AMC 10 mock test problems pending

Q1. Let N=1×2×3×⋯⋯×n, where n is a positive integer. Find the largest possible value of n for which N is not divisible by some two-digit number.

S1:
First thought - for n>=5 10 will divide each factorial so what is the question even asking?
But reading it carefully, we note that it's asking something else.
Find the largest n! which is not divisible by at least one 2-digit number.

4! is not divisible by any(hence at least one) 2 digit number but it's not the largest.
Largest such number could possibly be 99!. But it is divisible by all 2-digit numbers including 99.
98! is divisible by all except 99 but 11*9 = 99 so not correct.
97! is also divisible by 99 and 98(49*2).
96! is not divisible by 97 which is a prime. Hence 96!.

 
Q2. Real numbers a and b satisfy a+b=7 and ab=−3. What is the value of  a^3/b^2 + a +b^3/a^2 + b?
S2:
a^3/b^2 + a +b^3/a^2 + b
= (a + b) + (a^5 + b^5)/(a^2b^2)
We know the values of ab, a+b so we just need to focus on a^5 + b^5 which we can progressively generate. How?

a^2 + b^2 = (a+b)^2 - 2ab = 55
a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 406
(a^5 + b^5) = (a^3 + b^3)(a^2 + b^2) - a^2.b^2(a+b)
Eventually we will get:
22330/9

Also note that
a^n + b^n = (a^(n-1) + b^(n-1)).(a + b) - ab(a^(n-2) + b^(n-2))
We can use this to simplify it as well.

Q3. While riding a bicycle a girl reads in the speedometer 14941 which is a palindrome i.e when she reads the number from backward it is same when she reads the number from forward. After 5 hours the digital speed ometer shows the next higher palindrome. What will be the average speed in kilometers per hour during this 5 hours ? 
S3.
Next palindrome = 15051 => speed  = 110/5 = 22km/h = answer

Q4. What is the median of the following list of 2020 numbers:
1.2....1010,1^2,2^2...1010^2
S4.
961 = 31^2 < 1010 < 32^2
So there are [1...961] + 31 numbers are which are less than 1010.
So 992 numbers are less than 1010.
We need 1010th and 1011th numbers.
961 + 18 = 979 is the 1010th term and 980 is the 1011th term.
Median = 979.5

Q5. Narayan, Rishita, and Praveen were roaming around the town and were wondering how far away the nearest town was. When Narayan said, "We are at least 11 kilometers away," Rishita replied, "We are at most 5 kilometers away." Praveen then remarked, "Actually the nearest town is at most 6 Kilometers away." It turned out that none of the three statements were true. Let d be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of  d ?

S5.
Basically invert each answer and find intersection.
First statement: d>=11 invert => d < 11
Second: d <= 5 => d > 5
Third: d <= 6 => d > 6
Intersection of all => d belongs to (6,11) Answer.

Q6.

In rectangle ABCD, AB=10 and BC=8. Points E and F lie on AB, points G and H lie on BC, points I and J lie on CD, and points K and L lie on DA so that AE=FB<5 and the convex octagon EFGHIJKL has equal side lengths. The length of a side of this octagon can be expressed in the form i + sqrt(j)​, where i,j are integers and j is not divisible by the square of any prime. What is j−i?

S6.
Common mistake to avoid: don't assume it's a regular octagon.
Question only says all sides are equal and doesn't say anything about all angles being equal.

Claim: All vertical corner cut offs are equal and so are all horizontal ones.
BG = CH = KD = LA
AE = FB = CI = DJ
Proof:
Let the octagon side be x = EF
AE = FB = (10-x)/2
By Pythagoras: LA^2 = x^2 - AE^2 = GB^2
=> LA = BG => KD = CH => DJ = IC  => DJ = IC = AE = BF
IC = FB => CH = BG => BG = CH = KD = LA

Now: FB = (10-x)/2 , BG = (8-x)/2
=> x^2 = 1/4[(10-x)^2 + (8-x)^2]
=> 4x^2 = 100 + x^2 - 20x + 64 + x^2 - 16x
=> 2x^2 = 164 - 36x
=> x^2 + 18x - 82 = 0
=> x = [-18 +- sqrt(324 + 328)]/2 = -9 +- sqrt(81 + 82) = -9 +-sqrt(163)
=> i = -9 , j = 163
=> j-i = 172 = answer

Q8.

Seven circles of radius 2 inch are cut from a big circular metal plate, as shown. Neighboring circles are tangent, and all except the center circle are tangent to the edge of the metal plate. The leftover scrap is reshaped to form another circle of the same thickness. What is the radius in inches of the scrap circle?

S8.
It looks like R = 3r where R is the outer circle radius and 'r' for inner circle.
So Pi*(R^2 - 7.r^2) = Pi*(rx^2)
=> rx^2 = 2.r^2 = 8 => rx = 2.sqrt(2)

But why is R = 3r?
Let's form a regular hexagon with side = 2r by joining the centers of 6 smaller circles tangent to the outer circle.
So the center of this hexagon is equidistant from all these 6 centers. And it is same as that of the 7th inner circle since that too is 2r away from each of the 6 centers.
So C_h = C_7

Now look at the center of the big circle : C_B.
Draw a radius from it going through one of the 6 circles' center and meeting the outer circle.
C_B's distance from that center is R - r and it's same for each of the 6 circles.
So C_B is equidistant from those 6 points.
But that point is already C_h = C_7.
So R - r = 2r => R = 3r H.P.

Q10.

S10.
Using angle bisector theorem on ABC
ABG/ACG = 100/20
ABG + ACG = 240
=>
ABG = 200
ACG = 40

Again apply on ADE
AFD/AFE = 100/20
AFD + AFE = AED = 240/4  = 60(since DE || BC and DE = BC/2)
=> AFD = 50
AFE = 10
ECGF = ACG - AEF = 40 - 10 = 30
FDBG = 240 - AED - ECGF = 240 - 60 - 30 = 150 = Answer

Q11.

