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A is a natural number. When it's divided by 12 the remainder is 9 and when it's divided by 16 the remainder is 13. What is the least possible value of A?

DonAntonio
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Ally
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3 Answers3

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Hints:

$$\begin{align*}A&=12k+9\\A&=16m+13\end{align*}\implies 4(4m-3k+1)=0\implies 4m-3k=-1\ldots$$

DonAntonio
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  • The number should be positive – Ally Jan 04 '14 at 12:32
  • @Ally, another way to look at it is $3k-4m=1$. What are the smallest values of $k$ and $m$ that satisfy this? – JRN Jan 04 '14 at 13:11
  • Of course, @Ally: choose the minimal positive $;m,k;$ that fulfill the rightmost equation and you'll get the minimal positive $;A;$ . – DonAntonio Jan 04 '14 at 16:39
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Hint: In the first case the number may be $9$, $21$, $33$ an so on. What are the possible numbers in the second case? Write them down and find the smallest number that appears in both cases.

Michael Hoppe
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HINT:

So, the number $\displaystyle n\equiv9\pmod{12}-3\iff 12$ divides $n+3$

and $\displaystyle n\equiv13\pmod{16}\equiv-3\iff 16$ divides $n+3$

$\displaystyle\implies n\equiv-3\pmod{\text{lcm(12,16)}}$

or $n+3$ will divisible by lcm$(12,16)$

lab bhattacharjee
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