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1969 AHSME Problems/Problem 24

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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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