| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1830.1 | Olympic rings | FRASER::FRASER | Jim Fraser | Wed Dec 29 1993 12:30 | 10 |
| 1. There are 9 regions inside the 5 rings of the Olympics. Put a
different whole number from 1 to 9 in each so that the sum of the
numbers in each ring is the same.
Here's one solution:
9 4 1 8 3 2 5 6 7
------ ----------- ------
----------- -----------
13 13 13 13 13
|
| 1830.2 | | RTL::GILBERT | | Wed Dec 29 1993 16:10 | 35 |
| > 4. On each of three cards was written a whole number from 1 to 10.
> These cards were shuffled and dealt to three people who recorded the
> numbers on their respective cards. The cards were collected, and the
> process was repeated again. After a few times, the three people
> computed the totals of their numbers. They turned out to be 13, 15,
> and 23. What were the numbers on the cards?
Let the numbers be a, b, and c. We have: 13+15+23 = 51 = d*(a+b+c),
where d is the number of deals. Since 51 = 3*17, d = 1, 3, 17, or 51.
'A few times' implies d > 1. If d were 17, then a+b+c=3, and so
a=b=c=1, so the sums should be equal, which they aren't. If d were
51, a+b+c=1, which is can't. So d = 3. And a+b+c = 17.
Given that a+b+c=17, and 1 <= a,b,c <= 10, there are 15 ways the
cards could be numbered (ignoring permutations). Examine
ways/whether the sums 13, 15, and 23 can be produced, and we
see that the cards were numbered 9, 5, and 3.
Possibilities:
10 6 1 6+6+1=13
10 5 2 5+5+5=15
10 4 3 10+10+3=23
9 7 1 7+7+1=13 9+7+7=23
9 6 2 9+2+2=13
9 5 3 5+5+3=13 9+3+3=15 9+9+5=23 <== Solution!
9+3+3=13
9 4 4
8 8 1
8 7 2 8+8+7=23
8 6 3 8+6+3=15
8 5 4 5+4+4=13 5+5+5=15
7 7 3 7+3+3=13
7 6 4 7+4+4=15
7 5 5 5+5+5=15
6 6 3
|
| 1830.3 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Wed Dec 29 1993 18:41 | 67 |
| > 3. Peter bought some pigs at $590 each and some horses at $670 each.
> He bought more horses than pigs. He paid with a $10,000 bill and got
> back some $100 bills and some $10 bills. Had the numbers of pigs and
> horses bought been interchanged, so would the numbers of $100 and $10
> bills obtained as change. How many pigs and how many horses did Peter
> buy?
This can be formulated as
p = number of pigs
h = number of horses
a = number of $100 bills
b = number of $10 bills
590p + 670h + 100a + 10b = 10000 = 590h + 670p + 100b + 10a
h > p
h,p,a,b are positive integers
Modulo 9 the long double equation is
5p + 4h + a + b == 1 == 5h + 4p + b + a
or
h - p == 0 (mod 9)
So not only did he buy more horses than pigs, but the difference
is a multiple of nine.
Nine horses cost $6030; twice that already exceeds $10000 so the
difference is exactly nine. Four more horses and pigs would again
put the price over $10000, so there are at most three pigs.
Plugging h = p + 9 into the above yields
590p + 670h + 100a + 10b = 10000 = 590h + 670p + 100b + 10a
590p + 670(p + 9) + 100a + 10b = 10000
1260p + 6030 + 100a + 10b = 10000
126p + 10a + b = 397
and
590(p + 9) + 670p + 100b + 10a = 10000
1260p + 5310 + 100b + 10a = 10000
126p + 10b + a = 469
Subtracting the first from the second gives
9(b - a) = 72
or
b - a = 8.
Plugging b = a + 8 into 126p + 10a + b = 397 yields
126p + 10a + a + 8 = 397
126p + 11a = 389
As p = 0,1,2,3 this yields 11a = 389, 263, 137, 11. Only
the last is a multiple of 11, so we have p = 3 and a = 1,
with h = 3 + 9 = 12 and b = a + 8 = 9, i.e.,
Peter bought 3 pigs and 12 horses.
Dan
|
| 1830.4 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Wed Dec 29 1993 19:16 | 67 |
| > 5. Each of A, B, C, D, and E was told in secrecy a different whole
> number from 1 to 5. The teacher asked A: "Whose number is largest?" A
> said: "I don't know." The teacher then asked B: "Is C's number larger
> than yours?" B said: "I don't know." The teacher asked C: "Is D's
> number larger than yours?" C said: "I don't know." The teacher then
> asked D: "Is B's number larger than yours?" D's answer was not
> recorded. Finally, the teacher asked B again: "Is C's number larger
> than yours?" B said: "No." From the above information, it is possible
> to deduce which number was told to whom, and was D's answer "Yes", "No"
> or "I don't know"?
Recall the numbers are different, so one student has a one,
one student has a two, one student has a three, one student
has a four, and one student has a five. There are 5! = 120
such permutations.
