4. BE and CF are the altitudes from B and C on the opposite sides of a triangle ABC. If P is the mid point of BC, show that PE=PF. (Hint: Show that B,C,F,E are concyclic and BC is a diameter).
Special mention should be made of Darlington Hove of Kutama, who attempted all except Problem 4. He wins the prize for Problem 6. Joshua Sarai, Alfred Kumbirai and Jacob Ziwahwa all gave good solutions to Problems 2, 3, 5 and 6. Chari Zebron was the only one who was able to correctly answer Problem 1. The prizes for Problems 1, 2, 3, and 5 go, respectively, to Chari Zebron of Hippo Valley High School, Shepherd Chikoro and Alfred Kumbirai, both also of Hippo Valley High, and Thomas Masiwa of Rufaro Sec. School, Neshuro, who produced a brilliant solution to Problem 5. Sadly, no lady submitted any solutions to these problems. Are you ladies nervous about brain-challengers?
1. Five letters are written to five different persons, and their addresses (all different) are typed on five envelopes (one address on each). In how many ways can letters be placed in the envelopes so that no letter is placed in the correct envelope?
Solution: (given by Zebron Chari of Hippo Valley High School,
Chiredzi)
Suppose the letters are numbered 1, 2, 3, 4 and 5, and the
corresponding envelopes labelled A, B, C, D, E. The following table
shows the different ways of placing the letters in the wrong
envelopes.
| 1 | A | 2 | 2 | 2 | 2 | 3 | 4 | 5 | |||||||||||||
| 2 | B | 1 | 3 | 4 | 5 | 1 | 1 | 1 | |||||||||||||
| 3 | C | 5 | 4 | 4 | 1 | 5 | 5 | 5 | 1 | 4 | 4 | 1 | 2 | 5 | 4 | 2 | 5 | 5 | 2 | 4 | 4 |
| 4 | D | 3 | 5 | 5 | 5 | 1 | 3 | 1 | 5 | 3 | 1 | 3 | 5 | 2 | 5 | 5 | 3 | 2 | 3 | 2 | 3 |
| 5 | E | 4 | 3 | 1 | 4 | 4 | 1 | 3 | 3 | 1 | 3 | 4 | 4 | 4 | 2 | 3 | 2 | 3 | 4 | 3 | 2 |
| Total | 2 | 3 | 3 | 3 | 3 | 3 | 3 |
| 3 | 4 | 4 | 5 | 3 | 5 | ||||||
| 4 | 3 | 5 | 4 | 5 | 3 | ||||||
| 1 | 2 | 5 | 5 | 1 | 2 | 2 | 1 | 2 | 1 | 4 | 4 |
| 5 | 5 | 1 | 2 | 2 | 1 | 3 | 3 | 1 | 2 | 1 | 2 |
| 2 | 1 | 2 | 1 | 3 | 3 | 1 | 2 | 4 | 4 | 2 | 1 |
| 8 | 8 | 8 |
Therefore there are 2+3+3+3+3+3+3+8+8+8=44 ways of placing the letters in
wrong envelopes.
Here is another way of seeing this: the 1st letter can be put in wrong
envelope in 4 different ways - e.g. C; for EACH of these the
following possibilities exist for the other letters:
· Either the letter (3rd) that belongs where the first has gone,
goes into A (where the first should have gone); then the three others
(B, D, E) have
only 2 different ways of going all wrong.
· Or the letter that belongs where the first has gone, goes
somewhere else, which it can do in 3 different ways - e.g. 3rd
letter to B;
and then the letter that belongs where that letter went (in this case the
2nd) can go into any of the remaining 3 envelopes (in this case A, D, E - say 2nd goes into E), leaving the last 2 to their fully
determined fate (in our case 4th to A and 5th to D).
That is, for each of the 4 different ways of placing 1st letter there are 2 + (3 ×3)=2+9=11 possibilities for placing the rest, making in all 44 ways.
2. A machine numbers the pages of a book starting with 1 and uses 3189 digits in all. How many pages does the book have?
Solution: (A very concise solution given by Shepherd Chikoro of Hippo Valley High School, Chiredzi)
| Page Range | No. of Pages | Digits Per Page | Total Digits |
| 1 - 9 | 9 | 1 | 9 |
| 10 - 99 | 90 | 2 | 180 |
| 100 - 999 | 900 | 3 | 2700 |
| 1000 - 1074 | 75 | 4 | 300 |
| Totals | 1074 | 3189 |
Therefore the book has 1 074 pages.
(This was also solved by Alfred Kumbirai, Jacob Ziwahwa, and Joshua Sarai, of Hippo Valley High School, Chiredzi, and by Darlington Hove of Kutama College.)
3. What is the largest integer not exceeding 1000 which leaves remainder 3 when divided by 5, remainder 5 when divided by 7, and remainder 7 when divided by 9?
Solution: Correct solutions were given by Darlington Hove of Kutama College; Alfred Kumbirai, Shepherd Chikoro, Joshua Sarai and Jacob Ziwahwa, all of Hippo Valley High School. However, since their solutions were rather long, except for Shepherd who seemed to just `guess' the number, we provide the following simpler solution:
Let x be the required number. Then
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So 5(a+1)=7(b+1)=9(c+1). The LCM of 5, 7, 9 is 5×7×9=315.
And the largest multiple of 5, 7, 9 below 1 000 is 945=315×3.
Therefore to obtain x we simply use any one of the above expressions
e.g.
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5. Solve completely the following system of equations:
Solution: (given by Thomas Masiwa of Rufaro Secondary School, Neshuro)
We have x-2=[ 4/(y-2)] and z-3 = [ 9/(y-3)]. Hence (z-3-1)(x-2-2)=16 gives us
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(Other good solutions were given by Joshua Sarai, Alfred Kumbirai, and Jacob Ziwahwa, of Hippo Valley High, and S. Shoko of Kadoma)
6. The 29th of February in the year 2000 fell on a Tuesday, for your information. Show that in the whole of the present century the 29th of February will fall on a Tuesday only thrice. Give the three years when this will happen.
Solution: (The following is an edited version of T Masiwa's solution)
Between any two `leap days' - i.e. February the 29th - there are 4×365+1=1461 days. Denote Tuesday by 0, Wednesday by 1, Thursday
by 2,
Friday by 3, Saturday by 4, Sunday by 5 and Monday by 6.
So in 2 000, the 29th of February had value 0.
And in the next leap year 2 004, the same 29th will have value
1461=5 mod 7.
So in that order, the next 29th of February to have value 0 will occur
after 7 leap years from 2 000, i.e. 28 years.
Hence there can only be 3 leap Tuesdays in this century; namely in the years
2028, 2056 and 2084.