Consider the equation 15x + 14y = 7. Find the largest four-digit intyeger x for which there is an integer y so that the pair (x, y) is a solution.
2006-09-26 00:21:19 · 6 answers · asked by Sasha 2 in Science & Mathematics ➔ Mathematics
By rearranging it, you get 14y = 7-15x, so you would like 7-15x to be divisible by 14 in order to have a solution.
If it is divisible by 14 then also by 7, and 7-15x is divisible by 7 only if x is divisible by 7 (because 7 is divisible by 7 and 15 is not).
The largest four-digit multiple of 7 is 9996 (=7*1428), however, it does not work if you plug it in (you get y=-10709.5, which is not an integer), so you plug in the second largest multiple, which is x=9989 (=7*1427), and it works with y=-10702.
2006-09-26 00:32:26 · answer #1 · answered by ted 3 · 2⤊ 0⤋
I would re-write the equation as
15*x = 7*(1-2y)
which tells me
x== 0 mod 7 and,
1-2y == 0 mod 15 or,
y == 8 mod 15
The highest 4 digit multiple of 7 is 9996. Try that and progressively lower multiples of 7 until the equation is satisfied. The most you would have to try is 15 (so that y == 8 mod 15). Fortunately, the second (x) works giving x = 9989, y = -10702.
2006-09-26 11:20:27 · answer #2 · answered by Joe C 3 · 0⤊ 0⤋
x = (7-14y)/15
y = (7-15x)/14
2*7*3*5 = 210 (needed factor of the solution in order to have an [interger, interger] soulution)
9999/210 = 47.6142857142857..... = 47 129/210
therefore the largest could be 4 digit x = 9999-129 +7= 9877
the ordered pair would be (9877, -10582)
2006-09-26 07:45:14 · answer #3 · answered by Brian F 4 · 0⤊ 0⤋
I'm with Sergio on this one - I derived the same answer (9989, -10702)
2006-09-26 08:23:24 · answer #4 · answered by MollyMAM 6 · 0⤊ 0⤋
well i tried it by trial and error method and i got following solution:
(9989, -10702 )
i started from x=9999 and started calculating.
luckily, i got it at 9989!!!!
2006-09-26 07:37:16 · answer #5 · answered by pragyp 2 · 0⤊ 0⤋
x= 9989
y= -10702
2006-09-26 07:31:46 · answer #6 · answered by Sergio__ 7 · 0⤊ 0⤋