Mathiit mock test 5 practice questions pending
12 ways to choose the first edge.
Each of the 2 vertices of this edge had 2 more edges to it. So they are excluded from future selections.
7 edges and 6 vertices remain.
Out of the 6 vertices, 2 have 3 edges each and 4 have 2 edges each.
Total ways to select 2 edges from 7 = 7C2 = 21.
From this we have to subtract those edge pairs which share a vertex.
Consider a vertex with degree 3.
It gives us 3C2 such pairs. And there are 2 such vertices. Total 6 such edge pairs.
Similarly vertices with degree 2 give us 2C2*4 = 4 such pairs.
21 - 6 - 4 = 11 valid edge pairs.
So total: 12*11 = 132 valid pairs.
But in the first step we could have any of the 3 edges as the first edge.
Later we just counted pairs.
So 132/3 = 44 = answer.
Last 2 digits of 1^100 + 2^100 ... 2026^100.
S14.
Last 2 digits => mod 100.
By CRT, we know that we can break it down in mod 25 and mod 4.
Any even number square mod 100 = 0 mod 4
Any odd number square mod 100 = 1 mod 4
There are 1013 odd numbers => 1013 mod 4 = 1 mod 4.
Mod 25.
Any number which is multiple of 5 to the power 100 => 0 mod 25.
x = r mod 5 => x = 5k + r
(5k + r)^100 upon expansion => all terms will give 0 mod 25 except 100C1.(5k)^1.r^99 and 100C0.r^100.
So only last term would remain.
1^100 = 1 mod 25
2^100 = 1024^10 = 24^10 = (-1)^10 mod 25 = 1
3^100 = 27^33.3 = 2^33.3 = 24^3.24 = 1 mod 25
4^100 = 1024^50 = 24^50 = -1^50 = 1 mod 25
So we get 4*505 + 1 = 2021 mod 25 = 21
How many numbers less than 100 which are 21 mod 25:
21, 46, 71, 96
Which of them is 1 mod 4:
21
Answer = 21
Q16. How many non-equilateral isosceles triangles have all three vertices among the vertices of a regular dodecagon?
S16.
Isosceles triangles are counted by their apex.
12 possible apex.
5 pairs of vertices for each.
For e.g. apex = 1 => 2,12 3,11 4,10 5,9 6,8
Total: 12*5 = 60
4 equilateral: 1,5,9 2,6,10 3,7,11 4,8,12
Each of them counted thrice when each vertex used as apex.
60-12 = 48 = answer.
Q17.
49,1,0 => 3! * 4(-+7,-+1) = 24
25,16,9 => 6*8 = 48
25,25,0 => 3*4 = 12
Answer: 84
Q19. The integers (1,2,...,2026) stand in a circle. Starting with (2), every second remaining integer is removed, and the process continues cyclically until one integer remains. Find the number formed by the last two digits of the survivor.
Let's say what happens with a power of 2 like 4.
1,2,3,4
1,3
1
Survivor: 1
Non power of 2, for e.g. 5
1,2,3,4,5
1,3,5
3,5
3
Survivor increased by 2 when 4 increased by 1 to become 5.
This is known as Josephus problem.
Express the number as 2^m + k.
And answer is 1 + 2.k
2026 = 2^10 + 1002
1 + 2.k = 1 + 1002.2 = 2005
Last 2 digits = 05
Q20. In how many ways can the numbers (1,2,...,9) be placed in a (3x3) array so that entries increase from left to right in every row and from top to bottom in every column?
Young's Hook Formula
9!/5.4.3.4.3.2.3.2.1 = 42
Q21. The six faces of a cube are coloured using all three colours red, blue and green. Two colourings are considered the same if one can be obtained from the other by rotating the cube. How many distinct colourings are there?
S21.
Answer: 30
Detailed solution here.
Q22. Positive integers x <= y <= 100 satisfy x^2+y^2+1=3xy. Find the sum of all possible values of (x).
S22.
If we treat it as a quadratic in x(or y by symmetry):
x^2 - 3xy + y^2 + 1 = 0
If (x1,y1) is one solution then sum of roots x1 + x2 = 3y1 => x2 = 3y1 - x1
So(x1,y1) is a solution => (x1,3y1-x1) is also a solution.
Start with a simple solution assuming x = y => x^2 = 1 => x = y = 1 is the first solution.
(1,1) => (1,3*1-1) = (1,2) is also a solution.
(1,2) => (1,1) is also a solution.
See what happened here.
(x1,y1) => (x1,3x1-y1) => (x1,y1)
So applying the transformation twice yields back the original solution. So we can apply it only once.
But by symmetry(x1,y1) => (y1,x1) is also a solution.
(1,1 ) => (1,2)
(2,1) => (2,5)
(5,2) => 5,13
13,5 => 13,34
34,13 => 34,89
89,34 => x>100 so stop.
