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| Integer Equations (Posted on 2007-03-25) |
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Define [n] as the greatest integer less than or equal to n.
Given that x is a positive integer, determine analytically all possible solutions to each of the following equations:
(a) [x/3]+ [x/5] + [x/7] = 66
(b)[x/7]+ [x/11] + [x/13] = 245
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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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(a) We know that:
n-1< [n]<= n
Hence,
(x/3 - 1)+ (x/5 - 1)+(x/7 - 1)<= 66< x/3+ x/5 + x/7
Or, 71x/105< 69 and 71x/105>= 66
Or, 97.606<= x< 102.042
Since x is a positive integer, it follows that x= 98,99,100,101,102
For x=98, LHS of the given equation = 65, a contradiction.
For x=99, LHS = 66
For x>= 100; LHS>= 100, a contradiction.
Hence, x=99 is the only possible solution to the given equation.
(b) (x/7 - 1)+ (x/11 - 1)+(x/13 - 1)<= 66< x/7+ x/11 + x/13
Or, 311x/1001< 248 and 311x/1001>= 245
Or, 788.56913<= x< 798.225
Since x is a positive integer, it follows that x= 789, 790, ......, 798
For x<= 790, LHS of the given equation = 243, a contradiction.
For x=791, LHS = 244, a contradiction
For x=792, LHS = 245
For x>= 793; LHS>= 246, a contradiction.
Hence, x=792 is the only possible solution to the given equation.
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