Mathiit mock test 5 practice questions pending
Q13.
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.
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.
How many sets of three edges of a cube can be chosen so that no two chosen edges share a vertex?
S13.
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.
Q14. 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.
Now let's modify the original problem.
What if we need to count any pair of edges from a cube which don't share a vertex?
Each vertex has degree 3.
So 8*3C2 = 24edge pairs share a vertex.
Total edge pairs = 12C2 = 66
66 - 24 = 42 = 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.
How many ordered triples of integers ((x,y,z)) satisfy
x^2+y^2+z^2=50?
S18.
49,1,0 => 3! * 4(-+7,-+1) = 24
25,16,9 => 6*8 = 48
25,25,0 => 3*4 = 12
Answer: 84
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.
S20.
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?
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?
S20.
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
The pattern is this:
For e.g. with 4 elements(row) and 4 inversions(column), do this:
Go up to previous row in same column, add 4 elements to the left including the current one.
So 1 + 2 + 2 + 1 = 6
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.
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
The pattern is this:
For e.g. with 4 elements(row) and 4 inversions(column), do this:
Go up to previous row in same column, add 4 elements to the left including the current one.
So 1 + 2 + 2 + 1 = 6
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.
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