Hello! My name is Azat, I am a 3rd year student at the HSE Faculty of Computer Science. A few days ago a friend from the HSE Economics contacted me and asked for help with solving the problems of the entrance exam at the School of Economics. My classmate Daniil and I looked at the tasks, they seemed to us rather difficult, but very interesting, I wanted to break my head over them. As a result, we solved 1 of the options for 2019 and want to show our solutions to the world.
Task 1
Fill in the third column of the matrix\ frac {1} {6} \ left (\ begin {array} {ccc} 5 & -2 &? \\ - 2 & 2 &? \\ - 1 & -2 &? \ end {array} \ right)
\ frac {1} {6} \ left (\ begin {array} {ccc} 5 & -2 &? \\ - 2 & 2 &? \\ - 1 & -2 &? \ end {array} \ right)
if it is known that this is the matrix of orthogonal projection onto some plane.DecisionA, :
A2=A :
A2=136(5β2xβ22yβ1β2z)(5β2xβ22yβ1β2z)=136(29βxβββ14βyβββ1βzββ)
=16(5β2xβ22yβ1β2z)=A
2 3 , , .
:
{29βx=5β
6β14βy=β2β
6β1βz=β1β
6β{29βx=30β14βy=β12β1βz=β6β{x=β1y=β2z=5
, 3 ,
A=16(5β2β1β22β2β1β25)
Task 2
What can you say about the convergence (absolute or conditional) of a series sum inftyn=1(n+2019)anif it is known that the series sum inftyn=1(nβ2019)an converges (a) absolutely, (b) conditionally?Decision:
S=ββn=1(n+2019)an,T=ββn=1(nβ2019)an.
Sβ²=ββn=1|n+2019||an|,Tβ²=ββn=1|nβ2019||an|
A=ββn=1an,Aβ²=ββn=1|an|
(1).
ββn=1(n+a)an βAβ²=ββn=1an βaβZ.
ββn=1(n+a)ann+a (, , -a).
,
ββn=1Ξ±nΞ²n,
Ξ±n=1n+a,Ξ²n=(n+a)an.
n=max(1,βa+1), . . :
1.
Bn=βnk=1Ξ²k ,
ββn=1Ξ²n .
2.
Ξ±nβ©ΎΞ±n+13.
limnββΞ±n=0, . , .
a) T ,
Tβ²=ββn=1|nβ2019||an| . :
Tβ²=2018βn=1|nβ2019||an|+ββn=2019|nβ2019||an|=β¦
2018 , :
β¦=2018βn=1(nβ2019)|an|+ββn=2019(nβ2019)|an|=ββn=1(nβ2019)|an|
, (1),
ββn=1|an| .
2018 ( , ) :
Sβ²βTβ²=ββn=2019((n+2019)β(nβ2019))|an|=4038ββn=2019|an|
,
Aβ²=ββn=1|an| . ,
Sβ²=Tβ²+Aβ² β 2 . .
)
T ,
T ,
Tβ² β .
,
S .
,
T (1)
A. , ,
S T, ,
S . ,
Sβ² .
.
Sβ² . ,
2018 Sβ² Tβ², , :
ββn=2019|nβ2019||an|β€ββn=2019|n+2019||an|
|n+2019|β₯|nβ2019|βnβ₯2019.
, , .
Tβ² , .
Task 3
Alena loves algebra very much. Every day, going to her favorite algebraic forum, she is likely frac14 finds there a new interesting problem about groups, and with probability frac110An interesting puzzle about rings. With probability frac1320new tasks on the forum will not be. Let beX- this is the minimum number of days for which Alena will have at least one new task about groups and at least one about rings. Find the distribution of a random variableX. Only compact expressions (not containing signs of summation, dots, etc.) should participate in the answer.DecisionP[X=k]. ,
X=k. β
kβ1 , ,
k- . β
kβ1 , ,
k- . ,
kβ1 . :
P[x=k]=((1320+14)kβ1β(1320)kβ1)β
110+
((1320+110)kβ1β(1320)kβ1)β
14
Task 4
Dan array textA[1:n] real numbers, sorted in ascending order, as well as numbers p, q, r. Suggest an algorithm building an array textB[1:n]consisting of numbers px2+qx+rwhere x in textAalso sorted in ascending order. Time limit isO(n), for additional memory - O(n).DecisionA=[x1,β¦,xn],
x1β€β¦β€xn.
,
p>0.
.

