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Difference between revisions of "2000 PMWC Problems"
(→‎Problem T1)
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== Problem T1 ==
 
== Problem T1 ==
 +
A box contains <math>4000</math> to <math>6000</math> candies. When the candies are evenly distributed to <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, or <math>9</math> children, there is always one candy left. If the candies are packaged into small bags each having the same number of candies, what is the largest number of candies below <math>4000</math> in each bag so that no candies are left?
  
 
[[2000 PMWC Problems/Problem T1|Solution]]
 
[[2000 PMWC Problems/Problem T1|Solution]]

Revision as of 20:29, 21 April 2014

Problem I1

$y$ is a number that has $8$ different factors (including the number $1$ and itself). What is the smallest possible value of $y$?

Solution

Problem I2

As far as we know, the greatest prime number is $2^{6972593}-1$. What is the remainder when $2^{6972593}-1$ is divided by $5$?

Solution

Problem I3

How many whole numbers less than $1000$ contain at least one $2$ but no $3$?

Solution

Problem I4

Given that $A^4=75600\times B$. If $A$ and $B$ are positive integers, find the smallest value of $B$.

Solution

Problem I5

In a language college, $72\%$ students can speak Chinese, $65\%$ students can speak English, and $10\%$ students can speak neither Chinese nor English. Find the percentage of students who can speak both Chinese and English.

Solution

Problem I6

$p$ is the product of two $4$-digit numbers formed by the digits $1,2,3,4,5,6,7,8$ without any repetition. Find the largest value of $p$ in the form of $\Box\Box\Box\Box\times\Box\Box\Box\Box$. (You are not required to multiply the numbers).

Solution

Problem I7

$a$ and $b$ are two numbers that have prime factors $3$ and $5$ only. $a$ has $12$ factors ($1$ and itself are included), $b$ has $10$ factors ($1$ and itself are included), and their HCF (Highest Common Factor) is $75$. What is the LCM (Least Common Multiple of $a$ and $b$?

Solution

Problem I8

A circle and a triangle are placed on intersections of the grid. The circle and the triangle are not allowed to lie on the same vertical and horizontal line. How many total possible ways are there of placing the circle and the triangle? (the following is an example).

[asy] for(int i = 0; i < 4; ++i){ if(i < 2){ draw((i,0)--(i,3)); }else{ draw((i,0)--(i,2)); } } for(int i = 0; i < 4; ++i){ if(i < 3) draw((0,i)--(3,i)); else draw((0,i)--(1,i)); } filldraw(Circle((0,3),0.3),white); filldraw((2-0.3*cos(pi/2),1+0.3*sin(pi/2))--(2-0.3*cos(7pi/6),1+0.3*sin(7pi/6))--(2-0.3*cos(11pi/6),1+0.3*sin(11pi/6))--cycle,white); //Credit to chezbgone2 for the diagram[/asy] Solution

Problem I9

Using only odd digits, all possible three-digit numbers are formed. Determine the sum of all such numbers.

Solution

Problem I10

In the sum $\frac{a}{b}+\frac{c}{d}$, each letter represents a distinct digit from $1$ to $9$. The sum is as close as possible to $1$ without being greater than or equal to $1$. What is the sum?

Solution

Problem I11

You have a pack of cards, among which are $20$ red, $20$ yellow, $20$ green and $10$ blue ones. How many cards would you need to draw out to ensure that you have $12$ cards of the same colour?

Solution

Problem I12

During the rest hour, one of five students ($A$, $B$, $C$, $D$, and $E$) dropped a glass of water. The following are the responses of the children when the teacher questioned them:

  • $A$: It was $B$ or $C$ who dropped it.
  • $B$: Neither $E$ nor I did it.
  • $C$: Both $A$ and $B$ are lying.
  • $D$: Only one of $A$ or $B$ is telling the truth.
  • $E$: $D$ is not speaking the truth.

The class teacher knows that three of them NEVER lie while the other two ALWAYS lie. Who dropped the glass?

Solution

Problem I13

In the figure, the squares $ABCD$ and $DCEG$ both have the same area of $64cm^2$. $EFG$ is a semicircle. The point $F$ is the mid-point of the arc $EFG$. Find the area of the shaded part. (Assume $\pi=3.14$)

[asy] void unitsquare(pair p){ draw(p--(p.x+1,p.y)--(p.x+1,p.y+1)--(p.x,p.y+1)--cycle); return; } fill(buildcycle(arc((2,0.5),(2,0),(2,1)),(2,1)--(0,1)--(2.5,0.5)),gray); unitsquare((0,0)); draw((1,0)--(2,0)--(2,1)--(1,1)); draw(arc((2,0.5),(2,0),(2,1))); label("$A$",(0,1),NW); label("$B$",(0,0),SW); label("$C$",(1,0),S); label("$D$",(1,1),N); label("$E$",(2,0),S); label("$F$",(2.5,0.5),E); label("$G$",(2,1),N); //Credit to chezbgone2 for the diagram[/asy]

Solution

Problem I14

A copy machine has a following enlargement/reduction buttons:

\[\begin{tabular}{|l|l|l|l|l|l|l|} \hline 250\% & 200\% & 128\% & 125\% & 100\% & 50\% & 10\% \\ \hline \end{tabular}\]

The buttons $\boxed{50\%}$, $\boxed{100\%}$, and $\boxed{250\%}$ are out of order and cannot be used any more. Sam wants to make a copy that is the same size as the original document by using the remaining buttons. When he presses a button, he has to pay $\textdollar 1$. What is the minimum amount he has to pay?

