Mark:
10-DIGIT NUMBER
3816547290
(That guy is the greatest!)
There are 10! (or 3,628,800) permutations of digits in a 10 digit number where each digit is only used once. We can use well known divisibility rules to drastically reduce the number of possibilities.
1. The even digits must be b, d, f, h, and j; and so the odds will be a, c, e, g, and i.
2. Since 5|abcde, e=5, and since 10|abcdefghij, j=0.
3. Since 4|abcd, 4|cd. Remembering that c is odd and not 5 and d is even and not 0, we have that cd=12, 16, 32, 36, 72, 76, 92 or 96. Notice that d=2 or 6.
4. Since 6|abcdef, 3|abcdef, which means that 3|(a+b+c+d+e+f). Also, 3|abc, so 3|(a+b+c). So also 3|(d+e+f). Remembering that d=2 or 6, e=5, and f is even not 0, we have def=258 or 654. Notice that f=4 or 8.
5. Since 8|abcdefgh, 8|fgh. Remembering that f=4 or 8, g is odd not 5, and h is even not 0, we have fgh=416, 432, 472, 496, 816, 832, 872, or 896. Notice that h=2 or 6.
6. Now, because we have that both d and h are either 2 or 6, none of the other even numbers can be 2 or 6. Since we already know that j=0, b=4 or 8.
7. We already observed that 3|(a+b+c). Also, a and c are odd not 5, and b=4 or 8, so we have abc=147, 183, 189, 381, 387, 741, 783, 789, 981, or 987.
From here we can find the solution by "brute force". Using the numbered items above in the order 7, 4, 5, we fill in the digits from left to right by choosing from the possibilities listed for abc, def, and fgh, making sure that no digits are repeated. Of course j is always 0, and i is chosen as the only digit that remains. We have reduced our possible digit combinations to just ten.
Of these possibilities, only one meets all of the divisibility requirements! The others work with everything except the requirement that 7|abcdefg.
a b c d e f g h I j
1 4 7 2 5 8 9 6 3 0
1 8 3 6 5 4 7 2 9 0
1 8 9 6 5 4 3 2 7 0
1 8 9 6 5 4 7 2 3 0
3 8 1 6 5 4 7 2 9 0
7 4 1 2 5 8 9 6 3 0
7 8 9 6 5 4 3 2 1 0
9 8 1 6 5 4 3 2 7 0
9 8 1 6 5 4 7 2 3 0
9 8 7 6 5 4 3 2 1 0
SBOANNGTAITNLAE
banana in song title
YOTURRUSSETLF
trust in yourself
CHALOILFOIRDNAIYA
holiday in California
"smeteltoumies"
I can only come up with;
Sometimes lute(in). Well, it does use up all the letters.
Over in Merry Olde England, two friends, Alex and Bob, go to a bookshop, together with their sons Peter and Tim. All four of them buy some books; each book costs a whole amount in shillings. When they leave the bookshop, they notice that both fathers have spent 21 shillings more than their respective sons. Moreover, each of them paid per book the same amount of shillings as books that he bought. The difference between the number of books of Alex and Peter is five.
Who is the father of Tim
Impossible you say? I think not!
PWROOGRREKSS
YHEOURDBEGTES
