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Difference between revisions of "2005 iTest Problems/Problem 41"

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1. False. <math>13</math> is also such a number. It can be represented as <math>(\frac{2}{3})^3+(\frac{7}{3})^3</math>.  
 
1. False. <math>13</math> is also such a number. It can be represented as <math>(\frac{2}{3})^3+(\frac{7}{3})^3</math>.  
  

Revision as of 09:50, 14 October 2025

This was a problem on the chain reaction round, where later answers in a problem depend on earlier answers.

Problems

1A. The iTest, by virtue of being the first national internet-based high school math competition, saves a lot of paper every year. The quantity of trees saved (“$a$”) is determined by the following formula: $a = x^2 + 3x + 9$, where $x$ is the number of participating students in the competition. If $x$ is the correct answer from short answer problem 22, then find $a$. (1 point)

1B. Let $q$ be the sum of the digits of $a$. If $q = b! - (b-1)! + (b-2)! - (b-3)!$, find $b$. (2 points)

1C. Find the number of the following statements that are false: (4 points)

1. $q$ is the first prime number resulting from the sum of cubes of distinct fractions, where both the numerator and denominator are primes. 2. $q$ is composite. 3. $q$ is composite and is the sum of the first four prime numbers and $1$. 4. $q$ is the smallest prime equal to the difference of cubes of two consecutive primes. 5. $q$ is not the smallest prime equal to the product of twin primes plus their arithmetic mean. 6. The sum of $q$ consecutive Fibonacci numbers, starting from the $q^{th}$ Fibonacci number, is prime. 7. $q$ is the largest prime factor of $1bbb$. 8. $q$ is the $8^{th}$ largest prime number. 9. $a$ is composite. 10. $a + q + b = q^2$. 11. The decimal expansion of $q^q$ begins with $q$. 12. $q$ is the smallest prime equal to the sum of three distinct primes. 13. $q^5 + q^2 + q^1 + q^3 + q^5 + q^6 + q^4 + q^0 = 52135640$. 14. $q$ is not the smallest prime such that $q$ and $q^2$ have the same sum of their digits. 15. $q$ is the smallest prime such that $q$ = (the product of its digits + the sum of its digits).

Solution 1

1A. Since the solution to the short answer problem 22 was (spoiler) $x=20$, we can plug it in to get that $\boxed{a=469}$.

1B. The sum of the digits of $469$ is $4+6+9=19$. Since $5$ is clearly going to be too large for $b$, as $5!$ is equal to 120, we check if $4!$ satisfies the equation. Since $4!-3!+2!-1!=19$, our answer for $b$ is $\boxed{4}$.

1C.

1. False. $13$ is also such a number. It can be represented as $(\frac{2}{3})^3+(\frac{7}{3})^3$.

2. False. 19 is prime.

3. False. 19 is prime.

4. True, as 19 is the difference of the cubes of the two smallest primes.

5. True. Twin primes are two primes with difference two, and the smallest such pair is $(3,5)$. Since $3\cdot5+\frac{3+5}{2}=19$, this is true.

6. False. We consider each Fibonacci number's remainder when divided by 3. It repeats: $(0,1,1,2,0,2,2,1,0,1,1,2,0,2,2,1\cdots)$. We can quickly see that the 19th Fibonacci number has a remainder of 2 when divided by 3. Therefore, the sum of the desired 19 consecutive Fibonacci has the same remainder as $2+0+2+2+1+0+1+1+2+0+2+2+1+0+1+1+2+0+2+2=24$ when divided by $3$,and thus has a remainder of $\boxed{0}$ when divided by 3.

7. True. Since $b=4$, we are searching for the largest prime factor of $1444=2^2\cdot19^2$, so 19 is its largest prime factor.

8. True. 19 is the 8th largest prime number.

9. True. 469 is equal to $7\cdot67$.

10: False. We can plug in our $a, q,$ and $b$ values.

11: True. Using the binomial theorem, we can evaluate the first few terms of $19^{19}=(10+9)^{19}=10^19+\frac{19\cdot18}{2}\cdot10^9\cdot9+\binom{19}{3}10^8\cdot9^2+\binom{19}{4}\cdot10^7\cdot9^3...$, which begins $1978...$. As the remaining terms are too small to affect the first two digits, this statement is true.

12. True. For a sum of 3 primes to be odd, the 3 primes must be odd. $3+5+7$ does not work; its sum is $15$. We can now try increasing any of the primes: the smallest combination that works is $3+5+11=19$, so this statement is true.

13: True. Plug in 19 for $q$ to confirm this.

14. False. Plug in all the smaller prime numbers to show that 19 is the smallest possible.

15. True. Plug in all the smaller prime numbers to prove this.

All in all, we see that exactly $\boxed{6}$ statements are false.

See Also

2005 iTest (Problems, Answer Key)
Preceded by:
Problem 40 		Followed by:
Problem 42
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
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