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| Geometric Integers (Posted on 2006-11-09) |
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The number 201 is divided by a positive integer N. It is observed that the quotient, remainder and divisor (that is, N itself), but not necessarily in this order, are in geometric sequence.
What can N be?
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Submitted by K Sengupta
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Rating: 4.0000 (1 votes)
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Solution:
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Let 201 = dq+r, where d,q an r respectively denote the divisor, quotient and remainder.
Now, r < d and r can be either the first (the lowest) or second element of the geometric sequence.
If r is the second element, then r^2 + r = 201, which does not yield any integer value of r.
Hence, r is the lowest element. Let the common ratio be m/n. Then, it follows that:
r*m^2/n^2 must be an integer; so that r = k*n^2
Accordingly, (d,q) is a permutation of (kmn, k*m^2)
So 201 = dq+r, gives:
201 = (kmn)(k*m^2)+ k*n^2; so that, n is a factor of 201, so that :
n= 1, 3, 67, 201
If n = 201, then, 1 = k^2*m^3 + 201k, which is not feasible.
If n = 67, then, 3 = k^2*m^3 + 67k, which is not feasible.
If n =1, then, 201 = k^2*m^3 + k , which is not feasible.
If n = 3, then, 67 = k^2*m^3 + 3k, which gives (m,k) = (4,1), so that :
N = 16, 12.
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