Guest Posted September 27, 2007 Report Share Posted September 27, 2007 The sum of the ages of a father and his son is 66. The father's age is the son's age reversed. Furthermore, the father's age is a multiple of the son's age. How old could they be? Father: 60 years old Son : 6 years old Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 9, 2008 Report Share Posted July 9, 2008 (edited) 6 and 60 06+60=66 6 into 66 is 11 therefore 66 is a multiple of 6. Edited July 9, 2008 by my2monkeys524 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 6, 2008 Report Share Posted December 6, 2008 let f = 10*ft + fo let s = 10*st + so (Note that we must reasonably constrain f and s to be positive integers in the range [0,66], and ft,fo,st, and so to be positive integers in the range [0,9].) Constraint #1: f + s = 66 => 10*ft + fo + 10*st + so = 66 Constraint #2: f is "the reverse" of s, which I will take to mean: ft = so fo = st Note that this means the tens digit of the father's age is equal to the one's digit of the son's age, and vice versa. Constraint #3: f = n*s Here n is a positive integer. Apparently, some people on this forum do not think the phrase "x is a multiple of y" should mean that x is an INTEGER multiple of y, which of course is its generally accepted meaning. It should be obvious that every real number is some real "multiple" of every other real number, so if the word multiple meant a real multiple, it would become useless, or at least superfluous, to state this. I take this constraint to mean an integer multiple. This gives 4 equations and 5 unknowns (ft, fo, st, so, and n), which gives us infinite solutions. Fortunately, they are all constrained to be positive integers and confined within small ranges. Thus we can use a quick trial-and-error computation to find the solutions that are positive integers within their reasonable ranges. Therefore, we now proceed to solve the set of equations for ft,fo,st, and so, but in terms of n. The four equations are summarized as follows, in a form conducive to using matrices: 10*ft + 1*fo + 10*st + 1*so = 66 1*ft + 0*fo + 0*st - 1*so = 0 0*ft - 1*fo + 1*st + 0*so = 0 10*ft + 1*fo - 10*n*st - n*so = 0[/codebox] I have calculated the determinant of this matrix to be: N = -99*(n + 1) Thus, all values of n will give a non-zero determinant, and are thus valid, except for n = -1. But we have already constrained n to be a positive integer, so this irrelevant. Using Cramer's Rule, the solutions are as follows: ft = so = (2 - 20*n) / (-3*n - 3) fo = st = (2*n - 20) / (-3*n - 3) Note that ft = so, and fo = st, as was given by constraint #2. This is a useful check on our work. You can also, with a little algebra, check that 10*ft + fo + 10*st + so = 66, as per constraint #1. This also verifies our results thus far. Various values of n can be tested to see if these will give positive integers for ft,fo,st,so,f, and s. I have written a simple C program to test all the positive integers in the range [0,66], which I felt would give all the relevant results we might be interested in. Actually, I discovered values of n above 10 are all uninteresting, and so I only will reproduce the results for n in the range [0,10], below. I have also given my C program if you are interested. You can confirm for yourself that f + s always equals 66 for these results. [codebox]// program #include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h> int main(int argc, char *argv[] ) { // main program int n; double f,s, ft,fo, st,so; for (n = 0; n <= 10; n++) { ft = so = (2.0 - 20.0*n) / (-3.0*n - 3.0); fo = st = (2.0*n - 20.0) / (-3.0*n - 3.0); f = 10.0*ft + fo; s = 10.0*st + so; printf("\nn == %d:----------\n", n ); printf("ft == %8.5f, fo == %8.5f => f == %8.5f\n", ft, fo, f ); printf("st == %8.5f, so == %8.5f => s == %8.5f\n", st, so, s ); } // end for } // end main() // output n == 0:---------- ft == -0.66667, fo == 6.66667 => f == 0.00000 st == 6.66667, so == -0.66667 => s == 66.00000 n == 1:---------- ft == 