How it Works › Forums › Introductory Problems (AMC 8 Problems 1020) › Unique Properties of Integers Introductory Problems Set 2
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Hey Math Circle Students!Here are the introductory problems for the next $3$ days! These are more difficult than AMC 8 Problems 1015 but they fit the range of $1621$. Each of these problems are worth $5$ points to your leaderboard score! If you don’t understand one of the questions, feel free to post in the questions forum here
1. How many positive integers $x$ less than $56$ leave a remainder of $1$ when divided by $4$ and a remainder of $2$ when divided by $7$? Try to make mod equations if possible.
2. In continuation to problem 1, how many $x$ less than $56$ leave a remainder of $0$ when divided by $4$ and a remainder of $2$ when divided by $8$?
3. Find the remainder when $S=12+34+…+9596$ is divided by $10$.
4. How many primes less than $50$ can be written as the sum of two composite numbers. How many primes can be written as the sum of two other prime numbers!
Remember that you can post your solutions below to these problems and I will keep a running leaderboards for the top 10 students who participate the most! I especially appreciate the same problem being solved in different ways!
AntonGuest1. First, we can try to find a value that satisfies this condition. We can add 4 to 1 a few times and see if it leaves a remainder of 2 when divided by 7.
1,5 don’t work, but 9 works. After seeing that 9 works, we know that we can just add 28, and the resulting value will have the same remainders since 28 mod 4 =0, 28 mod 7 = 0. So, we get 9,37,65. It asks for values below 56, so there are 2.2.Since we are told that the number is 0 mod 4, that means that it must be a multiple of 4. However, when any multiple of 4 is divided by 8, the only possible remainders are 0 or 4, so there are none that satisfy this condition.
3.We can group all of these into pairs, and see that there are 96/2 = 48 pairs. Each pair equals 1, so the total is 48 We divide this by 10 to get a remainder of 8 = 2mod 10, so the remainder is 2.
BhavyaGuest1. 2 numbers (9 and 37)
2. None, if it is divisible by 4 then when divided by 8 there has to be a remainder of 4 or 0.
3. a remainder of 2
4. No prime can be written as the sum of two composite numbers because one is not prime or composite and every composite number is even. Prime numbers are odd and two evens do not make an odd.
Only five number under 50 can be written as the sum of two primes (5, 13, 19, 31, 43). 2 has to be added to that number because two odds make an even.. non prime. If two has to be added, the prime number before the answer has to be 2 lower than it. There are only five such numbers (shown above).AyushGuestGreat work Anton and Bhavya! Anton you receive 15 points and Bhavya you receive 20 points. However, Anton try to find a more efficient solution to problem 1. Also, try problem 4 as well.
Bhavya, your solution to number 4 is incorrect in the composite case. Note that 15 is composite yet it is still odd. Thus, not all composite numbers are even.
ArchitGuestQuestion#1: ( 9, 37), The requirements for the value of x is that the number has to leave a remainder of 1 when divided by four, and a remainder of 2 when divided by 7. By the statements I knew that the value had to be greater than 7, so the first answer I got was 9. I decided to find the other one by creating an equation: 9*4+1= 37. If you do the same thing for the divisor 7, you get 65, but the number has to be less than 56, so there are only two answers.
Question#2: (No solution), 4 and 8 come in the same table, so any number can’t equal to the remainders of both.
Question#3: (2), There are 96 numbers, and 48 pairs. Each pair equals 1, so S= 48/10. 48 mod 10= 2.
Question#4: (5, 13, 19, 31, 43) Actually it is true that two composite numbers do equal prime numbers, here are some examples, (25+6=31), (20+21=41), (25+12=37), (15+4=19), (39+8=47), (39+4=43), and many more. The solutions shown above subtracted by 2, and then added by that number equals to the numbers.
nimalanGuest#1 22 integers because for ex. 4+1=5divided by 4 = 1 remainder 1, for 7 ex. 7+2=9divided by 7 = 1 remainder 2 so you have to add 1 to four’s multiples and 2 to seven’s multiples to get the positive integers.
nimalanGuestthere is negativity to problem # 3, because 12= negative 1+3=24= negative 2 and so on, so by the time you get to 96 it will be negative 48 because 48 is half of 96 so negative 48 divided by 10 = negative 4.8 = 4 and remainder 8 is the answer.
nimalanGuestthe answer to number 2 is 19 integers because 4 times 14 = 56, and 8+2= 10 times 5=50 because 56 is less than 60
nimalanGuestthe answer to problem four is 25 numbers for the first Question, and 25 numbers for the second part. Because half of fifty is composite and the other half are prime.
AyushGuestGreat work Archit and Nimalan! Archit your solutions are very clear and accurate, you have been awarded 25 points!
Nimalan, I just had a few comments:
Problem 1. You are on the right track, but the numbers have to both leave a remainder of 1 when divided by 4 and 2 when divided by 7 at the same time, not individually.
Problem 2: Try to actually find any of the numbers, you will start to see the answer.
Problem 3: Great work, you know that 48 is the equivalent form of S in mod 10, but think about what 48 is in mod 10, when you divide by 10 your remainder is 8 not 8.
Problem 4: I think you were thinking about even/odd not prime/composite
AnthonyGuest1) 2
I made two lists, each with one requirement, then I looked for common terms.
2) 0
If a mod 4 = 0, then a mod 8 will equal 0 or 4, not 2.
3) 2
You can group it into 48 groups of 1. 48 mod 10 = 2.
4) 11 for two composites, 7 for two primes.
If it is the sum of two composites, then it will have to be an even plus an odd (primes cannot be even). If it is the sum of two primes, same thing, and the only even prime is 2. That means that only groups of primes like {2,5,7} or {29,31} should be counted. You can make a list using these two tips.nimalanGuestthe answer to problem number 1 is 6 integers nimalan
adminKeymasterNimalan good try, but I think you may have made a silly mistake.

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