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| An Integer Digit Puzzle (Posted on 2007-02-10) |
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Find all possible solutions to the system of equations:
(i) n(p) + n(q) = p;
(ii) p+q+ n(r) = r;
(iii) n(p) + n(q) + n(r) = q - 4;
where, each of p, q and r are positive integers and n(p), n(q) and n(r) respectively denote the number of digits of the integers p, q and r.
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Submitted by K Sengupta
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Rating: 5.0000 (1 votes)
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Solution:
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(p, q,r) =(2,8,12) is the only possible solution to the given problem.
EXPLANATION:
(1)+(2)-(3): r=2q- 4, which means that :
n(r)=n(q) or n(r)=n(q)+1......(A)
(3)-(1) gives: p=q- 4- n(r)
Now, n(r)=n(q) combined with p = q-4-n(r) yields :
p=q-4-n(q), so that:
n(p)=n(q-4-n(q))
Accordingly, n(p) = n(q) or n(q)-1.
Again, n(r)=n(q)+1 combined with p = q-4-n(r) yields :
p=q-4-n(q)-1; so that:
n(p)=n(q-5-n(q))
Accordingly, n(p) = n(q) or n(q)-1.
Consequently, n(p)=n(q) or n(p)=n(q) - 1
Substituting these back to (1) or (3); we obtain:
q=3n(q)+k where k=3 or 4 or 5 ......(B)
So, in view of the relations in (A) and (B); there are six possible values for q ; which are
q = 6,7,8,9,10,11
So, n(q) = 1 whenever q = 6, 7, 8, 9......(C)
and n(q) = 2 whenever q = 10, 11
But n(q) = 2 gives n(p) = p-2 in (i); which does not have a solution for positive integer p.
n(q) = 1 gives n(p) = p-1; which possesses the unique solution p=2
So, (p, q,r) = (2, 6, 8); (2, 7, 10); (2, 8, 12); (2, 9, 14)
Now, each of the triplets with the exception of (2,8,12) violate (iii).
Consequently, (p,q,r) = (2,8,12) constitute the only possible solution to the problem.
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Comments: (
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Subject |
Author |
Date |
 | No Subject | puneeth | 2007-03-10 19:30:54 |
 | Solution | atheron | 2007-02-10 17:17:02 |
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