| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 729.1 | | IJSAPL::PTTCS_TONY | | Fri Jul 10 1987 07:53 | 10 |
|
I imagine that there is some significance in the statement relating
to the sum of the ages but it eludes me entirely!
A fairly simple numerical task yeilds 4 possible answers to this
problem, but again, I think I'm maybe missing the WHOLE point.
|
| 729.2 | | IJSAPL::PTTCS_TONY | | Fri Jul 10 1987 08:05 | 11 |
|
Should have added : -
Each of the four possible solutions yield unique values on summation
of the three integers, so knowing the house number would provide
the questioner with a single, correct solution...BUT
There has got to be a more elegant method of solving this problem!!!
|
| 729.3 | Clarification? | SQM::HALLYB | Like a breath of fresh water... | Fri Jul 10 1987 10:34 | 7 |
| "Eldest" implies strictly greater than, right? (I mean, you could
have three 7-year-olds and one could still be "eldest...)
Ages and house numbers are positive integers, right? (E.g., you
don't wish to admit ages of 10, 7�, and 3 and an address of 20�).
John
|
| 729.4 | | CLT::GILBERT | eager like a child | Fri Jul 10 1987 12:20 | 5 |
| Yes, assume that the ages are all positive integers less than 200,
that "eldest" implies a unique greatest integer age, the house number
is a positive integer, the census taker knows the house number, and
needs to know the ages, and that the census taker and resident tell
no lies.
|
| 729.5 | "I am the cleverest????" | KEEPER::DEHOLLAN | | Fri Jul 10 1987 19:08 | 6 |
| Re .4
Your "clarification" sure messed up my dimwit solution:
Take "I am the eldest" in the same sense as when Muhammed Ali
says "I am the greatest," then the house number stuff is just
hokum and the resident talking is 225 years old, the other two
are 1 each.
|
| 729.6 | | SSDEVO::LARY | | Fri Jul 10 1987 21:47 | 4 |
| Another assumption which helps is that the census taker is very smart -
i.e. the census taker did not ask about ages the third time because he/she
was dim.....
|
| 729.7 | I think I got it | BANDIT::MARSHALL | hunting the snark | Mon Jul 13 1987 17:48 | 28 |
| here tis...
the factors of 225 are 5,5,3,3,1; all combinations below:
A B C SUM
--- --- --- ---
5*5 3 3 31
5*3 5 3 23
3*3 5 5 19
5*5*3 3 1 79
5*3*3 5 1 51
5*5 3*3 1 32
5*3 5*3 1 31
5*5*3*3 1 1 227
knowing the house number still left the census taker confused, so
the correct combination must be one of the two that sum to 31.
knowing that A is the eldest (and not a twin) yields the unique
solution of:
25, 3, 3
/
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/
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| 729.8 | solved | VINO::JMUNZER | | Tue Jul 14 1987 09:13 | 3 |
| .7 is right.
John
|