can 120 be expressed as the sum of two positive integers, one of which is divisible by 11 and the othe by 17?
if yes, show it in algebra way please..
thanks
2007-01-27 13:59:48 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You're asking if there exist positive integers p,q such that
11p+17q=120.
This is a simple case of ax+by=c and it is well-known (see any textbook on elementary number theory) that a solution is given by ax=c(mod b). But, in this case,
120(mod 17)=1 and with a=11 there is no integer value for p which will solve the congruence. Therefore, there are no such positive integers p,q such that 11p+17q=120.
Doug
2007-01-27 14:26:02 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
11x + 17y = 120
y = -11x + 120
y = -11x/17 + 120/17 =
Start plugging in values for x and see which y values come out to integers. I didn't find any (my data table is below), so the answer is "No".
xy
07.058823529
16.411764706
25.764705882
35.117647059
44.470588235
53.823529412
63.176470588
72.529411765
81.882352941
91.235294118
100.588235294
11-0.058823529
2007-01-27 22:07:24 · answer #2 · answered by Tim P. 5 · 1⤊ 0⤋
Let 120 =11n +17m, where n and m are integers.
Then,
n = (1/11)( 120 - 17m).
We give values to m=1,2,3,..... 7, till it gives n integer.
Doing it , we see, no value of m satisfies this condition.
The answer is : NOT POSSIBLE.:
2007-01-27 22:56:16 · answer #3 · answered by Anonymous · 0⤊ 0⤋
120 = 17x + 11y
Calculate this:
120 - 11
120- 22
120- 33
And find which of them is a multiple of 17
Ana
2007-01-27 22:08:13 · answer #4 · answered by Ilusion 4 · 0⤊ 0⤋
11x +17x = 120
28x = 120
x = 4.28
Since the answer is not a whole number, the answer is NO.
2007-01-27 22:05:29 · answer #5 · answered by ironduke8159 7 · 2⤊ 2⤋
Who says "X" has to be the same number, genius??
2007-01-27 22:07:10 · answer #6 · answered by $$ Profit of Doom $$ 2 · 3⤊ 3⤋
No
2007-01-27 22:07:55 · answer #7 · answered by bruinfan 7 · 0⤊ 0⤋