Compute the smallest positive integer k such that the fraction
(7k+100)/(5k-3)
is reducible.
The smallest value of k is 209.
The expression is reducible for all values of k generated by the following expression:
209 + j*521 for all integers j
In each case, the gcd of the numerator and denominator is 521
Not surprisingly, in the reduced fraction, the numerator increases by 7 and the denominator increases by 5 for each increment.
-2917 -20319/-14588 = -39/-28, the gcd is 521
-2396 -16672/-11983 = -32/-23, the gcd is 521
-1875 -13025/-9378 = -25/-18, the gcd is 521
-1354 -9378/-6773 = -18/-13, the gcd is 521
-833 -5731/-4168 = -11/-8, the gcd is 521
-312 -2084/-1563 = -4/-3, the gcd is 521
209 1563/1042 = 3/2, the gcd is 521
730 5210/3647 = 10/7, the gcd is 521
1251 8857/6252 = 17/12, the gcd is 521
1772 12504/8857 = 24/17, the gcd is 521
2293 16151/11462 = 31/22, the gcd is 521
2814 19798/14067 = 38/27, the gcd is 521
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for k in range(-3000,3000):
num = (7*k+100)
den = (5*k-3)
g = gcd(num,den)
if g == 1:
continue
print(k, '{}/{} = {}/{}, the gcd is {}'.format(num,den, int(num/g),int(den/g), g))
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Posted by Larry
on 2024-11-17 15:36:00 |
also brute force
