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| A Realmebon Problem (Posted on 2006-11-11) |
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In the planet Realmebon, the inhabitants use base 6 in their daily transactions. One of the inhabitants noticed that if he split his 6-digit ID number into two 3-digit numbers and squared each of them, their sum would equal his ID number.
How many possible ID numbers could he have been assigned, and what are they?
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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 the possible ID numbers assigned to the inhabitant be respectively denoted by R1, R2, ….., Rp.
Accordingly, we obtain:
Ri = 216*m + n for i= 1,2,…, p; where 36< = m,n< 216 …………….(#)
And, m^2 + n^2 = 216*m + n ……..(##)
From (##), we obtain,:
(2m – 216)^2 + (2n -1)^2
= 46657 = (3^2 + 2^2)(6^2 + 1^2)(9^2+4^2) --------(I)
Noting the algebraic identity:
(a^2+b^2) (c^2+d^2)
= (ac+ bd)^2 + (bc – ad)^2 = (ac – bd)^2 + (bc + ad)^2;
it follows that:
(2m - 216, 2n -1) = (204,71), (71, 204), (144, 161), (161, 144), (216,1), (1,216), (84, 199), (199, 84).......(II)
Imposing restriction (#) on (II), we obtain:
(m,n) = (210, 36),(180, 81),(36, 81),(150, 100),(66, 100)
Transforming each of the above values of m and n to base - 6, we obtain:
(m,n) = (550, 100),(500, 213),(100, 213),(410, 244),(150, 244);
giving p = 5
Hence:
(R1, R2, R3, R4, R5) = (550100, 500213, 410244, 150244, 100213)
Consequently, there are five possible ID numbers that could have been assigned to the inhabitant and these are: 550100, 500213, 410244, 150244 and 100213.
NOTE: By allowing for leading zeroes we would obtain three additional numbers in the decimal system, namely 0, 1 and 1332 and converting each of these into 6 digit base six numbers, we would obtain
000000,000001 and 010100 , thus giving three more ID numbers.
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