Divisible by 11,111 (Posted on 2008-10-02)

	Submitted by K Sengupta
Rating: 3.2500 (4 votes)
Solution:	(Hide)
The sum of the digits A to J = 0+1+2+.....+ 9 = 45, so that: the number represented by ABCDEFGHIJ is divisible by 9. Since 9 and 11,111 are relatively prime, it follows form the given conditions that ABCDEFGHIJ is divisible by 99,999. Thus, ABCDEFGHIJ (mod 99,999) = 0 Or, (ABCDE+ FGHIJ) (mod 99,999) = 0 ......(i) Since, each of the letters denote a distinct digit, it follows that each of ABCDE and FGHIJ is less than 99,999, and accordingly ABCDE+ FGHIJ < 199,998, and so that in terms of (i), we must have: ABCDE + FGHIJ = 99,999 ..... (ii) Now, each of the digits cannot exceed 9, so that there is no carryover in the lhs of (ii), and accordingly: F = 9 - A G = 9 - B H = 9 – C I = 9 - D J = 9 - E Since A is nonzero, there are 9 ways to choose A, whereby F will be known. This leaves 10-2 = 8 ways to choose B, whereby G will be known. C can thus be chosen in 6 ways, whereby H will be known. This leaves 4 ways to choose D, whereby I will be known. Finally, there will be 2 choices for E, whereby J will be known. Consequently, the required number of all the positive integers ABCDEFGHIJ that satisfy the given conditions = 9*8*6*4*2 = 3456.

	Subject	Author	Date
	My Python Solution	jcbar	2019-07-20 15:53:26
	winnowed	ed bottemiller	2008-10-02 17:59:41
	perl(computer) solution	Pete	2008-10-02 13:25:08
	re: Trivial Solution	Charlie	2008-10-02 12:48:37
	Trivial Solution	ed bottemiller	2008-10-02 12:41:09
	computer solution (spoiler)	Charlie	2008-10-02 12:40:46
	answer	Dej Mar	2008-10-02 11:30:48