In Florida I met a peculiar couple. His age digits were a reverse of her age digits. They were relaxed and open about their age differences and said that last year he was exactly twice as old as she was. What are their ages?
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Answers accepted all day long on Friday May 13th and Saturday May 14th, on our Family Puzzle Marathon. They will be hidden until Sunday morning (EST) and everyone who solved it will get a puzzle point. Please, explain your answer.
15 comments:
She is 37 and he is 73. Last year she was 36, and he was 2x her age -- 72.
The ages are 73 and 37.
If we let The man's age be 10x + y and the woman's age be 10y + x, then we have 10x + y - 1 = 2(10y - x - 1) from last year.
This simplifies to 8x - 19y = -1, which only has integer solutions between 0 and 9 when x = 7 and y = 3.
So, the man is 73 and the woman is 37.
He is 73, and last year he was 72 while she was 36, now 37.
I couldn't think any "formula" for this, and just found it iteratively (is that a word?). Guessing, I suppose, and I started guessing in the 60s.
Since his age doubled hers last year, his age then was an even number. That didn't help much however.
She is 37, he is 73. Last year, they were 36 and 72.
Not too many options, since his age must be two digits (so she's under 50), odd (so she is a teen or 30-something) and are adults (ruling out the teen, since 81 and 91 are clearly way more than double). So given her first digit is 3, doubling is 6, plus 1 is 7: 37 and 73.
She's 37, he's 73, and last year was 36 and 72.
FWIW - I just kinda searched on this in my head, thinking that the 1st digit of her age must be less that 5, and the 2nd digit greater than 5.
I'm thinking that people who read this blog also watch the Big Bang Theory (I know SO many people like Sheldon, and Howard). In any event, Sheldon sees 73 as the perfect number:
"The best number is 73, stating that 73 is the 21st prime number and its mirror, 37, is the 12th prime number. And 12's mirror, 21, is the product of 7 and 3. It is also a palindromic binary number 1001001."
Yeah, I'm a geek. In my defense, someone else is a geek too, because I Bing'd it and found at least 10 hits :)
A nice little algebraic puzzle before the weekend!
Given the information, we have:
10a + b - 1 = 2 (10b + a - 1)
and by simplifying, we get:
b = ( 1 + 8*a ) / 19
Therefore, we just need to find the right 'a' that is divisible by 19.
Since we only have 9 numbers to try, we find that a = 7 yields b = 3
Therefore, the ages are 73 and 37 with last year, the ages being 72 and 36 (72 = 2 * 36)
Wang
Hi,
I did it by elimination so I do not come out with nice formulas...
Last year, the guy's age was even, then the woman's age should begin by an odd number: 1 or 3 or 5 or 7 or 9.
It can not be 5 or above because the guy would be over 100 years old.
Then, the more realistic pick is the number 3.
Finally, I came out with 37 and 73 years old.
I think of Florida every winter, we used to escape Chicago long winter vacationing over there...
If he was twice her age last year, then his age needs to be an even number, so this year it is odd, so her age needs to have an odd first digit. If his age digits are a reverse of hers, there must be 2 digits, so he is 99 or younger, so she has to be younger than 50. Her first digit must be 1 or 3. If she is in her thirties, then her age doubled would make him in his sixties or seventies, so her second digit has to be 6 or 7. Last year she was 36 and he was 72, and this year she is 37 and he is 73.
He is 73, She is 37 ( A year ago he was 72 and she was 36)
Let his age be XY, her age is YX. X and Y are just the digits 0 thru 9
The VALUE of his age is 10 * X + Y; The VALUE of his age a year ago would be 10 * X + Y - 1. This value is equal to twice her age a year ago OR
10 * X + Y - 1 = 2 * (10 * Y + X - 1 ) ,
10 X + Y - 1 = 20Y + 2X - 2,
8X = 19Y - 1
Let Y be 0 thru 9; X MUST be an integer
When Y = 3, we get 56 for the right-hand side, an even multiple of 8 and yielding the answer of X = 7
Your neighbor John (Sam's Dad)
She is 37 years old and he is 73 years old. Last year, she was 36 years old and he was double her age at 72 years old. I solved this puzzle a lazy way using an Excel spreadsheet. The first column was her age last year; the second column was her last year's age doubled to reflect his age last year. Third column was her current age and the fourth was his current age. I scanned down the last two columns to find reverse digits at 37 and 73. Sometimes you do the math and sometimes you let the computer do the math.
Let the man's age be 10a+b. (a,b integers between 1 and 9)
Then the woman's age is 10b+a.
Last year the man's age was 10a + b -1; woman's age was 10b + a - 1
10a + b -1 = 2* (10b +a -1)
Solving, we get 8a = 19b - 1
Obviously, b is odd (otherwise 19b - 1 will be odd)
b = 1 => 19b - 1 = 18
b = 3 => 19b -1 = 56
b = 5 => 19b - 1 = 94
Of these, 56 is divisible by 8.
Hence 8a = 56 => a = 7; b = 3
Man's age is 73. Woman's age is 37.
(actually this was solved by my 11 year old daughter)
The husband is 73 and his wife is 37. Since their digits were the same this year, only transposed, I knew their age differential was a multiple of 9. For example 54 transposed is 45, 63 is 36, 72 is 27 etc. The difference is always a multiple of 9.
Then I tried all the combinations of numbers that had a 9 multiple difference AND where one number was twice of another. The possibilities I came up with were 36-18, 54-27, 72-36 and 90-45. The only one that shared the same digits one year later was 72-36, which became 73-37.
I've only been to Florida a couple of times but I remember more than a few couples that fit this same description!
mathmover
Houston, we have a problem.
It looks like we have two puzzle solvers named Pat. Or is it one that provided two different explanations on two different days (split personality)? Thank you for the Big Bang reference on the perfection of the number 73. This is cool.
If we have two Pats, we will need to find a fair way of splitting puzzle points. And you have to sign with two initials in the future.
You all figured out the right ages, one way or another: 73 and 37. There is a formula that gets you half way to the solution and many of you listed it above. Congratulations on another puzzle point for Ellina, SteveGoodman18, Tom, Bean, Pat^2, Wang, anne-maria, Andree, John W, rajagosh, mathmover.
Hey Maria, I posted my solution fairly quickly, but it seems to have disappeared from the list. It's a little disappointing. Blogspot ate my homework!
Sorry to hear this. As we saw they had some issues on Friday the 13th. I couldn't publish the puzzle for half a day. Don't even know what to suggest. You can alays feel free to just email the solutions to me via contact link at the bottom of each page.
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