Your university! (2)
(C) WLODEK BLACK
Drawings - (C) Soop (Alex Antipov)
... Hello dear sixth grader ;-)!
It is time to deal with the decision of the competition
problems. As we promised, upon the expiration of sending work
to the competition in the Palace Pioneers tasks are published.
Who did not - that, alas, too late ...
Task 1.
1. On the lake the twentieth day
blossomed a whole lake poured. Each wing tsvetaden number mi.
At what flowers Udval-day flowers
ivalos, and covered half of wine lake?
Addressing Strictly speaking, this problem - a typical geometric
progression with factor 2 and an initial value of 1. But even
without the knowledge of such abstruse terms you might guess,
that if each subsequent day the flowers grew in 2 times more
than in the previous day, then every day before they had the
same times smaller. If the twentieth day of the entire lake
became overgrown, half of the lake was filled the previous day,
ie 19 th day.
Objective 2.
2. Find the last digit of the number 14 ^ 14 (14 degree 14).
Addressing Every even level of 14 ends in June, and the odd -
4. Do not have to prove? Calculate, probably, too!
Task 3.
3. Figures some nomu. Can sumchisla recorded in small equal 999?
reverse order. Can she ravPoluchennoe number nyatsya 9999?
added to the exodus
Addressing first give an answer to the second question: 9999 to
get Of course, you can. To do this, ensure the condition: the
sum of the first and the last digit must be equal to 9, and the
sum of the second and third numbers must also be equal to 9.
Difficult even to count the total the number of options: 1098
8901, 1188 8811, 1278 and 8721, etc. to 8901 1098 - only 80
combinations. Formally, their 100: From 0099 9900 0099 to 9900,
but the normal mathematical number must not start from scratch.
Consider now the 999. At first glance, it suffices to note that
there is a positive answer would recognize that the average
nine is the sum of two identical digits (because, obviously,
when you turn the three-digit number average figure will remain
in place). But how real programmers ;-), we are obliged to
conjecture: what if the average nine is the sum of the two
quads plus a carry flag from the category of units? Oh, the
denial of such approval is not so obvious! Sensitive Tense
gyrus: if such a combination of numbers exist, it should
provide in the discharge of units nine (nine different way in
the discharge of units simply have nowhere to take) and still
pass the unit transfer to the discharge of tens. Alas, even 9
+9 is equal to only 18, that is, in the discharge of units -
eight. Get nine and transfer simultaneously is impossible. Here
and proved.
Task 4.
4. If 30 people sit down in
movie-theater, at least in
one row would be no less
2 people. If in the hall
seating for 26 people, then
at least 3 rows will be
empty.
How many rows in the hall?
Addressing If two out of 30 people inevitably find themselves
sitting on the same row, then maybe, Series 28 +1 = 29. If 26
people sit on one in a row and at the same 3 lines blank -
perhaps all three rows of 26 = and 29. Apparently, the rows in
the cinema actually 29, although the condition of the problem
is formulated so that the exact response still need to give is:
"not more than 29 series." Because both conditions restrict the
number of rows only from the peak, see for yourself!
Task 5.
5. In Petit, Vasey and Coley together
there were 120 sweets. First Peter
gave candies and Basil Cole -
each for as long as
Each of those were. Then
Vasya gave candy Cole and Pete -
each for as long as
each of those before this
was. Finally, Kohl gave
Candy Basil and Pete - each
as much as each of
those were at the time. In
result, all got
equally. How many candies were in
everyone in the beginning?
Addressing Unlike earlier, this task is 100% serious,
that is, there is not a trick, no obvious solutions. Solve it
best at the end. So, the last stage of each of the boys turned
on 40 candies, while Vasya and Petit is formed by adding the
number of the same amount of candy, what they had before this,
ie 20. That is, on the penultimate stage, Vasey and Petit had
20 sweets. And Coley, therefore, 80 (remember: only 120).
Rolled back by another step back: 80 Colin 20 Petin candy
turned by Basil, who gave Cole and Pete so much candy, how much
each of those before this stage was. This means that Kolya was
80 / 2 = 40, and Petya - 20 / 2 = 10 pieces. Vasya, therefore,
120-10-40 = 70 candies. And finally, in the first stage gave
Peter Kohl and Basil to chocolates so much, How many of those
have been, and was obtained by: Kolya - 40 / 2 = 20, while
Vasey - 70 / 2 = 35 candies. In Petit, it turns out, 120-35-20
= 65 candies. Can check, here's the scheme:
Kohl Petya
1. 20 35 65
2. 40 70 10
3. 80 20 20
4. 40 40 40
It all comes true?
