2022 AMC 10B Problems/Problem 17
- The following problem is from both the 2022 AMC 10B #17 and 2022 AMC 12B #15, so both problems redirect to this page.
Contents
Problem
One of the following numbers is not divisible by any prime number less than Which is it?
Solution 1 (Modular Arithmetic)
For modulo Thus, is divisible by
For modulo Thus, is divisible by
For modulo Thus, is divisible by
For modulo Thus, is divisible by
Therefore, the answer is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MrThinker (LaTeX Error)
Solution 2 (Factoring)
We have We conclude that is divisible by , is divisible by , is divisible by , and is divisible by .
Since all of the other choices have been eliminated, we are left with .
~not_slay
Solution 3 (Elimination)
Mersenne Primes are primes of the form , where is prime. Using the process of elimination, we can eliminate every option except for and . Clearly, isn't prime, so the answer must be .
Note: A organization of computer scientist and mathematicians named Great Internet Mersenne Prime Search (GIMPS) search for the worlds biggest prime numbers. More information can be found on https://www.mersenne.org/primes/
Solution 3a (Elimination)
We examine option E first. has a units digit of (Taking the units digit of the first few powers of two gives a pattern of ) and has a units digit of (Taking the units digit of the first few powers of three gives a pattern of ). Adding and together, we get , which is a multiple of , meaning that is divisible by 5.
Next, we examine option D. We take the first few powers of added with :
We see that the odd powers of added with 1 are multiples of three. If we continue this pattern, will be divisible by . (The reason why this pattern works: When you multiply by , you obtain . Multiplying by again, we get . We see that in every cycle of two powers of , it goes from to and back to .)
Next, we examine option B. We see that has a units of digits of (Taking the units digit of the first few powers of two gives a pattern of ). Adding to , we get . Since has a units digit of , it is divisible by .
Lastly, we examine option A. Using the difference of cubes factorization , we have . Since (Every term in the sequence is equivalent to ), is divisible by .
Since we have eliminated every option except C, is not divisible by any prime less than .
~arjken (+ minor LaTeX edits ~TaeKim)
Solution 3b (Elimination + Number Theory)
We know that the prime numbers less than 10 are and . We can start by testing if any of the answer choices are divisible by . We see that they are all sums of one even number and one odd number, which is simply odd. So, we cannot exclude any answer choices so far. Now, let's check divisibility by . We can use the fact that to our advantage:
So, we eliminate choices A and D from divisibility by 3. Now, we move onto divisibility by 5. We can use cycling of powers to find useful remainders. Let's start with choice .
We see that the remainders of powers of when divided by cycle in a pattern: . Since the pattern cycles every 4 terms, we use modulo 4 to simplify 606, getting that . So, we get that is congruent modulo 5 to the second term of our pattern, so . Thus, . We now eliminate choice B and compare choices C and E.
Looking at choice E, we see that we have to do similar cycling for powers of . We get a pattern of . Since this pattern also cycles every 4 terms, we use modulo 4 to simplify 607, getting that . Both the power of 3 and the power of 2 have an exponent of 607, so we use the third term (since we just found that ) in each corresponding 4 term pattern to get that . We eliminate choice E, and we are left with the correct answer: choice
~TaeKim
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Digit Cycles
~ pi_is_3.14
Video Solution(1-16)
~~Hayabusa1
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.