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2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2

Revision as of 17:42, 21 August 2023 by Tikachaudhuri (talk | contribs) (→‎Solution)

Problem

Determine all positive integers $a$ such that $a < 100$ and $a^3 + 23$ is divisible by $24$.

Solution

$\{ 1,25,49,73,97 \}$

If ${a^3}+23$ is divisible by $24$, then so is ${a^3}-1$. ${a^3}-1$ is the same as $(a-1)$$({a^2}+a+1)$, and it is divisible by $24$. There are many options: $(a-1)$ is divisible by $24$, $({a^2}+a+1)$ is divisible by $24$, $(a-1)$ is divisible by $8$ and $({a^2}+a+1)$ is divisible by $3$, and $(a-1)$ is divisible by $3$ and $({a^2}+a+1)$ is divisible by $8$.

Case 1: $(a-1)$ is divisible by $24$

This means that $a$ is congruent to $1$ $mod$ $(24)$. Satisfying the range, the following integers that satisfy are:

{$1,25,49,73,97$}

Case 2: $({a^2}+a+1)$ is divisible by $24$

Or $a(a+1)$ is in the form $23$ $mod$ $(24)$. This means that $a(a+1)$ can be {$23,47,71,95$}, in the list, the first three numbers are prime, and The fourth can be factorized into two non-consecutive primes. No results from this case.

Case 3: $(a-1)$ is divisible by $8$ and $({a^2}+a+1)$ is divisible by $3$

$a$ is congruent to $1$ $mod$ $(8)$ or $a$ = $8$ $k$ + $1$ where $k$ is a positive integer. This means that ${(8k+1)^2}$ + $8k+1$ + $1$ = $64$ ${k^2}$ + $24$ $k$ + $3$ which has to be divisible by $3$. That means so does ${k^2}$, or $k$ itself is divisible by $3$. The maximum it can be $12$, because $a<100$ or $a$ = $8k+1$. However, for the available values that can be inputted (0,3,6,9,and 12),the same list results from Case 1. No new values.

Case 4: $(a-1)$ is divisible by $3$ and $({a^2}+a+1)$ is divisible by $8$

$a$ is congruent to $1$ $mod$ $(3)$ or $a$ = $3$ $k$ + $1$ where $k$ is a positive integer. This means that ${(3k+1)^2}$ + $3k+1$ + $1$ = $9$ ${k^2}$ + $9$ $k$ + $3$ which has to be divisible by $8$. That means so does ${k^2}$ + $k$ + $3$. Checking k modulo eight for all values might result in a value of k which can narrow down search values.

Sub case 1: k is congruent to -4(mod 8)

${(-4)^2}$ + $-4$ + $3$ = 7(mod 8)

Sub case 2: k is congruent to -3(mod 8)

${(-3)^2}$ + ($-3$) + $3$ = 1(mod 8)

Sub case 3: k is congruent to -2(mod 8)

${(-2)^2}$ + ($-2$) + $3$ = 5(mod 8)

Sub case 4: k is congruent to -1(mod 8)

${(-1)^2}$ + ($-1$) + $3$ = 3(mod 8)

Sub case 5: k is congruent to 0(mod 8)

${(0)^2}$ + $0$ + $3$ = 3(mod 8)

Sub case 6: k is congruent to 1(mod 8)

${(1)^2}$ + $1$ + $3$ = 5(mod 8)

Sub case 7: k is congruent to 2(mod 8)

${(2)^2}$ + $2$ + $3$ = 1(mod 8)

Sub case 8: k is congruent to 3(mod 8)

${(3)^2}$ + $3$ + $3$ = 7(mod 8)

In no scenario is ${k^2}$ + $k$ + $3$ divisible by $8$, and by working backward, neither can $({a^2}+a+1)$. This means that the list noted in Case 1 are all the numbers possible that satisfy the condition. Our answer is $\boxed{\textbf{(1,25,49,73,97)}}$

See also

2018 UNM-PNM Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNM-PNM Problems and Solutions
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