Solution

Problem I15

The sum of several positive integers is $20$. Find the largest product that can be formed by these integers.

Solution

Problem T1

A box contains $4000$ to $6000$ candies. When the candies are evenly distributed to $5$, $6$, $7$, $8$, or $9$ children, there is always one candy left. If the candies are packaged into small bags each having the same number of candies, what is the largest number of candies below $4000$ in each bag so that no candies are left?

Solution

Problem T2

[asy] import three; currentprojection=orthographic(1,1/2,0.4); draw(surface((5,1,1)--(5,1,2)--(5,2,2)--(5,2,1)--cycle),black,nolight); draw(surface((5,1,3)--(5,1,4)--(5,2,4)--(5,2,3)--cycle),black,nolight); draw(surface((5,3,1)--(5,3,2)--(5,4,2)--(5,4,1)--cycle),black,nolight); draw(surface((5,3,3)--(5,3,4)--(5,4,4)--(5,4,3)--cycle),black,nolight); draw(surface((5,2,2)--(5,2,3)--(5,3,3)--(5,3,2)--cycle),black,nolight); draw(surface((1,5,1)--(1,5,2)--(2,5,2)--(2,5,1)--cycle),black,nolight); draw(surface((1,5,3)--(1,5,4)--(2,5,4)--(2,5,3)--cycle),black,nolight); draw(surface((3,5,1)--(3,5,2)--(4,5,2)--(4,5,1)--cycle),black,nolight); draw(surface((3,5,3)--(3,5,4)--(4,5,4)--(4,5,3)--cycle),black,nolight); draw(surface((2,5,2)--(2,5,3)--(3,5,3)--(3,5,2)--cycle),black,nolight); draw(surface((1,1,5)--(1,2,5)--(2,2,5)--(2,1,5)--cycle),black,nolight); draw(surface((1,3,5)--(1,4,5)--(2,4,5)--(2,3,5)--cycle),black,nolight); draw(surface((3,1,5)--(3,2,5)--(4,2,5)--(4,1,5)--cycle),black,nolight); draw(surface((3,3,5)--(3,4,5)--(4,4,5)--(4,3,5)--cycle),black,nolight); draw(surface((2,2,5)--(2,3,5)--(3,3,5)--(3,2,5)--cycle),black,nolight); for(int i = 0; i <= 5; ++i) draw((i,0,5)--(i,5,5)); for(int i = 0; i <= 5; ++i) draw((0,i,5)--(5,i,5)); for(int i = 0; i <= 5; ++i) draw((5,i,0)--(5,i,5)); for(int i = 0; i <= 5; ++i) draw((5,0,i)--(5,5,i)); for(int i = 0; i <= 5; ++i) draw((i,5,0)--(i,5,5)); for(int i = 0; i <= 5; ++i) draw((0,5,i)--(5,5,i)); //Credit to chezbgone2 for the diagram[/asy]

Solution

Problem T3

[asy] fill(buildcycle((0,2)--(1,0),(2,2)--(0,1),(2,0)--(1,2),(0,0)--(2,1)),gray); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((0,2)--(1,0)); draw((2,2)--(0,1)); draw((2,0)--(1,2)); draw((0,0)--(2,1)); label("A",(0,2),NW); label("B",(2,2),NE); label("C",(2,0),SE); label("D",(0,0),SW); label("P",(1,2),N); label("Q",(2,1),E); label("R",(1,0),S); label("S",(0,1),W); //Credit to chezbgone2 for the diagram[/asy] Solution

Problem T4

[asy] import geometry; draw((0,0)--(3,0)--(3,2)--(0,2)--cycle); draw((0,1)--(1.5,2)); draw((2.3,0)--(2.3,0.7)--(3,0.7)); draw(ellipse((1.5,1),0.6,0.5)); //Credit to chezbgone2 for the diagram[/asy] Solution

Problem T5

Solution

Problem T6

[asy] fill((1,2.5)--(0.9,0.72)--(1.82,1.46)--cycle,black); draw((0,5)--(-3,0)--(4,0)--(0,5)--(0,0)); draw((0,0)--(2.5,2)); draw((0,5)--(1,2.5)--(0.9,0.72)); draw((1,2.5)--(1.82,1.46)); label("$A$",(0,5),N); label("$B$",(-3,0),W); label("$C$",(4,0),E); label("$D$",(0,0),S); label("$E$",(2.5,2),NE); label("$F$",(1,2.5),E); label("$G$",(0.9,0.72),S); label("$H$",(1.82,1.46),S); //Credit to chezbgone2 for the diagram[/asy] Solution

Problem T7

Solution

Problem T8

Solution

Problem T9

Solution

Problem T10

Solution

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