Brain-Teaser 31: [Birthday weddings]
From The Sunday Times, 22nd October 1961 [link]
Jane said:
My mother and I were each married on our birthday and each at the same age. I was born four years and a day after my mother’s marriage, and my twins were born four years and a day after my marriage to John, who is three years older than I am, and who shares a birthday with my mother.
If you write a date, say Christmas Day this year, in a continuous line of figures it looks like this – 25121961. Very well, if you write down our five dates of birth in that way and add the resultant numbers, the total is 8829685.
When was her wedding day?
This puzzle was originally published with no title.
[teaser31]
S2T2 
Jim Randell 8:18 am on 18 April 2021 Permalink |
I was wondering when exactly “a year and a day” after 29th February was. It’s probably easiest to imagine there is a zero length 29th February in non-leap years, and then it is easier to calculate these “year and a day” offsets. In the analysis it turns out there is only one possible set of dates anyway.
Assuming the 5 years of birth are all before 2000, we see that their sum must be a 4-digit number, i.e. 9685. And there is no carry into the sum of the day and month digits, which are 882. So we can deal this these three columns separately.
As there are 5 birthdays they cannot all be in the same month (as the final digit of the day/month sum would end in 5 or 0). In order for the final digit to be 2, twins birthday must at the beginning of a month, Jane at the end of the previous month, and John and Mum the day before that.
So the twins could be born on:
And only 1st Mar (twins), preceded by 29th Feb (Jane), and 28th Feb (John, Mum), gives the sum 882.
If the twins were born (on 1st Mar) in year y, then the year of Jane’s marriage is 4 years and 1 day earlier. And Jane is married on her birthday, which is 29th Feb, so (y − 4) (and hence y) must be a leap year.
If Jane is married at age x years, then her birth date must be (y − x − 4) (also a leap year, so x must be a multiple of 4).
And John is born on 28th Feb, 3 years earlier, i.e. in year (y − x − 7).
Jane’s Mum was married exactly 1 year before that, i.e. in year (y − x − 8).
So she was born in year: (y − 2x − 8).
And the sum of the 5 birth years is 9685:
Looking at leap years, going backwards from 1960:
So the only viable candidate for (y, x) is (1960, 24) (as x must be a multiple of 4).
So the dates we are interested in are:
Adding the 5 birth dates converted to integers we get:
Solution: Jane’s wedding day was: 29th February 1956.
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