[Guest]

I have a ten digit number featuring 0-9, each digit appearing only once. The 10-digit number is perfectly divisible by 10. Now remove the 10th digit to get a 8-digit number.The 9-digit number is perfectly divisible by 9. Now remove the 9th digit to get a 8-digit number. The 8-digit number is perfectly divisible by 8.Now remove the 8th digit to get a 7-digit number. The 7-digit number is perfectly divisible by 7. So on..... What is the number?

Use divisibility rules for 3,4 and 6.....

[Thalia]

3816547290

Edited July 19, 2011 by Thalia

[Guest]

I have a ten digit number featuring 0-9, each digit appearing only once. The 10-digit number is perfectly divisible by 10. Now remove the 10th digit to get a 8-digit number.The 9-digit number is perfectly divisible by 9. Now remove the 9th digit to get a 8-digit number. The 8-digit number is perfectly divisible by 8.Now remove the 8th digit to get a 7-digit number. The 7-digit number is perfectly divisible by 7. So on..... What is the number?

Use divisibility rules for 3,4 and 6.....

"Now remove the 10th digit to get a 9-digit number" instead "Now remove the 10th digit to get a 8-digit number".

[Guest]

9876543210

I made some guesses that were right for all but 1 digit, and then realized that there were only 576 possibilities and started to just put them all in a sheet and then stopped and tried one total guess, bingo!

[Thalia]

I wonder how many different possible solutions there are.

[Guest]

Correction to problem: "....The 10-digit number is perfectly divisible by 10. Now remove the 10th digit to get a 8-digit 9-digit number."

3816547290

3816547290 modulo 10 = 0

381654729 modulo 9 = 0

38165472 modulo 8 = 0

3816547 modulo 7 = 0

381654 modulo 6 = 0

38165 modulo 5 = 0

3816 modulo 4 = 0

381 modulo 3 = 0

38 modulo 2 = 0

3 modulo 1 = 0

[Guest]

Shoot, I thought my number was right but stupid excel rounded it and formatted without the decimal. I made a sheet to find all possibilities using mod and it did not catch my number because their should have been a decimal for 7. I just found the one solution you both found. But my sheet also had a mistake in the combinations so a few were left out. The odds of another solution in those few is remote though.

[Guest]

3816547290

[Guest]

I wonder how many different possible solutions there are.

There is only one solution.

In order for a number to be divisible by 10, the number must end in 0, hence the tenth digit is 0.

In order for a number to be divisible by 5, the number must end in 0 or 5. As the 0 is the tenth digit, the fifth digit must be 5.

As an even number must be divisible by a even number, the even numbered positions must each be an even numbered digit, i.e., of the set {2, 4, 6, 8, 0}, which leaves only odd digits for the odd positions.

For a number to be divisible by 3, the sum of the digits must also be divisible by 3.

With digits 5 and 0 already positioned as the fifth and tenth digits, with odd positioned digits as odd and even positioned digits as even, and without duplicate digits, there are only the following 3-digit numbers that are divisible by 3: 123, 129, 147, 183, 189, 321, 327, 369, 381, 387, 723, 729, 741, 783, 789, 921, 927, 963, 981, and 987.

In order for the number to be divisible by 4, the number formed by the concatenation of the last 2 digits must be divible by 4. Given the list of 3-digit numbers, the numbers divisible by 4 without duplicating digits and noting that the end position must be even would be: 1236, 1296, 1472, 1476, 1832, 1836, 1892, 1896, 3216, 3276, 3692, 3812, 3816, 3872, 3876, 7236, 7296, 7412, 7416, 7832, 7836, 7892, 7896, 9216, 9276, 9632, 9812, 9816, 9872 and 9876.