, :
1.
βx:x>βq2p f(x) .
2.
βx:xβ€βq2p f(x) .
«»
f , .
O(logn) A, . reverse .
f. 2 .
merge O(n) .
p<0 βf, reverse
f(x) -1. .
p=0 q.
1.
q>0βf(xi)β€f(xi+1)βi2.
q<0βf(xi)β₯f(xi+1)βi.
q<0 O(n) reverse.
qβ₯0 .
Task 5
Real-valued function f defined on the segment [a;b] (ba geqslant4)and differentiable on it. Prove that there is a pointx0 in(a;b), for whichfβ²(x0)<1+f2(x0).
Decision.
βxβ(a;b):fβ²(x)β₯1+f2(x).
, :
fβ²(x)=1+f2(x)
dfdx=1+f2ββ«df1+f2=β«dx
arctg(f)=x+Cβf(x)=tg(x+C)
g(x)=tg(x+C). ,
fβ²(x)β₯gβ²(x)βxβ(a;b)βf(x)βf(a)β₯g(x)βg(a)βxβ(a;b). ,
f , , ,
g.
g C. ,
g(a)=f(a). :
f(x)βf(a)β₯g(x)βg(a)βf(x)β₯g(x)
,
bβaβ₯4.
(a;b) xβ² ,
xβ²+C=Ο2+Οk (
Ο<4).
g(xβ²)=+β. ,
f(xβ²)β₯g(xβ²)βf(xβ²)=+β.
, -
(a;b) . , , . .
Task 6
Square real matrix A such that A textT=p(A)where p(x)- polynomial with nonzero free term. Prove thatAreversible. Is it true that for any operator varphi: mathbbRn to mathbbRn there is a polynomial p(x) and some basis in which the matrix varphi satisfies the condition A textT=p(A)?Decision, ,
A=p(AT), :
A=(AT)T=(p(A))T=(pnAn+β¦+p1A+p0E)T
=(pn(An)T+β¦+AT+p0E)=(pn(AT)n+β¦+AT+p0E)=p(AT)
,
A=p(AT)=p(p(A)).
1. .
A .
xβ 0 ,
Ax=0. ,
xTAx=0. :
0=xTAx=xTp(AT)x=xT(pn(AT)n+β¦+p1AT+p0E)x
=pn(xTAT)(AT)nβ1x+β¦+p1xTATx+p0xTEx
,
xTAT=(Ax)T=0:
0=β¦=p0xTx=p0βxβ
,
p0β 0ββxβ=0βx=0. .
2.
Ο A=(0100) .
B=Cβ1AC,
C β .
,
A2=0,
An=0βnβ₯2.
Bn=Cβ1AnC=0βnβ₯2.
BT=p(B). 1 ,
BT=Ξ±B+Ξ²E.
Ξ²=0, , , ,
B , (..
detB=detA=0).
,
B=p(BT)=p(p(B)).
:
B=Ξ±(Ξ±B)=Ξ±2Bβ(1βΞ±2)B=0.
:
1.
Ξ±=β1:
:
BT=p(B)=βBβB+BT=0
2, :
{b11+b11=0b12+b21=0b21+b12=0b22+b22=0β{b11=b22=0b12=βb21
detB=detA=0. :
detB=0β
0βb12(βb12)=b212=0βb12=b21=0βB=0
.
.
Ο , .
2.
Ξ±=1:
BT=p(B)=B.
BTB=B2=0.
(BTB)ii=nβk=1(Bki)2=0βiβBki=0βk,iβB=0
.
3.
Ξ±β Β±1.
,
(1βΞ±)2B=0βB=0.
A Ο ,
AT=p(A). .
Task 7
Dan count with 30peaks. It is known that for any5 vertices in the graph there is a cycle of length 5containing these vertices. Prove that there is10 peaks pairwise connected by edges with each other.Decision,
diamGβ€2.
2
u v 3 . , 5 . , 5 .
u v 2 ( ).
diamGβ€2.
vβG. ,
v 1, 2. , Β«β Β»β
2 :

.
a,b,cβL2.
xβG. , . , ,
v a,
b c 1, .

,
|L2|β€2β|L1|β₯30β1β2=27,
27.
(, β , ) 10. 10
G β , ( , ) 10
Β―G.
Β―G.
G v degvβ₯27βvβG,
degΒ―vβ€2βΒ―vβΒ―G.
«» (1) «» (2). -
degΒ―vβ€2.

(1)
βm2β, (2) β
βm2β.
k β ,
k1 β «». :
|I|=k1βi=1βmi2β+kβi=k1+1βmi2β kβi=1mi=30
, , , . , , 10 3. .
|I|β₯10βi=1β32β=10
,
Β―G 10 ,
G 10 . .
Task 8
Find the limitlimnββ5nβk=nCnβ1kβ1(15)n(45)kβn.
Decision.
:
ΞΎn=# n
β
15.
P[ΞΎn=k] :
P[ΞΎn=k]=Cnβ1kβ1(15)n(45)kβn
Cnβ1kβ1 β
nβ1 ( 1 ).
(15)n β .
(45)kβn β .
limnββP[ΞΎnβ€5n].
,
ΞΎnβ€5n -. ,
5n β₯n .
:
S5n=# 5n
:
S5n=5nβi=1Ξ·i,Ξ·i=I{i }
ES5n=5nβi=1EΞ·i=5nβ
15=n
S5n. :
limnββP[ΞΎnβ€5n]=limnββP[S5nβ₯n]=limnββP[S5nβnβ₯0]=
limnββP[S5nβnΟβnβ₯0]=P[Ξ·β₯0],Ξ·βΌN(0,1)
, :
limnββP[ΞΎnβ€5n]=P[Ξ·β₯0]=12
Conclusion
In general, the exam is quite complicated. My friend complained that preparing is not easy. This is really so - you need to not only know the vast mathematical theory, but also have the skill to solve olympiad problems, in the ShAD they give just such. Therefore, to prepare you need to train a lot, remember the theory and fill your hand.If you have other ideas for solving problems or any comments, feel free to write to me in telegrams @ Azatik1000 . Always happy to answer!Azat Kalmykov, curator at ShAD Helper