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Difference between revisions of "1973 AHSME Problems/Problem 3"

(Solution to Problem 3)
 
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<math> \textbf{(A)}\ 112\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 92\qquad\textbf{(D)}\ 88\qquad\textbf{(E)}\ 80 </math>
 
<math> \textbf{(A)}\ 112\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 92\qquad\textbf{(D)}\ 88\qquad\textbf{(E)}\ 80 </math>
  
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==Solutions==
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==Solution (Trial and Error)==
  
 
We can guess and check small primes, subtract it from <math>126</math>, and see if the result is a prime because the further away the two numbers are, the greater the difference will be.  Since <math>126 = 2 \cdot 3^2 \cdot 7</math>, we can eliminate <math>2</math>, <math>3</math>, and <math>7</math> as an option because subtracting these would result in a composite number.
 
We can guess and check small primes, subtract it from <math>126</math>, and see if the result is a prime because the further away the two numbers are, the greater the difference will be.  Since <math>126 = 2 \cdot 3^2 \cdot 7</math>, we can eliminate <math>2</math>, <math>3</math>, and <math>7</math> as an option because subtracting these would result in a composite number.
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==See Also==
 
==See Also==
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{{AHSME 35p box|year=1973|num-b=2|num-a=4}}
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{{AHSME 30p box|year=1973|num-b=2|num-a=4}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Latest revision as of 13:30, 1 January 2024

Problem

The stronger Goldbach conjecture states that any even integer greater than 7 can be written as the sum of two different prime numbers. For such representations of the even number 126, the largest possible difference between the two primes is

$\textbf{(A)}\ 112\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 92\qquad\textbf{(D)}\ 88\qquad\textbf{(E)}\ 80$

Solution (Trial and Error)

We can guess and check small primes, subtract it from $126$, and see if the result is a prime because the further away the two numbers are, the greater the difference will be. Since $126 = 2 \cdot 3^2 \cdot 7$, we can eliminate $2$, $3$, and $7$ as an option because subtracting these would result in a composite number.

If we subtract $5$, then the resulting number is $121$, which is not prime. If we subtract $11$, then the resulting number is $115$, which is also not prime. But when we subtract $13$, the resulting number is $113$, a prime number. The largest possible difference is $113-13=\boxed{\textbf{(B) } 100}$.

See Also

1973 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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
All AHSME Problems and Solutions
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