A Variable for Each DigitDate: 11/28/2001 at 02:26:53 From: Joe Smith Subject: Inverse ages A husband's age (h) is the inverse of his wife's age (w). His age is also greater than her age. Also 11(h-w) = h+w. Some example ages are 82 and 28 or 54 and 45. Is there a way to solve this without simple trial and error? I have tried to establish some relation for inverse numbers, but the only thing I know of is the divisible by 11 thing. Date: 11/28/2001 at 12:41:21 From: Doctor Peterson Subject: Re: Inverse ages Hi, Joe. Apparently you are using the term "inverse" to mean "reversal of the digits"; this is not the usual meaning of the term. The best way to work with problems involving digits is to use a separate variable for each digit, so that h = "ab" = 10a + b and w = "ba" = 10b + a Now the sum and difference are h+w = 10(a+b) + (a+b) = 11(a+b) h-w = 10(a-b) + (b-a) = 9(a-b) (Notice what this tells us about divisibility of the sum and difference of reversed 2-digit numbers.) You are told that 11(h-w) = h+w, which means that 11[9(a-b)] = 11(a+b) which simplifies to 99a - 99b = 11a + 11b 88a = 110b 44a = 55b 4a = 5b Since a and b are both single digits, and we see that a must be a multiple of 5 and b must be a multiple of 4, the only solutions of this are a = 0, b = 0 a = 5, b = 4 The first of these is not allowed, so we have h = "ab" = 54 w = "ba" = 45 11(h-w) = 11(9) = 99 h+w = 99 So this works. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/