MathFun4Kids

MATHCOUNTS Exercises – 18

Posted on January 30, 2020 by kevin

How many 4-digit integers between 1000 and 9999 have their digits in non-decreasing order, such as 3447? Click here for solutions.

Solution 1 – Brute Force Denote $f(m, n)$ as the number of $m$-digit integers with their digits in non-decreasing order and the first digit value as $n$, where $0\le n\le 9$. We have the following relation: $$ \begin{array}{rcl} f(1,\ n)&=&1 \\ f(m,9)&=&1 \\ f(m, n)&=&f(m-1, 9) + … + f(m-1, n+1) + f(m-1, n)\\ &=&f(m, n+1) + f(m-1, n) \end{array} $$ Then we have the following table $$ \begin{array}{c|rrrrrrrrrr}
f(m,n)& & & & & n & & & & & \\ \hline
m & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ \hline
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
2 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
3 & 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & 45 & 55 \\
4 & 1 & 4 & 10 & 20 & 35 & 56 & 84 & 120 & 165 & 220 \\
5 & 1 & 5 & 15 & 35 & 70 & 126 & 210 & 330 & 495 & 715 \\
\hline \end{array} $$

The answer to the question is equivalent to $f(5, 1)=495$, i.e. number of 5-digit integers with their digits in non-decreasing order and the first digit as $1$.
Note Do you notice the Pascal’s triangle in the above table?

Solution 2 – Use Stars and Bars Theorem This problem can be translated into inserting 4 identical balls in the boxes after 9 ordered digits as the following, where the number of balls in a box indicates the number of consecutive occurrences of the proceeding digit in forming a 4-digit number: $$\fbox{1}\Box{ }\fbox{2}\Box\fbox{3}\Box{ }\fbox{4}\Box{ }\fbox{5}\Box{ }\fbox{6}\Box{ }\fbox{7}\Box{ }\fbox{8}\Box{ }\fbox{9}\Box{ }$$ For example, 3447 will have 1 ball in the box after 3, 2 balls in the box after 4, and 1 ball in the box after 7 in the above figure. Therefore, using the formula, the answer to the question is $$ {n+k-1\choose n} = {4+9-1 \choose 4} = {12 \choose 4} = \dfrac{12\times 11\times 10\times 9}{4\times 3\times 2\times 1} = 495$$

Posted in Combinatorics, MATHCOUNTS | Comments Off

Crazy Jumping Frog

Posted on January 29, 2020 by kevin

Crazy Jumping Frog can jump forward at a distance either 1 foot or 2 feet. The probability it jumps 1 foot forward is $p$, and the probability it jumps 2 feet is $1-p$. Assume that each jump is independent and the frog has infinite amount of energy 🙂

What the expected distance after the frog making $n$ jumps?
If there is an uncovered well $n$ feet away from the frog, what is the probability that the frog fells into the well if it jumps forward trying to pass the well?

Click here for the solutions.

Posted in Algebra, Probability | Comments Off

MATHCOUNTS Exercises – 17

Posted on January 28, 2020 by kevin

In rectangle $ABCD$, shown here, point $M$ is the midpoint of side $BC$, and point $N$ lies on $CD$ such that $DN:NC=1:4$. Segment $BN$ intersects $AM$ and $AC$ at points $R$ and $S$, respectively. If $NS:SR:RB=x:y:z$, where $x$, $y$ and $z$ are positive integers, what is the minimum possible value of $x + y + z$? Source: MATHCOUNTS 2012 State Sprint Round. Click here for solutions.

Solution 1 – Using Similar Triangles Please visit Art of Problem Solving for using Similar Triangles and Proportional Reasoning to solve the problem.

Solution 2 – Using Mass Points This may be the simplest method found without drawing extra lines.

$$ \because DN:NC=1:4 $$ $$ \therefore CN:AB=CN:CD=NC:(DN+NC)=4:(1+4)=4:5$$ $$ \because AB \parallel CD \ \ \ and \ \ \angle{ASB}=\angle{CSN} $$ $$ \therefore \triangle{ASB} \sim \triangle{CSN}$$ $$ \therefore NS:BS=CS:AS=CN:AB=4:5\tag{1}$$ For $\triangle{ABC}$, let $A_{mass}=4$, $C_{mass}=5$, then, $$S_{mass}=A_{mass}+C_{mass}=4+5=9$$ Because $M$ is the midpoint of $BC$, $B_{mass}=C_{mass}=5$, and $$M_{mass}=B_{mass}+C_{mass}=5+5=10$$ Finally, we have balanced mass at $S$: $$S_{mass}=A_{mass}+M_{mass}=B_{mass}+S_{mass}=14$$ Therefore $$RB:BS=9:14\tag{2}$$ $$SR:BS=5:14\tag{3}$$ Combine (1), (2) and (3), we have: $$NS:SR:RB=(NS:BS):(SR:BS):(RB:BS)=(4:5):(5:14):(9:14)$$ $$= (56:70):(25:70):(45:70)=56:25:45=x:y:z$$ Therefore, $x+y+z=56+25+45=126$