> The teacher asked A: "Whose number is largest?" A said: "I
> don't know."
So A does not have the five, otherwise A would have known that
A had the largest number. 24 possible permutations are thus
ruled out, leaving 96 possible permutations.
> The teacher then asked B: "Is C's number larger than
> yours?" B said: "I don't know."
So B has neither the one nor the five, otherwise B could have
answered "Yes" or "No" to the question. 54 possible
permutations remain.
> The teacher asked C: "Is D's number larger than yours?" C
> said: "I don't know."
So C also has neither the one nor the five, for the same
reason as B. 24 possible permutations remain.
> The teacher then asked D: "Is B's number larger than yours?"
> D's answer was not recorded. Finally, the teacher asked B
> again: "Is C's number larger than yours?" B said: "No."
If D has the one or two, then D answered "Yes." If D has the
four or five, then D answered "No." If D has the three then
D answered "I don't know."
If D answered "I don't know." then D has the three, and the
only possibilities for the numbers for ABCDE are 12435 and
14235. In either case B knows if C's number is larger, and
since B answered "No." then the sequence must be 12435.
If D answered "Yes." then the possibilities include 13425 and
43215 and so if B has a three then B cannot know if C's number
is larger. If B has a two then D has the one and so B knows
C's number to be larger. So for B to know that C's number is
not larger then B must have the four, and the possibilities
are 14325 24315 and 34215.
If D answered "No." then again for B to know that C's number
was not larger, B must have had the four, and this leaves the
possibilities 14253 14352 24351 and 34251.
Since we are given that the answers were sufficient to rule
out all but the actual possibility, then the numbers were
given out as A - 1, B - 2, C - 4, D - 3, E - 5, and D's answer
was "I don't know."
Dan
|
| 1830.5 | partial answer to Q 1. | AIAG::MOORE | Jim Moore | Thu Dec 30 1993 14:15 | 53 |
| re .1
> Here's one solution:
>
> 9 4 1 8 3 2 5 6 7
> ------ ----------- ------
> ----------- -----------
> 13 13 13 13 13
>
Hers's the other ( the smallest total )
A B C D E F G H I
____________________________________________________
9 2 5 4 6 1 7 3 8
------ ----------- ------
----------- -----------
11 11 11 11 11
Since
A + B =
B + C + D =
D + E + F =
F + G + H =
H + I = X
then
A + 2B + C + 2D + E + 2F + G +2H + I = 5X
the lower limit would be if B,D,F and G were 1-4
5+6+7+8+9 = 2(1+2+3+4) = 55 = 5X
X = 11
the upper limit would be B,D,F, and G were 6-9
1+2+3+4+5 + 2(6+7+8+9) = 75 = 5X
X = 15
11 < X < 15
but X = 15 is impossible as using 6-9 at the intersections
forces 9 6 or 7 8
--------- ---------
n n
which requires n = 0
Lastly, X = 12 or X = 14 can not be accomplished.....
...
...
left as an exercise :-)
|
| 1830.6 | | RTL::GILBERT | | Thu Dec 30 1993 14:19 | 12 |
| 2. A country has only $6 and $7 bills. How many different amounts, in
whole numbers of dollars, cannot be paid for exactly, and what is the
highest such amount?
The amounts 6*n thru 7*n can be formed with n bills; that is:
0, 6..7, 12..14, 18..21, 24..28, 30..35, 36..42, 42..49, 48..56, ...
The only amounts that can't be formed are:
1..5, 8..11, 15..17, 22..23, and 29.
So 5+4+3+2+1 = 15 amounts can't be formed, and the largest of these
is 29.
|
| 1830.7 | | RUSURE::EDP | Always mount a scratch monkey. | Thu Dec 30 1993 15:34 | 7 |
| Re .5:
Ah, that explains why I had trouble interpreting the problem; they were
asking for the largest and smallest possible sums.
-- edp
|
| 1830.8 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Mon Jan 03 1994 14:09 | 12 |
|
I can pay for a 1$ item by giving a 7$ bill and getting back a 6$ bill in change.
Hence I can pay for an X$ item by repeating the 1$ scenario X times. So all
prices are possible to pay.
Am I missing something ? (Yeah, yeah, I know it would require having lots of
cash on hand, but this is math not reality so I assume the bills are available)
/Eric
|
| 1830.9 | | AUSSIE::GARSON | Hotel Garson: No Vacancies | Tue Jan 04 1994 16:36 | 13 |
| re .8
You are probably right that the wording leaves something to be desired
but then it may have lost something in the translation from Japanese. I
think you have to interpret the word "exactly" to mean that the amount
you hand over is exactly equal to the purchase price.
FWIW I vaguely recall that if (a,b)=1 then there exists a maximum
integer (dependent on a and b of course) that cannot be expressed as
ma+nb where m,n>=0 [EFTR] whereas, still with (a,b)=1, if m and n are not
required to be non-negative then it is obvious that all integers are so
expressible i.e. as you observe the problem would be uninteresting if
change (m or n negative) is allowed.
|