So the solutions are
1,1 and 1,2 and 2,5 and 5,13 and 13,34 and 34,89 since all these have x<=y<=100
Sum of x values = 56
Now the question says all possible values so we have to take it as disintct.
So answer = 55 since 1 appears twice.
Q24. An inversion of a permutation (a1,a2...a8) is a pair (i<j) with (a_i>a_j). How many permutations of ({1,2,....,8}) have exactly 10 inversions? Give the number formed by the last two digits.
S24.
Let's start with 1.
It has 0 inversions.
0
1 1
Adding 2 will give 0 or 1 inversions depending on whether it's added on right or left.
0 1
1 1
2 1 1
Add 3
0 1 2 3
1 1
2 1 1
3 1 2 2 1
Add 4
0 1 2 3 4 5 6
1 1
2 1 1
3 1 2 2 1
4 1 3 5 6 5 3 1
So if we already have permutations with inversion counts of 1,2,3 and we want 3 inversions with 1,2,3,4 then what are we going to do?
If we add 4 at the end, we don't get any new inversion.
If we add it just before the last number, we gain 1 inversion.
Similarly we can gain 2 or 3 more inversions by adding it 2 or 3 steps left.
So 0,1,2,3 new inversions can be gained. So the existing inversions from 3 which we can choose are 0,1,2,3.
So when you have to fill 4,3 you go up in that column to the previous row and add 4 elements backwards.
So 1 + 2 + 2 + 1 = 6
Similarly when you have to fill 6,10 you go up and start adding till you reach 5,4. You can't go further left since that will give you permutations of 5 with 3 inversions and adding 6 can't give more than 6 inversions, so you will get only 9.
Doing this we will get:
0 1 2 3 4 5 6
1 1
2 1 1
3 1 2 2 1
4 1 3 5 6 5 3 1
5 1 4 9 15 20 22 20 15 9 4 1
6 1, 5, 14, 29, 49, 71, 90, 101, 101, 90, 71
7 1, 6, 20, 49, 98, 169, 259, 359, 455, 531, 573
Now once we are done till 7, pause.
When we add 8, we can add 0,1,2..7 more inversions.
For exact 10 inversions we need to add up those cases from 7 elements which give 3,4...10 inversions.
49, 98, 169, 259, 359, 455, 531, 573
Adding them we get 2493.
So answer = 93.
Q25.
A function (f) on the nonnegative integers satisfies
f(m+n) + f(|m-n|) = 2f(m) + 2f(n) for all nonnegative integers (m,n), and (f(1)=1.
Find the number formed by the last two digits of f(2026).
S25.
m=1, n=0
f(1)+f(1) = 2f(1) + 2f(0) => f(0) = 0
m = n
f(2m) + f(0) = 4f(m)
=> f(2m) = 4f(m)
f(m+1) + f(m-1) = 2f(m) + 2
f(m+1) - f(m) = 2 + f(m) - f(m-1)
f(2) - f(1) = 2 + f(1) - f(0) = 3
f(3) - f(2) = 2 + f(2) - f(1) = 5
...
f(n) - f(n-1) = 2n - 1
Add all
f(n) - f(1) = 3 + 5 ... 2n-1
f(n) = 1 + 3 + 5 ... 2n+1 = n^2 (sum of first 'n' odd numbers)
f(2026) = 2026^2
mod 100 = 26^2 mod 100 = 625 + 50 + 1 = 76 mod
Q26. Two circles of radii 4 and 9 touch externally at (T). A common external tangent touches the smaller circle at (A) and the larger circle at (B). Find the radius of the circumcircle of triangle (ATB).
S26.
Let the center of smaller and bigger circle be O1, O2.
O1AT and O2BT are both isosceles.
Angle O1AT = O1TA = x
Angle O2BT = O2TB = y
Angle TBA = 90 - y
Angle TAB = 90 - x
=> Angle ATB = 180 - (90-x) - (90-y) = x + y
Also
O1TA + ATB + O2TB = 180 = x + x + y + y = 180 => x + y = 90 = ATB
So ATB is right angle triangle.
AB is the hypotenuse and twice of circumradius of ABT.
Now how to get AB?
It's the common external tangent. You may recall the formula. If not, here is the proof.
Extend O1A to O1A1 s.t. O1A1 = O2B.
Now O1A1BO2 is a ||gram.
And A1AB is a right triangle with right angle at A.
AB^2 + AA1^2 = BA1^2
=> AB^2 = (r1+r2)^2 - (r1-r2)^2 = 13^2 - 5^2 = 12^2
AB = 12, R = 6 = answer.
Q27.
A 4x4 array uses the symbols (1,2,3,4) once in each row and once in each column.