Solution 3 – Using Crossed Ladders Theorem Draw line $RT$ parallel to $AB$ and intersecting $BC$ at $T$, and line $ME$ parallel to $AB$ and intersecting $BN$ at $E$.
$$ \because DN:NC=1:4 $$ $$ \therefore AB:CN=CD:CN=(DN+NC):NC=(1+4):4=5:4 \tag{1}$$ $$ \because AB \parallel CD \ \ \ and \ \ \angle{ASB}=\angle{CSN} $$ $$ \therefore \triangle{ASB} \sim \triangle{CSN}$$ $$ \therefore BS:NS=AB:CN=5:4$$ $$BS:BN=BS:(BS+NS)=5:(4:5)=5:9 \tag{2}$$ Applying Crossed ladders theorem, we have $$ \because RT \parallel AB \parallel EM \parallel CD $$ $$ \therefore \frac{1}{RT}=\frac{1}{AB}+\frac{1}{EM}\tag{3}$$ Additionally $$ \because ME \parallel CN $$ $$ \therefore \triangle{BME}\sim\triangle{BCD} $$ Because point $M$ is the midpoint of side $BC$, therefore $$EM:NC=BM:BC=1:2$$ Apply (1), we have $$EM=\frac{1}{2}NC=\frac{1}{2}\cdot\frac{4}{5}AB=\frac{2}{5}AB\tag{4} $$ Apply (4) to (3), we have $$ \frac{1}{RT}=\frac{1}{AB}+\frac{1}{\frac{2}{5}AB}=\frac{7}{2}\cdot\frac{1}{AB}$$ $$\therefore RT:AB=2:7$$ Apply (1), we have $$RT:NC=(RT:AB) \cdot (AB:CN) = (2:7)\cdot(5:4)=5:14$$ $$ \because \triangle{BTR} \sim \triangle{BCN} $$ $$\therefore RB:NB=RT:NC=5:14\tag{5}$$ $$\therefore SR:NB=(BS-RB):NB = BS:NB-RB:NB=\frac{5}{9}-\frac{5}{14}=\frac{25}{126}\tag{6}$$ Additionally, $$NS:NB=NS:(NS+SR)=4:(4+5)=4:9\tag{7}$$ Combine (5), (6) and (7), we have: $$NS:SR:RB=(4:9):(25:126):(5:14)=56:25:45=x:y:z$$ Therefore, we reach the same answer: $x+y+z=56+25+45=126$

Posted in Geometry, MATHCOUNTS | Comments Off

MATHCOUNTS Exercises – 16

Posted on January 26, 2020 by kevin

A semicircle and a circle are placed inside a square with sides of length 4 cm as shown. The circle is tangent to two adjacent sides of the square and to the semicircle. The diameter of the semicircle is a side of the square. In centimeters, what is the radius of the circle?Source: MATHCOUNTS 2012 State Target Round. Click here for the solution.

Solution First, connect the center of the circle with the center of the semicircle. Then drop a perpendicular from the center of the circle to side AB, as shown.

Let $r$ be the radius of the circle. So, segment $IJ=2+r$, since the side of square $ABCD$ is 4. Also, segment $IK=4-r$ and $KJ=2-r$. Since $IK \perp KJ$, we can set up a Pythagorean:
$$(4-r)^2+(2-r)^2=(2+r)^2$$
This simplifies to $$16-8r+r^2+4-4r+r^2=4+4r+r^2$$ Simplifying further, we get a quadratic equation:
$$r^2-16r+16=0$$
Applying the quadratic formula, we get $r=8-4\sqrt{3}$.

Posted in Circles in a Square, Geometry, MATHCOUNTS | Comments Off

MATHCOUNTS Exercises – 15

Posted on January 22, 2020 by kevin

How many distinct unit cubes are there with two faces painted red, two faces painted green and two faces painted blue? Two unit cubes are considered distinct if one unit cube cannot be obtained by rotating the other. SOURCE: 2014 MATHCOUNTS Chapter Target Round. Click here for the solution.

Solution First, there is only one way to paint the cube with all opposite faces painted in the same color, as shown below, with the front face colored in red:

Second, there are 3 ways to paint the cube with one pair of opposite faces in the same color, while the other two pairs of opposite faces in different colors, as shown below, with the front face in the same color of the back face:

Finally, there are 2 ways to paint the cube without any pair of opposite faces in the same color, as shown below, with the front face colored in red:

Therefore, the final answer to the question is 6.

Posted in Combinatorics, Geometry, MATHCOUNTS | Comments Off

MATHCOUNTS Exercises – 18

Crazy Jumping Frog

MATHCOUNTS Exercises – 17

MATHCOUNTS Exercises – 16

MATHCOUNTS Exercises – 15

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