math problem
posted by Samantha on .
My ID number is quite remarkable.Its a 9 digit number with the digits 19 appearing only once. The entire 9 digit nuymber is divisible by 9. If you remove the last digit, the remaining 8 digit number is divisible by 8. If you remove the last digit again, you are left with a 7 digit number dicisible by 7. Again remove the last digit to have a 6 digit number divisible by 6. This process continues all the way down to one digit. What is my ID number?
987654321
I'm sorry, but that's not the correct answer, 7 does not divide 9876543, check for yourself. All the other numbers did work however.
Ok, we have the number 19 appearing once in some order. Let abcdefghi denote the number, then we're told
9abcdefghi
8abcdefgh
7abcdefg
6abcdef
5abcde
4abcd
3abc
2ab
1a
Here is what you should be able to tell by inspection alone:
(1) b,d,f, and h are all even since evens do not divide odds.
(2) since 0 in not included, a,c,e,g and i are the odds.
(3) e=5 since 5 only divides numbers ending in 5.
(4) if 8 divides a number, then it divides the last 3 digits of the number.
(5) 4 divides a number if and only if it divides the last 2 digits.
Here is what you can conclude right now: there are 4! ways to permute 1,3,7 and 9 for a,c,g and i; there are 4! ways to permute 2,4,6, and 8 in b,d,f and h. Thus there are 576 potential numbers to check.
I did a check of all of them and there is exactly one solution. That's all I'll give you right now.
Lets take the easy road and see where it takes us.
Divisibility rules:
Odd numbers can only be divided evenly by another odd number.
Even numbers can be evenly divided by either odd or even numbers.
2Numbers that end in 0, 2, 4, 6, or 8 are evenly divisible by 2.
3If the sum of a number's digits is evenly divisible by 3, the number is divisible by
4If the last two digits are both zero or they form a two digit number evenly divisible by 4, then the whole number is evenly divisible by 4.
5Any number ending in 5 or 0 is evenly divisible by 5.
6If the number is even and the sum of the digits is evenly divisible by 3, the whole number is divisible by 3.
7Double the last digit and subtract from the number without the last digit. If the result is evenly divisible by 7, so is the original number.
8If the last 3 digits are zero or if they form a number that is evenly divisible by 8, then the whole number is evenly divisible by 8.
9If the sum of the digits is evenly divisible by 3, the number is evenly divisible by 9.
10Any number ending in 0 is divisible by 10.
11 Add up the odd position digits. Add up the even position digits. Calculate the difference between the two sums. If the difference is divisible by 11, the original number is divisible by 11
Lets portray our number, with the digits identified, by
_.._.._.._.._.._.._.._.._.._
1..2..3..4..5..6..7..8..9.10
Clearly, the last digit must be 0 in order for the whole number to be divisible by 10.
Similarly, the 5th digit must be 5 in order for the 5 digit number to be divisible by 5.
This makes our number so far _.._.._.._..5.._.._.._.._..0
.............................................1..2..3..4..5..6..7..8..9.10
The sum of the digits 1  9 is 45, which makes any nine digit number using all nine digits divisible by 9. Thus, we need not concern ourselves with the 9th digit.
By the definition of our problem, all even numbers of digits must be divisible by an even number and all odd digit numbers must be divisible by an odd number, making the even digits even, and the odd digits odd.
The 2nd, 4th, 6th, and 8th digits can then only be 2, 4, 6, or 8 and the 1st, 3rd, 5th, 7th, and 9th digits can only be 1, 3, 5, 7, or 9.
For the 8 digit number to be divisible by 8, the last 3 digits must be divisible by 8, making these three 3 digit numbers start, and end, with 2, 4, 6, or 8. These three digits therefore form numbers in the 200+, 400+, 600+ or 800+ family, each of which is, itself, divisible by 8, thus forcing the 7th and 8th digits together to be evenly divisible by 8. The only possibilities for these two digits then becomes 16, 32, 72 and 96.
This makes our possible numbers so far....._.._.._.._..5.._..1..6.._..0
.............................................................. _.._.._.._..5.._..3..2.._..0
.............................................................. _.._.._.._..5.._..7..2.._..0
.............................................................. _.._.._.._..5.._..9..6.._..0
..............................................................1..2..3..4..5..6..7..8..9.10
From our divisibility rules, a number is divisible by 4, if, and only if, the last 2 digits are divisible by 4 forcing our 3rd plus 4th digit number to be divisible by 4 with the 3rd digit being odd. The possible 2 digit numbers are therefore 12, 16, 32, 36, 52, 56, 72, 76, 92 and 96. Since all of these candidates end in 2 or 6, the 4th digit can be safely said to be 2 or 6.
This makes our possible numbers so far....._.._.._..2..5.._..1..6.._..0
.............................................................. _.._.._..6..5.._..3..2.._..0
.............................................................. _.._.._..6..5.._..7..2.._..0
.............................................................. _.._.._..2..5.._..9..6.._..0
..............................................................1..2..3..4..5..6..7..8..9.10
From our divisibility rules, a number is divisible by 3 if the sum of its digits is divisible by 3. Similarly, a number is divisible by 6 if the sum of its digits is divisible by 3. SInce the first 3 digits must be divisible by 3, as must the first 6 digits, the number formed by the 4th, 5th, and 6th digits must be divisible by 3. Having the 4th and 5th digits, and knowing that the 6th digit must be even, we know that this 6th digit can only be 4 or 8.
This makes our possible numbers so far... _..8.._..2..5..4..1..6.._..0
..............................................................._..4.._..2..5..8..1..6.._..0
.............................................................. _..8.._..6..5..4..3..2.._..0
.............................................................. _..4.._..6..5..8..3..2.._..0
.............................................................. _..8.._..6..5..4..7..2.._..0
.............................................................. _..4.._..6..5..8..7..2.._..0
.............................................................. _..8.._..2..5..4..9..6.._..0
.............................................................. _..4.._..2..5..8..9..6.._..0
..............................................................1..2..3..4..5..6..7..8..9.10
However, the digits in 254 and 658 do not add up to a number divisible by 3.
This makes our possible numbers so far.. _..4.._..2..5..8..1..6.._..0 A
.............................................................. _..8.._..6..5..4..3..2.._..0 B
.............................................................. _..8.._..6..5..4..7..2.._..0 C
.............................................................. _..4.._..2..5..8..9..6.._..0 D
..............................................................1..2..3..4..5..6..7..8..9.10
We know that any number placed in the 9th digit will make the total number divisible by 9 so lets work with the 1st and 3rd digits.
The numbers remaining in the 4 cases are 379 for case A, 179 for case B, 139 for case C, and 137 for case D.
We have already pointed out from our divisibility rules that a number is divisible by 3 if the sum of its digits is divisible by 3. Which of the three candidate groups of 3 remaining digits will make our first three digits divisible by 3?
I think this final screening will lead you to the remaining possibilities, with the divisibility by 7 forcing the final answer to fall out.
Try 381654729
What is the answer 3 2/5  1 3/5.
Thanks

Your answer works, but it does not encompass all possible answers. The solution is far from unique.