How it Works › Forums › Introductory Problems (AMC 8 Problems 1020) › Unique Properties of Integers Introductory Problems!
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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. Solve the equation $2x+3y+5z=15$ for $x,y,z$ over the positive integers (This means that $x,y,z$ are positive integers). What if you were asked to solve over the reals ($x,y,z$ could be any real numbers)
2. What is the smallest positive integer that has $3$ distinct prime factors? What about $4$ distinct prime factors?
3. Let $k$ be a positive integer and the remainder of $2^k$ when divided by $3$ is $1$. How many such $k$ exist if $k \leq 100$
4. Let $S=1+2+….+24$, find the remainder when $S$ is divided by $25$ and do this in two different ways if possible. (Hint: What type of sequence is this? Also, can you pair the terms in a clever way?)
5. Students guess that Bob’s age is $24, 28, 30, 32, 36, 38, 41, 44, 47$, and $49$. Bob says, “At least half of you guessed too low, two of you are off by one, and my age is a prime number.” How old is Bob?
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!
Tejas1. x=2 y=2, z=1
If it were solved over reals, then I would know that I could use intercepts as ) is a real number.2. 10 k exist for that problem
3.The remainder would be 12 because the numbers can pair into 25 then 12 would be left.
4.Bob is 37 years old
Ridhi1. x=2 y=2 z=1
2. 3 distinct prime factors: 30 4 distinct prime factors: 120
3. 50 k exist
4. remainder=0
5. Bob is 37 years old
Anton1. Since we know that x,y,z are all positive, we can tell that 2x<15, 3y<15,5z<15, or else some other term wouldn’t be a positive integer. So, x<7.5,y<5,z<3. We can try out the values of z, since there are only two options, 1 and 2. If z = 1, 2x+3y+5(1) = 15. 2x+3y =10. Two values that work for this are x=2, y=2. So, a possible solution is (2,2,1). However, there are more than one possible solution, such as (1,1,2).
If we were asked to solve over reals, there would be an infinite number of solutions. We can isolate each variable to find the values.
x = 7/2 – 3y/2 – 5z/2
y = 5 2x/3 – 5z/3
z = 3 2x/5 – 3y/52. 3 distinct factors: 4. 1,2,4
4 distinct factors: 6. 1,2,3,63.We can look at the pattern for remainders when power of 2 are divided by 3.
2^1:2
2^2:1
2^3:2
2^4:1
So, half of the powers of 2 have a remainder of 1 when divided by 3. If k<=100, then and it has a remainder of 1 for all even exponents, then there must be 51, including 0.4. We can use the sum of numbers formula to get S = 25(24)/2 = 300, which is divisible by 25. So, the remainder is 0.
Another way to do this is to group the numbers: (1+24)+(2+23)+(3+22)….(12+13) = 25+25+25…+25. All of the terms are 25, so it must be divisible by 25.
5.Since at least half of them answered too low, we know that the age must be above half, which is 36. Since two of them are off by 1, that means that it has to be directly in between two integers, since no number is repeated. The only two pairs that have numbers directly in between them are 36 and 38, and 47 and 49. So, the possible numbers are 37 and 48. However, 48 is not a prime number, so he must be 37 years old.
Great job Ridhi and Anton, but Anton I would be careful about your solution to problem 2 as there is a slight error. Ridhi you get 25 points and Anton you get 20 points.
Prem Dhoot1. x=1 y=1 z=2
2. 3 distinct=30 4 distinct=120
3.
4.24*25/2 remainder=0
5.Age=37
Laasya1) x=2
y=2
z=12) smallest # with three prime factors: 30
smallest # with four prime factors: 2103) 50k exist
4) the remainder would be 12
5) 37
AyushGreat job Laasya! You get 20 points
AnthonyFor the first problem, I got z=1, x=2, y=1. z=1, x=1, and y=1 would work if two variables are allowed to be the same. If you solve over the reals, then you could use zero and negatives, and their would be a whole lot more correct answers, like x=9, y=1 and z=0.
For the second problem, I got 30 for three distinct prime factors and 210 for four distinct prime factors.
For the third problem, I got 50 k’s as if k is even, it will work.
For the fourth problem, I a remainder of 25. If you cleverly add it (1+24 + 2+23…) you get 25*12. Since there is a 25, the remainder is zero. Another way to do it is only looking at the ones place while adding. You get 1+2+3+4+5 = 5 if you only look at the ones digit. Then just add on and use the divisibility rule for 5.
For the fifth problem, I got 37 years old. 
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