Its first row and first column are both (1,2,3,4) in that order. How many such arrays are possible?
S27.
1 2 3 4
2
3
4
Second row has 6 options, let's try them one by one.
Case 1: 2134 not possible since Column 3 already has 3.
Case 2: 2143 has 2 options as shown below. Once you lock any one number, others are fixed.
1 2 3 4
2 1 4 3
3 4 1,2 1,2
4 3 1,2 1,2
Case 4: 2341 one choice
1 2 3 4
2 3 4 1
3 4 1 2
4 1 2 3
Case 5: 2431 not possible
Case 6: 2413 one choice
1 2 3 4
2 4 1 3
3 1 4 2
4 3 2 1
Q28.
Let k,1k2...k15 run over all primitive 15th roots of unity. Find the number formed by the last two digits of the integer product:
(2-k1)(2-k2)...(2-k15)
S28.
What is 15th root of unity?
All those numbers(real or complex) for which x^15 = 1.
As you recall e^i.2.Pie = 1
So
x1 = e^i.2.Pie.1/15
x2 = e^i.2.Pie.2/15
....
x15 = e^i.2.Pie.15/15
What are the primitive 15th roots of unity?
All those roots which have to be raised to at least power of 15 for them to become 1.
For e.g.
x3 = e^i.2.Pie.3/15
is not primitive since
it can be raised to ^5 to become 1.
So 3rd 5th, 6th, 9th, 10th, 12th and 15th roots have to be ignored and rest all multiplied.
These are non primitive since their gcd with 15 is not 1.
Also
x^15 - 1 = (x-x1)(x-x2)...(x-x15)
=>
2^15 - 1 = (2-x1)(2-x2)...(2-x15)
So what we want is:
(2^15 - 1)/(2-x3)(2-x5)(2-x6)(2-x9)(2-x10)(2-x12)(2-x15)
Now notice that
x3,x6,x9,x12,x15 are 5th roots of unity since raising them to ^5 gets 1.
So (2-x3)(2-x6)...(2-x15) = 2^5-1
Similarly,
(2-x5)(2-x10)(2-x15) = 2^3 - 1
Note that we used 2-x15 twice as a multiplier which is ok since it equals 1.
So final expression becomes:
(2^15 - 1)/(2^5 - 1)(2^3 - 1)
Now 2^15 - 1 = 2^5)^3 -1 = (2^5 - 1)(2^10 + 2^5 + 1)
So it becomes:
1024 + 32 + 1/7 = 1057/7 = 151
Answer: 51
Q29.
Let P be the product of all positive integers x <= 1000 satisfying x^5 = 1 mod 1001. Find the number formed by the last two digits of P.
S29.
x^5 = 1 mod 1001
1001 = 7.11.13
=>
x^5 = 1 mod 7
x^5 = 1 mod 11
x^5 = 1 mod 13
There is a theorem/lemma which says that number of distinct solutions of x mod p for x^n = 1 mod p is gcd(n,p-1) where p is a prime.
So
x^5 = 1 mod 7 has gcd(5,6) = 1 distinct solutions which is x = 1 mod 7
x^5 = 1 mod 13 has gcd(5,12) = 1 distinct solutions which is x = 1 mod 13
x^5 = 1 mod 11 has gcd(5,10) = 5 distinct solutions which are x = 1,3,4,5,9 mod 11
If you don't recall that theorem, it's fine to do it manually as well.
Now, we have that
x = 1 mod 7
x = 1 mod 13
=>
x = 1 mod 91
So x = m.91 + 1
Let's solve one by one.
m.91 + 1 = 1 mod 11
=> m.91 = 0 mod 11
=> m.3 = 0 mod 11 => m = 0 mod 11 => x = 91.0 + 1 = 1
m.91 + 1 = 3 mod 11 =>
3m = 2 mod 11 =>
12m = 8 mod 11 => 1.m = 8 mod 11 => m = 8 mod 11 => x = 8.91 + 1 = 729
m.91 + 1 = 4 mod 11 =>
3m = 3 mod 11 =>
m = 1 mod 11 => x = 1.91 + 1 = 92
m.91 + 1 = 5 mod 11 =>
3m = 4 mod 11 =>12m = 16 mod 11 => m = 5 mod 11
=> x = 5.91 + 1 = 456
m.91 + 1 = 9 mod 11 =>
3m = 8 mod 11 =>12m = 32 mod 11 => m = 10 mod 11
=> x = 10.91 + 1 = 911
Multiply all and take mod 100
911 * 456 * 92 * 729 * 1 mod 100 = 11.56.-8.29 mod 100 = 616.(-)232 = -16.32 = -512 = -12 = 88 mod 100
Answer: 88
Q30.
How many connected simple graphs can be formed on five labelled vertices? Give the number formed by the last two digits.
S30.
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