Difference between revisions of "1995 AIME Problems/Problem 10"

Revision as of 07:40, 16 August 2025

Problem

What is the largest positive integer that is not the sum of a positive integral multiple of $42$ and a positive composite integer?

Solution

The requested number $\mod {42}$ must be a prime number. Also, every number that is a multiple of $42$ greater than that prime number must also be prime, except for the requested number itself. So we make a table, listing all the primes up to $42$ and the numbers that are multiples of $42$ greater than them, until they reach a composite number.

$\begin{tabular}{|r||r|r|r|r|r|} \hline 1 & 43 & 85&&& \\ 2&44&&&& \\ 3&45&&&& \\ 5&47&89&131&173&215 \\ 7&49&&&& \\ 11&53&95&&& \\ 13&55&&&& \\ 17&59&101&143&& \\ 19&61&103&145&& \\ 23&65&&&& \\ 29&71&113&155&& \\ 31&73&115&&& \\ 37&79&121&&& \\ 41&83&125&&& \\ \hline \end{tabular}$

Since $\boxed{215}$ is the greatest number in the list, it is the answer. Note that considering $\mod {5}$ would have shortened the search, since $\text{gcd}(5,42)=1$ , and so within $5$ numbers at least one must be divisible by $5$ .

~minor edit Yiyj1

Afterword

Basically, we are looking for a number where when a multiple of 42 is subtracted from it, the result is a prime number. Any number that ends in a 5 is not prime, except for 5 itself. Since 42 keeps the parity the same and advances odd numbers in the unit digit, then we can conclude that the sought number is $5 \mod 42$ . Specifically, $5 * 42 + 5$ .

-jackshi2006

I've completely rewritten this answer, to make it closer to that of the given solution.)

Let us consider such positive integers separately, by their remainder modulo 42 .

So fix an r , where 1≤r≤42 . (We're picking 1≤r≤42

instead of the conventional 0≤r<42

, because we only care about positive numbers and this makes some arguments simpler.) Let us consider numbers n≡rmod42 .

That is, we're looking for an n

in the sequence

r,r+42,r+2(42),r+3(42),r+4(42),… such that n

is not the sum of a positive composite integer and a positive multiple of 42

. This is equivalent to saying that n , wherever it occurs in this sequence, has no composite number earlier in the sequence than it.

Therefore, the largest "good" n

in the sequence is the largest n
such that all earlier elements of the sequence are non-composite, which means that n
is the first composite number in the sequence.

So for each r , where 1≤r≤42 , we simply need to write down the sequence r,r+42,r+2(42),… , and record the first composite number that occurs in it. The largest of these is the answer. As a shortcut (as e.g. considering r=2

shows that the the answer happens to be at least 44

), we can choose to start with only prime r .

Solution 2

Let our answer be $n$ . Write $n = 42a + b$ , where $a, b$ are positive integers and $0 \leq b < 42$ . Then note that $b, b + 42, ... , b + 42(a-1)$ are all primes.

If $b$ is $0\mod{5}$ , then $b = 5$ because $5$ is the only prime divisible by $5$ . We get $n = 215$ as our largest possibility in this case.

If $b$ is $1\mod{5}$ , then $b + 2 \times 42$ is divisible by $5$ and thus $a \leq 2$ . Thus, $n \leq 3 \times 42 = 126 < 215$ .

If $b$ is $2\mod{5}$ , then $b + 4 \times 42$ is divisible by $5$ and thus $a \leq 4$ . Thus, $n \leq 5 \times 42 = 210 < 215$ .

If $b$ is $3\mod{5}$ , then $b + 1 \times 42$ is divisible by $5$ and thus $a = 1$ . Thus, $n \leq 2 \times 42 = 84 < 215$ .

If $b$ is $4\mod{5}$ , then $b + 3 \times 42$ is divisible by $5$ and thus $a \leq 3$ . Thus, $n \leq 4 \times 42 = 168 < 215$ .

Our answer is $\boxed{215}$ .

1995 AIME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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