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Two Digit Integers (Posted on 2006-07-14) Difficulty: 4 of 5
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.

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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.

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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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
123456789pat2006-11-22 18:27:51
123456789pat2006-11-22 18:27:26
AcknowledgementK Sengupta2006-08-31 05:39:32
Some AnalysisRichard2006-07-21 04:05:40
re(3): Solution - thanks RichardPatrick2006-07-19 08:49:54
re(2): SolutionRichard2006-07-18 23:06:36
re: SolutionPatrick2006-07-18 13:39:29
Some Thoughtsre: SolutionAdy TZIDON2006-07-15 09:44:47
SolutionSolutionDej Mar2006-07-14 13:36:07
A Program Helps Here (Numerical Spoiler)Richard2006-07-14 13:10:55
Computer SolutionDaniel2006-07-14 13:04:05
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