S11.

Q12.

S12.
Step 1: Prove that the intersection of EG and FH is the midpoint of diagonals.
Proof: midpoints of both EG and FH are 6,3.5 and since both pass through it it's their intersection point. Also this is the diagonal midpoint. Let it be T.
[TDG] = [TEG] by congruence = 1/2 * 7/2 * 3
TDH = TBF = 1/2 * 6 * 3
Total = 18 + 21/2 = 28.5

Q13. 4 letters. Wrong letter to each house.
S13.
Using derangements D4 = 9.
Formula:
Dn = (n-1)(D_{n-1} + D_{n-2})
D0 = 1, D1 = 0
Keep doing it to get D4 = 9.

Q13. Three black balls, two blue balls and one red ball are placed in line in random order. What is the probability that no two neighboring balls are the same color?
S13.
Assuming that balls of same color are identical, let's first place the black balls.
There are 4 possible ways:
B_B_B
_B_B_B
B_B_ _B
B_ _ B_B

Remaining balls can have 3 possible orderings:
Red-Blue-Blue
Blue-Red-Blue
Blue-Blue-Red

Try them in the empty spaces above you will get 10 valid ways.
Total ways to arrange them = 6!/1!2!3! = 60
Answer = 1/6

Important point to note here:
Even if we treat the balls of same color as distinct, we will get the same answer.
Why?
Since we are calculating probability here, numerator and denominator get multiplied by the same multipliers which get cancelled out.
How?
Each arrangement of black balls will have 3! variations. Same for Blue(2!) and Red(1!).

So total ways will become 60*12 = 720 which is nothing but 6!.
Similarly the selected ways will also get multiplied by 12 and hence it gets canceled out.

Q17.

Answer: 2.sqrt(6)
Why?
Area of this quadrilateral is twice [ABD].
= 2*1/2.height.base = 1/2*cube diagonal/2*face diagonal/2
= sqrt(3*4)/2*sqrt(2*4)/2
= 2.sqrt(6)

Q18.
Two friends agree that books are worth 400 dollars and that pens are worth 250 dollars. When one friend  owes the other money, he pays the debt in books or pens , with "change" received in the form of books or pens as necessary. (For example, a 550 dollar debt could be paid with two books , with one pen  received in change.) What is the amount of the smallest positive debt that can be resolved in this way?

Solution:

The problem can be expressed as in equation form 400x+250y=z

where x is the number of the books and y is the number of pens and z is the positive debt.

The problem asks us to find the lowest z possible. x and y must be integers, which makes the equation a Diophantine equation. Bezout’s Identity tells us that the smallest c for the equation  am+bn=c to have solutions is when c is the greatest common divisor of a and b.

Therefore, the answer is gcd(400,250), which is 50.

Q19.
For how many integers x, is x^4 - 41x^2 + 40 negative?
S19.
roots are -+sqrt(40),-+1

sqrt(40) = 6.something
Using wavy curve you get [-6,-2] and [2,6] as integer intervals for x when the expression is negative.

Q20.

x∈R is randomly chosen from the interval [0,1]. What is the probability(approx) that x is closer to 0.58 than  0.22?
S20.
Let's break the interval like this:
0___.22__.58__1

First interval is closer to .22 and third is closer to .58.
In the middle interval, the midpoint is .4. Between .4 to .58 again it's closer to .58.
So .4 to 1 it's closer to .58
.6/1 = answer.

 Q21. △PQR has all three side lengths as 4. If the area of the region containing all points that are outside the triangle but within 2 units from some point of the triangle is x+yπ, find x/y.

S21.

So total area = 3 rectangles + 3 arcs of 120 degree and radius 2.
= 3*8 + 3*pi*4/3 = 24 + 4pi => x/y = 6

Q22.
Let x = 2012^2 + 2^2012
What is the unit digit of x^2 + 2^x?
S22.
We need to compute mod 10.
x mod 10 = 2012^2 mod 10+ 2^2012 mod 10
2012^2 mod 10 = 2012 mod 10 * 2012 mod 10
2012 mod 10 = 2
2012^2 mod 10 = 4
2^2012 mod 10 =6 since powers of 2's unit digit cycles in 4.
x mod 10 = 4 + 6 = 0
x^2 mod 10 = 0*0 = 0

Now
2^x mod 10
x = 2012^2 + 2^2012 which is divisible by 4 so unit digit of 2^x will be 6 = answer.

Q23.

Given that 

m{n+1} = 1 + m1.m2....mn and m5 =37

Find the value of m6.

​S23.
m1.m2.m3.m4 = 36
m6 = 1 + 36.37 = 1333 = answer.

Q24.

A solid cubical block of wood has side length 3 unit. An 1× 1 square hole is scooped out of the center of each face of the block. The edges of each cut are parallel to the edges of the block, and each hole goes all the way through the block. What is the volume of the remaining solid?

S24.
Total volume = 3^3
Each hole volume = 1*1*3
Remaining = 27 - 3*3 = 18
But each hole has a center cube overlapping of 1x1x1 which is removed thrice in counting.
So 18 + 2 = 20 = answer.

Q25.

Let n be the smallest possible natural number that is divisible by 4 and 9, and the decimal representation have only 4's and 9's, with at least one of each. What are the last four digits of n?

S25.
For a number to be div by 4 last 2 digits should be div by 4.
So last 2 digits are 44.
For a number to be div by 9, sum of digits should be div by 9.
So in total 9 fours.
But we need at least 1 9 so let's place it just before the last 2 digits to make it smallest.
So n = 444-4444-9-44
Answer = 4944