Hoàng Việt Nguyễn
09/03/2017 at 21:32
Vũ Thị Hương Giang 09/03/2017 at 21:54
The current child age is:
44 x 1/4 = 11 (fresh)
The age when the father is 60 years old is:
11 + (60  44) = 27 (fresh)

Asuna Yuuki 09/03/2017 at 21:40
His age now is :
44 x 1 : 4 = 11 ( age )
His age when his father is 60 years old is :
60  ( 44  11 ) = 27 ( age )
=> His age is 27 .

FA FIFA Club World Cup 2018 16/01/2018 at 21:59
The age of the present child is :
44 x 1/4 = 11 ( age )
The age of child when his father 60 years old is :
11 + ( 6044)=27 ( age )
answer : his age when his father 60 years old is 27 years old .
[...]Kurosaki Akatsu Coordinator
09/03/2017 at 21:31
Answers
2
FollowGive P=ax2+bx+ca′x2+b′x+c′
Demonstrate if aa′=bb′=cc′
then P does not depend on x
[...]
mathlove 09/03/2017 at 22:26Set aa′=bb′=cc′=k
. We have k=ax2a′x2=bxb′x=cc′=ax2+bx+ca′x2+b′x+c=P
.
So that P=k,∀x
. So P does not depend on x
.
Kurosaki Akatsu selected this answer.
[...]
FA FIFA Club World Cup 2018 just finishedSet aa′=bb′=cc′=k. We have k=ax2a′x2=bxb′x=cc′=ax2+bx+ca′x2+b′x+c=P
.
So that P=k,∀x
. So P does not depend on x .
[...]Kurosaki Akatsu Coordinator
09/03/2017 at 21:15
Answers
4
FollowCalculate expression value :
a) 4xy3 + 13
xy8 + 25xy3+ (−13)xy8 when x = 23 ; y = −12
[...]
Vũ Thị Hương Giang 09/03/2017 at 21:28a) 4xy3 + 13
xy8 + 25xy3 + −13
xy8
= 4xy3 + 25
xy3 + 13xy8 + −13xy8
= 143xy3+0=143xy3
[...]
Kurosaki Akatsu Coordinator 09/03/2017 at 21:21a) 4xy3+13xy8+23xy3+(−13)xy8
=(4+23)xy3+[(−13)+13]xy8
=143xy3+0
=143.23.(−12)3=−718
[...]
FA FIFA Club World Cup 2018 just finisheda) 4xy3+13xy8+23xy3+(−13)xy8
=(4+23)xy3+[(−13)+13]xy8
=143xy3+0
=143.23.(−12)3=−71
[...]Linh Kute
09/03/2017 at 19:08
Answers
1
FollowA shool has 24 classes and there are 40 students in each class. 1/8 of the students wear glasses and 1/12 of those who wear glasses are boys. How many schoolgirls wear glasses in the schoool?
[...]
Ace Legona 09/03/2017 at 19:26it's very easy. you should do it yourself
[...]falcon handsome moderators
09/03/2017 at 12:34
Answers
6
FollowLet the answer be k . We count the total number of paris of students were on duty together in the k days. Since every pair of students was on duty together exactly once, this is equal to C152
x 1 = 105. On the other hand, since 3 students were on duty per day, this is also equal to C32
x k = 3k. Hence 3k = 105 and so k = 35
double counting
[...]
Ace Legona 09/03/2017 at 18:03What did you try ?
[...]
Pham Hoang Nam 18/04/2017 at 21:33What did you try ?
[...]
phanhuytien 11/08/2017 at 20:59What did you try?
[...]falcon handsome moderators
09/03/2017 at 12:27
Answers
6
FollowReal numbers are written in an m x n table. Is is permissible to reverse the signs of all the numbers in any row or column, Prove that after a number of these operations we can make them sum of the numbers along each line (row or column) nonnegative
games
[...]
Ace Legona 09/03/2017 at 19:22Let S
be the sum of all the mn numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most 2mn tables. So S can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then S increases. Since S has finitely many possible values, S
can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative.
[...]