3.00000, fo == 3.00000 => f == 33.00000 st == 3.00000, so == 3.00000 => s == 33.00000 n == 2:---------- ft == 4.22222, fo == 1.77778 => f == 44.00000 st == 1.77778, so == 4.22222 => s == 22.00000 n == 3:---------- ft == 4.83333, fo == 1.16667 => f == 49.50000 st == 1.16667, so == 4.83333 => s == 16.50000 n == 4:---------- ft == 5.20000, fo == 0.80000 => f == 52.80000 st == 0.80000, so == 5.20000 => s == 13.20000 n == 5:---------- ft == 5.44444, fo == 0.55556 => f == 55.00000 st == 0.55556, so == 5.44444 => s == 11.00000 n == 6:---------- ft == 5.61905, fo == 0.38095 => f == 56.57143 st == 0.38095, so == 5.61905 => s == 9.42857 n == 7:---------- ft == 5.75000, fo == 0.25000 => f == 57.75000 st == 0.25000, so == 5.75000 => s == 8.25000 n == 8:---------- ft == 5.85185, fo == 0.14815 => f == 58.66667 st == 0.14815, so == 5.85185 => s == 7.33333 n == 9:---------- ft == 5.93333, fo == 0.06667 => f == 59.40000 st == 0.06667, so == 5.93333 => s == 6.60000 n == 10:---------- ft == 6.00000, fo == -0.00000 => f == 60.00000 st == -0.00000, so == 6.00000 => s == 6.00000 The only two values of n for which all variables were positive integers were n = 1 and n = 10, giving f = s = 33, and f = 60, s = 6, respectively. Many of you have already figured out this result intuitively, which is excellent. It is always good to compare intuitive solutions to rigorous proofs as a means of confirming mathematical results. As a final note, many of you seem to be concerned with the problem of 6 being the "reverse" of 60. This is a well-founded concern, because the phrase "the reverse of x" is ambiguous. I have mathematically defined this phrase in such a way that 6 is in fact the reverse of 60, because the tens digit of 60 equals the ones digit of 6, and vice versa. But obviously, this is not aesthetically pleasing. I hope anyone who happens to discover this solution will find it helpful and enjoyable. I certainly did. (In case anyone is interested, I'm a 3rd Year Mechanical Engineering student at the University of Toronto. The education I have received so far in this program in the area of mathematical analysis has provided the basis for my work above. You can reach me at bgold12@gmail.com if you have any comments/questions.) Benji Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 21, 2008 Report Share Posted December 21, 2008 The sum of the ages of a father and son is 66. The father's age is the son's age reversed. Furthermore, the father's age is a multiple of the son's age. How old could they be? Father: 60 years old Son : 6 years old well it can be 24 and 42 also.. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 21, 2008 Report Share Posted December 21, 2008 well it can be 24 and 42 also.. Nope. 42 is not a multiple of 24. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 31, 2011 Report Share Posted October 31, 2011 Father 60 & Son 6 OR!! Father 33 and Son 33 if they were father/son by law (E.g Famililies wedded) Quote Link to comment Share on other sites More sharing options...
0 Half Fast Posted May 30, 2012 Report Share Posted May 30, 2012 42 and 24 Quote Link to comment Share on other sites More sharing options...
0 James_Puzzle Posted May 17, 2018 Report Share Posted May 17, 2018 I was gonna say 33, but then I asked it to a friend and she was immediately "60 and 6". Quote Link to comment Share on other sites More sharing options...
0 Who.Will.Ever.Know? Posted January 31, 2020 Report Share Posted January 31, 2020 I was thinking the Father could be 42 and his son 24, there for together makes 66 and 42 backwards is 24. the father must of had his son (if in fact it is his son?) at the age of 18. Quote Link to comment Share on other sites More sharing options...
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The sum of the ages of a father and his son is 66. The father's age is the son's age reversed. Furthermore, the father's age is a multiple of the son's age. How old could they be?
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