Decimals...
Algebra
Level
2
Find triplets of single digit numbers a, b, c such that ab^2/c^3 = 1.0ca0ca0ca0ca..., where c <> 0.
(a,b,c) =(9,5,6)
(a,b,c) =(10,7,6)
(a,b,c) = (7,2,3)
(a,b,c) =(8,0,3)
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Solution 1:
1) ab^2/c^3 = 1+(0ca/999) = 1+(ca/999) = 1+(10c+a)/999
2) (ab^2)/(c^3) = 1+(10c+a)/999
3) (ab^2 - c^3)/(c^3) = (10c+a)/999
4) 999 = (10c+a)(c^3)/(ab^2 - c^3)
5) (37*3^3)/1 = (10c+a)(c^3)/(ab^2 - c^3)
By using # 5 equation, we will form
5I) 37*3^3 = (10c+a)(c^3)
Solve equation 5I.. By using my reasoning, c^3 = 3^3... There fore, c = 3
5II) (10c+a)(c^3) = 37*3^3
Solve equation 5II...
37 3^3 = (10 3+a)(3^3)
30+a = 37
a = 7
5III) 1 = (ab^2 - c^3)
Solve equation 5III...
1 = (ab^2 - c^3)
1 = 7b^2 - 3^3
28 = 7b^2
4 = b^2
2 = b
Generalizing all the answers, the triplets (a,b,c) = (7,2,3)....
Solution 2:
We have
1000(ab^2/c^3)-ab^2/c^3 = 10ca-1 = 999+ca,
999(ab^2) = (c^3)(999+ca),
(*) 999(ab^2-c^3) = (c^3)(ca).
Thus 999 = (3^3)*37 divides (c^3)(ca), or 37 divides c^3 or ca. Since c is a digit, 37 does not divide c^3. Therefore 37 divides ca, i.e. ca = 37 or 74.
Assume ca = 37. Then (*) gives 27(7b^2-27) = 27, or b = 2. Therefore (a, b, c) = (7, 2, 3).
Assume ca = 74. Then ( ) gives 27(4b^2-343) = 343 2 = 686. But 27 does not divide 686, so there is no integer solution for b.
In all cases, (a, b, c) = (7, 2, 3) and ab^2/c^3 = 7*2^2/3^3 = 1.037037....