I was bored at work today (it being Monday morning) and I started thinking about my badge number.
It's nine digits long and contains all nine digits from 1 to 9. Remarkably, the whole number is divisible by 9 (no remainders!). If you remove the last digit from the number, it is divisible by 8. If you remove the next number, it is divisible by 7. And so on... Right down to the last number, which is, of course, divisible by 1.
What's my badge number?
I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!
Monday, August 31, 2009
Posted by Unknown at 10:24 AM
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Trial and error, though I'm sure there is a better way to figure it out. Well, I did know that the 5th digit had to be 5, and the 2nd, 4th, 6th, and 8th, had to be some combination of even numbers...but the rest was guess work for me. :-P
I think you can work out the second digit has to be 8 due to the fact that the first and third digits have to be odd, and all three of the fisrt digits have to add up to a number that is divisible by 3 (in order to be divisible by 3)ReplyDelete
Of course I'm working backwards after seeing your answer, but mathematically you should be able to see that the first 3 digits are 381 or 183.
After that, the third and fourth digit have to be divisible by 4 in order for the whole four digits to be divisible by 4, so we're looking at 381X or 183X, which gives us 3812, 3816, or 1832, 1836... and I don't really know where to go from there and it's lunch time.
So I created an excel program with some knowns:ReplyDelete
Even digits in place #: 2,4,6,8
Odd digits in place #: 1,3,7,9
A 5 must go in position 5
Then i created rules for the division, i.e.:
First 2 digits: even, divisible by 2
First 3 digits: odd, summation div by 3
First 4 digits: even, last two digits div by 4
First 5 digits: must end in 5
First 6 digits: even, summation div by 3
First 8 digits: even, last 3 digits div by 8
Then I figured the harder numbers would be more dificult to place than the smaller numbers so I started with 9.
Then 8, so far so good, then 7, looking good, then 6 all working then 5, success, then 4,3,2,1. All worked. So your badge number could be:
I guess I should have said the "larger" numbers would be more difficult to place, not the "harder" numbers.ReplyDelete
The number is 381654729.ReplyDelete
If the number is ABCDEFGHI:
Then B, D, F and H are even numbers since they are divisible by even numbers. The rest are odd.
ABCDE can be divided evenly by 5, so E is 5.
ABCD is divisible by 4, therefore, CD can also be divided evenly by 4, and since C is an odd number, D can only be 2 or 6.
ABCDEF can be divided by 6 (by 2 and by 3). Since ABC can be divided by 3, DEF can be also. So, DEF is 258 or 654.
Keep going from here to get the rest.
BTW, USAF Pilot: 987654321 comes close but doesn't work. 9876543/7 = 1410934.71. Every other number works, though.ReplyDelete
Dang excel -- rounded up that decimal. Shoul've looked better. good puzzleReplyDelete
I started with 5 in the 5th slot knowing that would be the only end number divisible by 9 if there is no 0 and worked my way up.ReplyDelete
I thought if 5 was in the 5th slot then what number in the 50's is divisible by 6. I got 54 and then continued on thinking what number in the 40's is divisible by 7. 42 and 49. I tried both and eventually discovered the last five digits by concentrating on only two digits at a time. Knowing the four digits left, 1 7 8 and 2, and knowing that the second number in the badge number had to be even to be divisible by two I tried out 2 and 8. Once I discovered 2 wouldn't be able to make a number divisible by three with the numbers I had left, I just worked my way up from the 8 in the 2nd position and got my final answer.
I tested it along the way using only the last two digits and it worked. 18 is divisible by 2, 87 is divisible by 3, and so on. But when I tried the entire badge number it failed once. 187/3=62.3333 arrg.
good puzzle though. Thanks!