Nguyễn Anh Tuấn 24/03/2017 at 17:49Let SS be the sum of all the mnmn numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most 2mn2mn tables. So SS can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then SS increases. Since SS has finitely many possible values, SS can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative.
[...]
Hà Minh Hiếu 18/06/2017 at 21:09Let SS be the sum of all the mnmn numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most 2mn2mn tables. So SS can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then SS increases. Since SS has finitely many possible values, SS can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative.
[...]
FA FIFA Club World Cup 2018 just finishedLet SS be the sum of all the mnmn numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most 2mn2mn tables. So SS can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then SS increases. Since SS has finitely many possible values, SS can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative
[...]falcon handsome moderators
09/03/2017 at 10:48
Answers
2
FollowEach member of a club has at most three enemies in the club. (Here enemies are mutual). Show that the members can be divided into two so that each member in each member in each group has at most one enemy in the group
games
[...]
Ace Legona 09/03/2017 at 20:09In the beginning, randomly divide the members into two groups. Let S be the sum of the number of the pairs of enemies in each group. If a member has at least two enemies in the same group, then the member has at most one enemy in the other group. Transferring the member to the other group, we will decrease S by at least one. Since S is a nonnegative integer, it cannot be decreased forever. So after finitely many transfers, each member can have at most one enemy in the same group.
[...]
Vũ Việt Vương 01/04/2017 at 16:55n the beginning, randomly divide the members into two groups. Let S be the sum of the number of the pairs of enemies in each group. If a member has at least two enemies in the same group, then the member has at most one enemy in the other group. Transferring the member to the other group, we will decrease S by at least one. Since S is a nonnegative integer, it cannot be decreased forever. So after finitely many transfers, each member can have at most one enemy in the same group.
[...]falcon handsome moderators
09/03/2017 at 10:26
Answers
2
FollowThere are three piles of stone numbering 19,8 and 9, respectively . You are allowed to choose two piles and transfer one stone from each of these two piles to the third piles . After several of these operations, is it possible that each of three piles has 12 stones ?
games
[...]
Ace Legona 09/03/2017 at 19:37No. Let the number of stones
in the three piles be a,band c,
respectively. Consider (mod 3) of these
numbers. In the beginning, they are 1,2,0. After one operation, they become 0,1,2 no matter which two piles have stones
transfer to the third pile. So the
remainders are always 0,1,2 in some
order. Therefore, all piles having 12stones are impossible.
[...]
Neymar Jr 08/04/2017 at 17:06i don't know.and you? [leuleu]
[...]falcon handsome moderators
09/03/2017 at 10:20
Answers
2
FollowFour x's and five o's are written around the circle in an arbitraty order . If two consecutive symbols are the same then insert a new x between them. Otherwise insert a new o between them. Remove the old x's and o's . Keep on repeating this operation .It is possible to get nine o's ?
games
[...]
Vũ Việt Vương 01/04/2017 at 16:56If we let x=1x=1 and o=–1o=–1, then note that consecutive symbols are replaced by their product. If we consider the product PP of the nine values before and after each operation, we will see that the new PP is the square of the old PP. Hence, PP will always equal 11 after an operation. So nine oo's yielding P=–1P=–1 can never happen.
[...]
Ace Legona 09/03/2017 at 20:12If we let x=1
and o=–1, then note that consecutive symbols are replaced by their product. If we consider the product P of the nine values before and after each operation, we will see that the new P is the square of the old P. Hence, P will always equal 1 after an operation. So nine o's yielding P=–1
can never happen.
[...]American
09/03/2017 at 09:51
Answers
2
FollowIn an 8x8 board, there are 32 white pieces and 32 black pieces, one piece in each square. If a player can change all the white pieces to the black and all the black to the white in any row or column in a single move, then is it possible that after finitely many moves, there will be exactly one black piece left on the board?
games
[...]
An Duong 10/03/2017 at 14:28The answer id NO.
At the begining there are 32 black pieces. We will prove that the number of black pieces left on the board is always even.
