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Difference between revisions of "1969 AHSME Problems/Problem 24"

(Created page with "== Problem == When the natural numbers <math>P</math> and <math>P'</math>, with <math>P>P'</math>, are divided by the natural number <math>D</math>, the remainders are <math>R</...")
 
(Solution to Problem 24)
 
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== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
 
  
== See also ==
+
The divisors are the same, so take each variable [[modulo]] <math>D</math>.
 +
<cmath>P \equiv R \pmod{D}</cmath>
 +
<cmath>P' \equiv R’ \pmod{D}</cmath>
 +
That means
 +
<cmath>PP’ \equiv RR’ \pmod{D}</cmath>
 +
Thus, <math>PP’</math> and <math>RR’</math> have the same remainder when divided by <math>D</math>, so the answer is <math>\boxed{\textbf{(E)}}</math>.
 +
 
 +
== See Also ==
 
{{AHSME 35p box|year=1969|num-b=23|num-a=25}}   
 
{{AHSME 35p box|year=1969|num-b=23|num-a=25}}   
  
 
[[Category: Introductory Number Theory Problems]]
 
[[Category: Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:06, 22 July 2018

Problem

When the natural numbers $P$ and $P'$, with $P>P'$, are divided by the natural number $D$, the remainders are $R$ and $R'$, respectively. When $PP'$ and $RR'$ are divided by $D$, the remainders are $r$ and $r'$, respectively. Then:

$\text{(A) } r>r' \text{  always}\quad \text{(B) } r<r' \text{  always}\quad\\ \text{(C) } r>r' \text{  sometimes and } r<r' \text{  sometimes}\quad\\ \text{(D) } r>r' \text{  sometimes and } r=r' \text{  sometimes}\quad\\ \text{(E) } r=r' \text{  always}$

Solution

The divisors are the same, so take each variable modulo $D$. \[P \equiv R \pmod{D}\] \[P' \equiv R’ \pmod{D}\] That means \[PP’ \equiv RR’ \pmod{D}\] Thus, $PP’$ and $RR’$ have the same remainder when divided by $D$, so the answer is $\boxed{\textbf{(E)}}$.

See Also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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