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Getting Squared With Dates (Posted on 2008-12-23) Difficulty: 3 of 5
During the course of the year 2000, two brothers Andrew and Brady set out to research the family tree and began by listing the dates of birth of all the family members born during the period covering January 1, 1900 to December 31, 1999 inclusively.

Each date was recorded as three pairs of digits, written in succession to form a positive six digit decimal integer with a zero preceding any single digit day, month or year. Andrew followed the dd-mm-yy format, while Brady followed the mm-dd-yy format. For example, May 7, 1998 would be written as 070598 and 050798 respectively by Andrew and Brady.

Comparing the two different integers representing Chloe’s birth date, they noticed that one of them was a perfect square, and the absolute difference between them was a nonzero perfect square. The same was true for their versions of the birth date of Chloe’s sister Denise, who is younger by between one and two years.

When was Chloe born? What was the birth date of Denise?

Note: Try to derive a non computer-assisted solution, although computer program/spreadsheet solutions are welcome.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution (mostly) non computer-assisted solution Comment 3 of 3 |

Call the month, day and year in which Chloe (or Denise) was born a, b, and c where a is between 1 and 12 (inclusive), b is between 1 and 31 (inclusive) and c is between 0 and 99 (inclusive). Then one of the two numbers is a*10^4 + b*10^2 + c and the other is b*10^4 + a*10^2 + c.

We're given that their difference is a perfect square. Well, that difference is:

(a-b)*10^4 + (b-a)*10^2 = 10^2 * (a-b) * (10^2 - 1)

= 10^2 * 3^2 * 11 * (a-b).

Since this number is a perfect square, (a-b) or (b-a) must be 11 times a perfect square. Since the maximum value of (a-b) is 30, the only candidate that fits
this requirement is 11*1^2 = 11 => |a-b| = 11. If a > b, then December 1 is the only possible date. Additionally, each month a has exactly one b (where b=a+11) that also applies. So there are a total of 13 possible month/day combinations for Chloe's (and Denise's) birthdays before looking at years.

We also have that one of the following equations must be true:
a*10^4 + (a+11)*10^2 + c = k^2 for some k OR
(b+11)*10^4 + b*10^2 + c = k^2 for some k OR
a*10^4 + (a-11)*10^2 + c = k^2 for some k OR
(b-11)*10^4 + b*10^2 + c = k^2 for some k


At this point it's fairly fast to consider each case in turn, noting that BOTH Chloe's and Denise's birth dates must be on the list. Also, since any time a=x,b=y a>b is a solution, then a=y,b=x b>a is also a solution since the two integers (of which one must be a perfect square) are the same pair in the opposite order. We can safely just consider cases when a < b and note the additional solution when swapping a and b's values still meets the range restrictions of the problem. That means we need only consider equations 1 and 4 above. But since |a-b| = 11, any (a,b) that solves eqation 1 is necessarily a solution to equation 4 so we can skip all of that and only consider equation 1 and the 12 possible
values of a.

a = 1 produces the solution => 011236 = 106^2
(note that this corresponds to TWO dates in 1936 -- December 1 and January 12 -- both are solutions but one format is the
square in one case and the other in the other case. It turns out not to matter, though -- see below)
a = 2 produces the solution => 021316 = 146^2
a = 3 produces no solution. 3*10^4 + 14*10^2 is more than 100 below the next perfect square (178^2)
a = 4,5, 6 also produce no solutions for the same reason
a = 7 produces the solution => 071824 = 268^2
a = 8,9,10 produce no solutions for the same reason as a=3
a = 11 produces the solution 112225 = 335^2
a = 12 produces no solutions

Of these, only the a = 11 and a = 7 solutions are less than two years apart and hence candidates for Chloe and Denise.

Chloe, being older, was therefore born July 18, 1924 and Denise a bit more than 16 months later on November 22, 1936.
12

 

[admission: I did use a calculator to calculate the square roots and squares when examining the a's above, but I feel that this is still in the spirit of a non computer assisted solution if not the letter. ]


  Posted by Paul on 2008-12-23 18:12:51
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