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 Two Digit Integers (Posted on 2006-07-14)
Find all possible two digit positive integers N for which the sum of digits of 10N - N is divisible by 170.

 Submitted by K Sengupta Rating: 3.0000 (2 votes) Solution: (Hide) The integers are 20,39,58,77 and 96. EXPLANATION: Method I (Given by Richard): Richard's method is provided in this location. ----------------------------------------------------------------------------------------------------------------------------------------------------------------- Method II: Let, f(x) = 10^x – x and g(x) = 99…….99( x-2 times). Then, f(N) = 100*g(N) + u(N); where u(N) = 100 – N. We proceed to analyse all the multiples of 170 ( including itself), satisfying conditions of the problem Now, 170 = 18 x 9 +8, giving N= 18+2 = 20, since s.o.d.(100 -20) = s.o.d.(80) =8 None of the other quotients and remainders obtained by dividing 170 by 9 would yield a valid N. For example, 170 = 9*17 + 17, but s.o.d.(81) = 9 which is not euqal to 17. Also, 340 =37 x 9 +7, giving N= 37+2 = 39, since s.o.d.(100 -39) = s.o.d.(61) =7 And also, 510 =56 x 9 +6, giving N= 56+2 = 58, since s.o.d.(100 - 58) = s.o.d.(42) =6 And also, 680 =75 x 9 +5, giving N= 75+2 = 77, since s.o.d.(100 -77) = s.o.d.(23) =5 And also 850 = 94 x 9 +4,giving N= 94+2 = 96, since s.o.d.(100 -96) = s.o.d.(4) =4 All the other multiples of 170 equal to or greater than 1020 will always yield three digit positive integers. This is a contradiction. It can easily be verified that None of the other quotients and remainders obtained by dividing 170*p ( for p= 2,3,4,5) by 9 would yield a valid N. Hence, the integers are 20,39,58,77 and 96. ------------------------------------------------------------------------------------------------------------------------------------------------------- Method III: Let, f(x) = 10^x – x and g(x) = 99…….99( x-2 times). Then, f(N) = 100*g(N) + u(N); where u(N) = 100 – N. Let, the minimum value of N, for 10< = N < = 99, be given by N= r. Then, f(r) = 100*g(r) + 10*s +t; where 10*s + t = 100 –r. So, s.o.d.(f(r)) = 9(r-2) + s+ t. Also, f(r+19) = 100*g( r+19) + 10(s-2)+(t+1); giving, s.o.d.(f(r+19)) = 9(r-2)+ 9*19 + s+t-1 = s.o.d.(f(r)) + 170. Hence, by induction s.o.d.(f(r + 19p)) = s.o.d.(f(r)) + 170*p, where p is a positive integer, whereby r+19p<=99-------------(#) Now, there can be at most 5 distinct values for p. Accordingly, the last digit of (100-r) is any one of 0, 1, 2, 3, 4, 5 and so, (#) will be satisfied only when the last digit of r is any one of 5,6,7,8,9,0. Hence, s.o.d.(f(r + 19*p)) is divisible by 170 iff s.o.d.(f(r)) is divisible by 170, whenever r+19*p <=99, for p = 0,1,2,---, and the last digit of r is any one of 5,6,7,8,9,0. But, 170 = 9*18 +8. Since, 10(10-8) +0 = 20 = 18+2; it follows that r=20, giving N = 20+19*p. Accordingly, N = 20, 39, 58, 77 and 96 are the only two digit integers satisfying all the conditions of the problem under reference.

 Subject Author Date 123456789 pat 2006-11-22 18:27:51 123456789 pat 2006-11-22 18:27:26 Acknowledgement K Sengupta 2006-08-31 05:39:32 Some Analysis Richard 2006-07-21 04:05:40 re(3): Solution - thanks Richard Patrick 2006-07-19 08:49:54 re(2): Solution Richard 2006-07-18 23:06:36 re: Solution Patrick 2006-07-18 13:39:29 re: Solution Ady TZIDON 2006-07-15 09:44:47 Solution Dej Mar 2006-07-14 13:36:07 A Program Helps Here (Numerical Spoiler) Richard 2006-07-14 13:10:55 Computer Solution Daniel 2006-07-14 13:04:05

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