6 is divisible by 3 and 2, therefore the sum of the digits must also be divisible by 3 and the last digit must be even (and, as it already reserved, not 0). With the fifth digit a 5, with no duplication of digits, the possibilities are then: 123654, 129654, 147258, 183654, 189654, 321654, 327654, 369258, 381654, 387654, 723654, 729654, 741258, 783654, 789654, 921654, 927654, 963258, 981654 and 987654.

The numbers to be tested for divisibilty by 7 are then the following 41 numbers: 1236547, 1236459, 1296543, 1296547, 1472583, 1472589, 1836547, 1836549, 1896543, 1896547, 3216547, 3216549, 3276541, 3276549, 3692581, 3692587, 3816547, 3816549, 3876541, 3876549, 7236541, 7236549, 7296541, 7296543, 7412583, 7412589, 7836541, 7836549, 7896541, 7896543, 9216543, 9216547, 9276541, 9276543, 9632581, 9632587, 9816543, 9816547, 9876541 and 9876543. Only 9 of them are divisible by 7, these are: 1236459, 1296547, 1472583, 3216549, 3816547, 7296541, 7836549, 9216543, and 9632581.

The eighth digit would be the remaining even digit, thus the possibilities would be:

12364598, 12965478, 14725836, 32165498, 38165472, 72965418, 78365492, 92165438, and 96325814. Of these, only 38165472 is divisible by 8. The 9th digit is then the remaining odd digit 9. (The 9-digit number passes the test that, if the sum of the digits is divisible by 9, then the number itself is divisible by 9).

[Guest]

since 10 digit number is divisible by 10, last digit must be zero. For a similar reason, 5th digit must be 5. Let's say number is: abcp5rxyz0.

Since, ab, abcp, abcp5r, abcp5rxy are divisible by 2, 4, 6 and 8 respectively, b, p, r and y must be even. And hence, rest are odd.

Let's work on p5r: Since abc is divisible by 3, abc000 will be divisible by 6 and hence p5r must be divisible by 6, in order to make abcp5r to be divisible by 6. So, p+5+r is divisible by 3 with p and r as even digits. Putting values, we can find, p5r can have only 4 values: 258, 456, 654 and 852. But, abcp and hence cp is divisible by 4 with c being odd digit. So, p cannot be 4 or 8. Hence, p5r = 258 or 654.

Work on rxy:

[Guest]

since 10 digit number is divisible by 10, last digit must be zero. For a similar reason, 5th digit must be 5. Let's say number is: abcp5rxyz0.

Since, ab, abcp, abcp5r, abcp5rxy are divisible by 2, 4, 6 and 8 respectively, b, p, r and y must be even. And hence, rest are odd.

Let's work on p5r: Since abc is divisible by 3, abc000 will be divisible by 6 and hence p5r must be divisible by 6, in order to make abcp5r to be divisible by 6. So, p+5+r is divisible by 3 with p and r as even digits. Putting values, we can find, p5r can have only 4 values: 258, 456, 654 and 852. But, abcp and hence cp is divisible by 4 with c being odd digit. So, p cannot be 4 or 8. Hence, p5r = 258 or 654.

Work on rxy: Since it is divisible by 8 with x being odd digit, it can be either 816 or 896 (when p5r = 258) or it can be 432 or 472 (when p5r = 654).

So, we have 4 possibility for p5rxy: 25816, 25896, 65432, 65472. And left over digits are: (3,4,7,9), (1,3,4,7), (1,7,8,9), (1,3,8,9). Of these left over digits even one will be value of b.

From first set, we don't find any value for abc. From second set we have 2 values for abc (147 and 741). Similarly from third set we have (987, 789, 981, 189), and from fourth we have (981, 189, 381, 183). Hence, in total we have 10 numbers for abcp5rx. Checking them for divisibility with 7, we are left with only one: abc = 381.

Only one digit is left, 9. So, y = 9. Hence number is 3816547290.

Edited July 19, 2011 by swapnil

[Guest]

I have got another possible solution

1472583690

[Guest]

I have got another possible solution

1472583690

14725836 is not divisible by 8.