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1. The faces of a standard die are numbered 1, 2, 3, 4, 5, and 6 such that the sum of the numbers on any two opposite faces is 7. Tia writes one number on each vertex of the die such that the number on each face of the die is the greatest common divisor of the numbers at the four vertices of that face. What is the smallest possible sum of the eight numbers Tia writes?

 

2. Richard cuts a  \(4\frac{1}{2} ~\text{ft} \times 7\frac{1}{2} ~\text{ft} \times 11\frac{1}{4}~ \text{ft}\) rectangular prism into congruent cubes without any part of the prism leftover. If he makes the cubes as large as possible, how many will he produce?

 

3.

N is a positive integer. When N is divided by 3, the remainder is 2. When N is divided by 4, the remainder is 3. What is the remainder when N is divided by 12?

4.

The zoo transferred 131 birds to the safari park over the course of many days. The zookeeper planned to send the same number of birds each day, except that there might be fewer on the last day, when he would send whatever was left. However, he mistakenly thought that there were 113 birds in total instead of 131, which caused the transfer to take 3 more days than he anticipated. Interestingly, the number of birds transferred during the last day was the same as he had originally planned. How many birds were transferred during the last day?

5.

N is a four-digit positive integer. Dividing N by 9, the remainder is 5. Dividing N by 7, the remainder is 3. Dividing N by 5, the remainder is 1. What is the smallest possible value of N?

 

6.

A box contains many candies. If 3 candies are taken away, the rest can be shared evenly among 7 people. If 6 are taken away, the rest can be shared evenly among 8 people. If 3 candies are added, then the box can be shared evenly among 5 people. What is the smallest number of candies in the box?

 Feb 23, 2019
edited by sudsw12  Feb 23, 2019
edited by sudsw12  Feb 23, 2019
 #1
avatar+5797 
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\(\#3)\\ N = 3m + 2 = 4n+3\\ m = \dfrac{4n+1}{3} \in \mathbb{Z}\\ 4n+1 \pmod{3} = 0\\ n \pmod{3} = -1\\ n\pmod{3} = 2\\ n = 2+3k\\ N = 4n+3 = 4(2+3k)+3 = 12k+11\\ N \pmod{12} = 11\)

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 Feb 23, 2019
 #2
avatar+5797 
0

#5)

 

\(N\pmod{9}=5\\ N\pmod{7}=3\\ N\pmod{5}=1\)

 

\(9m+5=7n+3\\ n=\dfrac{9m+2}{7} \in \mathbb{Z}\\ 9m+2 \pmod{7} = 0\\ 2m \pmod{7}=5\\ m \pmod{7} = 20 \pmod{7} = 6\\ m = 7k+6, ~k \in \mathbb{Z}\)

 

\(N=9(7k+6)+5 = 63k + 59\\ 63k+59 = 5j + 1\\ j = \dfrac{63k+58}{5}\in \mathbb{Z}\\ 63k+58 \pmod{5} = 0\\ 3k+3 \pmod{5} = 0\\ 3k \pmod{5} = 2\\ k \pmod{5} = 4\\ k = 5i + 4\)

 

\(N = 63(5i+4)+59\\ N = 315i + 311\\ i = 3 \text{ gets us the smallest 4 digit number}\\ N = 945 + 311 = 1256\)

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 Feb 23, 2019
 #3
avatar+5797 
0

#6) more diophantine equations

 

\(\text{Let the number of candies in the box = }N\\ N-3 \pmod{7}=0\\ N-6 \pmod{8} = 0\\ N+3 \pmod{5} = 0\)

 

\(N \pmod{7}=3,~N\pmod{8}=6\\ 7m+3 = 8n+6\\ m = \dfrac{8n+3}{7}\in \mathbb{Z}\\ 8n+3 \pmod{7}=0\\ n \pmod{7}=4\)

 

\(N=8(7k+4)+6 = 56k+38\\ 56k+38+3\pmod{5}=0\\ k + 1\pmod{5} = 0\\ k \pmod{5} = 4\\ N=56(5j+4)+38 = 280j+262\)

 

\(\text{we can't have negative candies so the smallest }N\\ \text{occurs when }j=0 \text{ i.e. when }N=262\)

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 Feb 23, 2019
 #4
avatar+5797 
0

2)

 

\(\text{we seek }d \text{ such that}\\ i d = \dfrac{18}{4},~j d = \dfrac{30}{4},~k d = \dfrac{11}{4}~ft,~i,j,k \in \mathbb{Z}\\ LCM(18,30,11) = 990\\ \text{so split up the longest side into }990 \text{ cubes}\\ d = \dfrac{30}{4} \times \dfrac{1}{990} = \dfrac{1}{132}~ft\\ \text{The prism will be split into }\\ \dfrac{\frac{18}{4}}{\frac{1}{132}}\times \dfrac{\frac{30}{4}}{\frac{1}{132}}\times \dfrac{\frac{11}{4}}{\frac{1}{132}}=\\ 595 \times 990 \times 363 = 213465780 \text{ cubes}\)

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 Feb 23, 2019
 #5
avatar+5797 
0

#4

 

\(\text{assume we transfer }n \text{ birds at a time and it was assumed that }\\ \text{transferring }113 \text{ birds would take }t \text{ days. }\\ r \text{ is the common odd number of birds for the last trip}\\ 113 = n(t-1) + r\\ 131 = n(t+2)+r\)

 

\(\text{subtract 1 from 2}\\ 18=3n\\ n=6\)

 

\(131 = 6(t+2)+r\\ 119 = 6t + r\\ \text{note that we are told }r < n\\ \left \lfloor \dfrac{119}{6} \right \rfloor= 19\\ 119=19 \cdot 6 + r\\ 119=114+r\\ r=5 \)

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 Feb 23, 2019

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