Number Puzzle #7 – A solution
February 22nd, 2009 by SplineGuy
I love these kinds of puzzles because there is only one solution but there are several ways to get there. It’s at least as interesting to hear the different approaches as it is to solve it.
Here was the puzzle posted last week:
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with 4-digit number in which each of two digits appears two times.
“Look, daddy!” she exclaims. “That number is evenly divisible by the age of each of us kids!”
“That’s right,” replies Mr. Jones, “and the last two digits just happen to be my age.”
What is the four digit number in the license plate?
The first thing I noted when I saw the problem is that the solution must be in one of the following forms: aabb, abba, or baba (where a and b are distinct digits). Also, I noticed that since the oldest child is 9, the sum of digits must be a divisible by 9. So that 2a + 2b = 9k for some k. Because 2a+2b is even, k must be even. We know that a and b are distinct digits between 0 and 9, inclusive, the largest 2a+2b could be is 34, so k must be 2. In other words, a+b=9.
Next thing I noticed is that Mr. Jones will have either 4 year old or an 8 year old or both. Either way, the 4-digit number is divisible by 4. This means the last two digits must be divisible by 4. We thus have the following possibilities: 9900, 5544, 1188, 3636, 6336, 2772, 7272. I arrived at these by thinking first of numbers of the form bb that are divisible by 4 (00, 44, 88) and providing the appropriate a to get aabb that is divisible by 9. Then I considered the numbers of the form ba, that are divisible by 9 and 4 (36, 72) and listed both abba and baba.
Notice that in this list only 9900 is divisible by 5 and 00 can’t possibly be the age of Mr. Jones so he must not have a child age 5. Only 5544 is divisible by the digits 1, 2, 3, 4, 6, 7, 8, and 9.
SOLUTION: 5544
A few folks emailed me their solutions and were all correct, but each had an approach that was not quite the same as mine. Well done to those!
My approach was similar:
I first noticed that there were 8 children aged 9 or under, so (ignoring age 0) there must be only 1 age between 1 and 9 not present. That age cannot be 1, 2, 3 or 4, since they are all required for more than one other age to be divisible into the 4-digit number, nor can it be 9, as one son is 9. Thus the number must be divisible by all but 5, all but 7 or all but 8. The prime factors you need for 1,2,3,4,6,9 divisibility are 2×2x3×3, giving 36. So the options are:
No 5: Multiple of 36×7x2=504
No 7: Multiple of 36×5x2=360
No 8: Multiple of 36×5x7=1260
Also, the number is ABAB, ABBA OR AABB. ABAB is divisible by 101, AABB is divisible by 11.
Since 101 is too large to yield a 4 digit number (360×101=36360), we can guess that 11 is a factor, and that the pattern is ABAB. 1260×11 will also be >9999.
360×11 = 3690
504×11 = 5544 (bingo)
Much less guessing.