In the single move, we change a row or column, assume that row or column has k black pieces and 8  k while pieces, after change all black to white and while to black, that row or column will has 8k black and k white. The difference between black pieces after a single move is (8  k)  k = 8  2k. The difference is a even (8  2k). Therefore, after every move, the black pieces left is alaways even and cannot be one black piece at any time.
[...]
Vũ Việt Vương 01/04/2017 at 16:57My answer is no.
At the begining there are 32 black pieces. We will prove that the number of black pieces left on the board is always even.
In the single move, we change a row or column, assume that row or column has k black pieces and 8  k while pieces, after change all black to white and while to black, that row or column will has 8k black and k white. The difference between black pieces after a single move is (8  k)  k = 8  2k. The difference is a even (8  2k). Therefore, after every move, the black pieces left is alaways even and cannot be one black piece at any time.
[...]American
09/03/2017 at 09:39
Answers
6
Follow(1974 Kiew Math Olympiad)
Numbers 1, 2, 3, ..., 1974 are written on the board. You are allowed to replace any two of these numbers by one number, which is either the sum or the difference of these two numbers. Show that after 1973 times performing this operations, the only number left on the board cannot be 0.
games
[...]
hghfghfgh 26/03/2017 at 20:16At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a  b, we see that:
+ if a odd and b even, or a even and b odd then a + b or a  b is still odd
+ if a and b are both even then a + b or a b is still even
+ If a and are both odd then a + b or a  b is even
So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even). [haha]
[...]
An Duong 10/03/2017 at 14:14At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a  b, we see that:
+ if a odd and b even, or a even and b odd then a + b or a  b is still odd
+ if a and b are both even then a + b or a b is still even
+ If a and are both odd then a + b or a  b is even
So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even).
[...]
xicor 24/08/2017 at 10:39coppy là gian nghe
[...]falcon handsome moderators
09/03/2017 at 12:43
Answers
6
FollowLet a0
= 1, a1=1 and an=4an−1−4an−2
for n>=2
Find a formula for an
in terms of n.
generating functions
[...]falcon handsome moderators
09/03/2017 at 12:55
Answers
1
FollowIn each cell of a 100 x 100 table, one of the integers 1,2,...,5000 is written. Moreover, each integer appears in the table exactly twice. Prove that one can choose 100 cells in the satisfying three conditions below.
(1) Exactly one cell is chosen in each row.
(2) Exactly one cell is chosen in each column.
(3) The numbers in the cells chosen are pairwise distinct
probabilistic method
[...]
demo acc 10/03/2017 at 14:42very diffcult [khocroi] [oe]
[...]falcon handsome moderators
09/03/2017 at 12:58
Answers
2
FollowGiven 2n points in a plane with no three of them collinear. Show that thay can be divided into n pairs such that the n segments joining each pair do not intersect
games
[...]
Ace Legona 09/03/2017 at 19:14In the beginning randomly pair the points and join the segments. Let S
be the sum of the lengths of the segments. (Note that since there are finitely many ways of connecting 2n points by n segments, there are finitely many possible values of S.) If two segments ABand CDintersect at O, then replace pairs ABand CDby ACand BD
. Since
AB+CD=AO+OB+CO+OD>AC+BD
by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease S
. Since there are only finitely many possible values of S
, so eventually there will not be any intersection.
[...]
Vũ Việt Vương 01/04/2017 at 16:59[banh]
In the beginning randomly pair the points and join the segments. Let SS be the sum of the lengths of the segments. (Note that since there are finitely many ways of connecting 2n2n points by nn segments, there are finitely many possible values of SS.) If two segments ABABand CDCDintersect at OO, then replace pairs ABABand CDCDby ACACand BDBD. Since
AB+CD=AO+OB+CO+OD>AC+BDAB+CD=AO+OB+CO+OD>AC+BD
by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease SS. Since there are only finitely many possible values of SS, so eventually there will not be any intersection. [leuleu] [eoeo] [bucqua] [gianroi] [undefined]
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In the name of love 11/03/2017 at 20:42
The age of the present child is :
44 x 1/4 = 11 ( age )
The age of child when his father 60 years old is :
11 + ( 6044)=27 ( age )
answer : his age when his father 60 years old is 27 years old .