Friday Puzzler -- abcdefghij

Re: Friday Puzzler -- abcdefghij

Larry Barrett

No answer, but some helpful 'rules':

Divisibility Test for 2: The last digit is 0, 2, 4, 6, or 8.

Divisibility Test for 3: The sum of the digits is divisible by 3.

Divisibility Test for 4: The last two digits are divisible by 4.

Divisibility Test for 5: The last digit is 0 or 5.

Divisibility Test for 6: The number is divisible by both 2 and 3.

Divisibility Test for 7: Cross off last digit, double it and subtract. Repeat if you want. If new number is divisible by 7, the original number is divisible by 7.

Divisibility Test for 8: The last 3 digits are divisible by 8.

Divisibility Test for 9: The sum of the digits is divisible by 9.

It seems easy to find a 6 digit number that follows these rules. For example,

a can be any digit; say 3

b can be 0,2,4,6,or 8; say 2

for c, a+b+c must be divisible by 3, so c could be 1 or 4; say 4

for d, c+d must be divisible by 4, so so this example d could be 0,4,8; say 4

e must be 0 or 5; say 5

At this point we have abcde = 32445.

To find f the rule says that 3+2+4+4+5+f must be divisible by both 3 and 2. To be divisible by 2, f could be 0,2,4,6,or 8. To be divisible by 3, 3+2+4+4+5+f must be divisible by 3. 3+2+4+4+5 =18, so f could be 3,6,or 9. So for t his example, f must be 6 and abcdef is 324456.

Finding g to satisfy the rule for divisibility by 7 is left for an exercise for the reader.

Re: Friday Puzzler -- abcdefghij

Larry Barrett

To reply to my own comment, the original problem says that each letter must be a different digit, so my example is just an example of how to apply the 'rules'.

Good start

Re: Good start

Larry Barrett

I have stumbled on one answer that seems to meet requirements - 1836547290. There may be more.

Here is some observations that help to get to the ballpark.

We know b,d,f,h,j must all be divisible by 2 so must be 0,2,4,6,8 in some order.

We know e must be divisible by 5, so must be either 0 or 5; but 0 is already taken so e must be 5.

I think we can also say that j must be 0. We know that the sum of the digits a thru i must be divisible by 9 and if those digits are 1 thru 9 in some order, that sum is 45 (n*(n+1)/2). If 0 is in the mix for a thru i and j is 2 or 4 or 6 or 8, then the sum of the digits a thru i would be 45-2 or 45-4 or 45-6 or 45-8 and none of those are divisible by 9.

A little trial and error, testing possibilities for a, b, c, and d led to the answer above.

Oops and a correction

Larry Barrett

I set up a spreadsheet using the 'divisibility by n' rules to quickly check possible answers. I just discovered an error in the spreadsheet for the divisibility by 7 rule, so my 'solution' is not a solution.

But after fixing that error I tried again, and this time I stumbled on a good solution (I hope) - 3816547290. There may still be others.

Correct

Alex Y

That solution is unique.

A solution not requiring silicon:

j is 0 since abcdefghij is divisible by 10 (your rationale for j=0 is correct, but this seems easier ;-) )

e=5

As you mentioned, every other letter is an even digit.

d+e+f is a multiple of 3 (skipping a step here, as I will in the future without explanation, but ask if not clear. a+b+c is a multiple of 3 as is a+b+c+d+e+f, so d+e+f is also a multiple of 3), so the only pairs of even digits for d and f are 2&8 and 4&6

abcd is a multiple of 4, and since c is odd, d is not a multiple of 4. Thus def is 654 or 258

fgh is a multiple of 8, and since f is a multiple of 4, gh is a multiple of 8. Since g is odd, h is not a multiple of 4.

So now we have (case1) a8c654g2i0 or (case2) a4c258g6i0

Since gh is a multiple of 8 and g+h+i is a multiple of 3, we have ghi = 321 or 729 for the first case, or 963 in the second case.

The second case completions then become 1472589630 and 7412589630. Using plain old long division on the first seven digits, we find that

1472589=6mod7 X

7412589=2mod7 X

(BTW, I've seen three rules for determining divisibility by 7, but they all seem to me to be at least as calculation-intensive as using long division, just tracking the remainders.)

The first case has four subcases, which check out as follows:

7896543 = 4mod7 X

9876543 = 5mod7 X

1836547 = 6mod7 X

3816547 = 0mod7 That's it! 3816547290