2 or 6
Digit nine: 1, 3, 7 or 9
Digit ten: 0

If digit eight is 6, then the seventh and ninth digits are (9,3), from the divisibility by 8 conclusions (digit seven could only be 1 or 9). Otherwise, digit eight is 2, and the seventh and ninth digits are (3,1), (3,7), (7,3), or (7,9).

Now let's look at the second digit. If it is 4, then the only possibilities for the first three digits are 147 or 741. Also, the sixth digit is 8, the fourth digit is 2, and the eighth digit is 6. The only possible numbers formed are 1472589630 and 7412589630. Neither satisfy the divisibility by 7 requirement.

Therefore, the second digit is 8. This means the sixth digit is 4, the fourth digit is 6, and the eighth digit is 2.

This will leave eight possible numbers:

1836547290
1896547230
1896543270
3816547290
7896543210
9816547230
9816543270
9876543210.

Of those, only 3816547290 satisfies the divisibility by 7 requirement. This is your answer (and it is unique!).

3 / 1 = 3
38 / 2 = 19
381 / 3 = 127
3816 / 4 = 954
38165 / 5 = 7633
381654 / 6 = 63609
3816547 / 7 = 545221
38165472 / 8 = 4770684
381654729 / 9 = 42406081
3816547290 / 10 = 381654729.

-End of proof. Phew! :) -

Answer 2

From the action picture "pi" : Restate my assumptions: One, arithmetic is the language of nature. 2, each and every thing around us could nicely be represented and understood by way of numbers. 3: in case you graph the numbers of any gadget, varieties emerge. hence, there are varieties everywhere in nature. data: The cycling of illness epidemics;the wax and wane of caribou populations; sunlight spot cycles; the develop and fall of the Nile