
It is known that \[2726,4472,5054,6412\] have the same remainder when they are divided by some two digit natural number \[m\]. Find the value of \[m\].
Answer
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Hint: In order to find the natural number that divides all the numbers but leaves the same remainder, we must apply Euclid’s division algorithm which is \[a=bq+r\]. Then we have to apply this algorithm to all of the numbers and then we obtain the respective equations. Upon solving them, we obtained the required natural number.
Complete step-by-step solution:
Now let us have a brief regarding the Euclid division algorithm. It is also called as Euclid division Lemma which states that \[a,b\] are positive integers, then there exists unique integers satisfying \[q,r\] satisfying \[a=bq+r\] where \[0\le r< b\].
Now let us find the natural number \[m\].
We know that Euclid division algorithm i.e. \[a=bq+r\]
By applying the division algorithm to the numbers, we get
\[\begin{align}
& 2726=bx+r\to \left( 1 \right) \\
& 4472=ax+r\to \left( 2 \right) \\
& 5054=cx+r\to \left( 3 \right) \\
& 6412=dx+r\to \left( 4 \right) \\
\end{align}\]
Now we should subtract equation \[\left( 1 \right)\] from \[\left( 2 \right)\],\[\left( 2 \right)\] from \[\left( 3 \right)\], \[\left( 3 \right)\] from \[\left( 4 \right)\] and \[\left( 4 \right)\] from \[\left( 1 \right)\].
Upon subtracting, we get the following equations.
\[\begin{align}
& 1746=\left( a-b \right)x \\
& 582=\left( c-a \right)x \\
& 1358=\left( d-c \right)x \\
& 3686=\left( d-a \right)x \\
\end{align}\]
We can express these equations numerically in the following way-
\[\begin{align}
& \Rightarrow 1746=2\times 3\times 3\times 97 \\
& \Rightarrow 582=2\times 3\times 97 \\
& \Rightarrow 1358=2\times 7\times 97 \\
& \Rightarrow 3686=2\times 19\times 97 \\
\end{align}\]
So from the above expansion, we can observe that \[97\]is the only two digit number which is in common for all four numbers.
\[\therefore \] The value of \[m\]is \[97\].
Note: Using the Euclid division algorithm we can also find the HCF of the numbers. We must have a point to note that the numbers must be positive in order to apply the Euclid division algorithm in order to obtain a unique quotient and remainder.
Complete step-by-step solution:
Now let us have a brief regarding the Euclid division algorithm. It is also called as Euclid division Lemma which states that \[a,b\] are positive integers, then there exists unique integers satisfying \[q,r\] satisfying \[a=bq+r\] where \[0\le r< b\].
Now let us find the natural number \[m\].
We know that Euclid division algorithm i.e. \[a=bq+r\]
By applying the division algorithm to the numbers, we get
\[\begin{align}
& 2726=bx+r\to \left( 1 \right) \\
& 4472=ax+r\to \left( 2 \right) \\
& 5054=cx+r\to \left( 3 \right) \\
& 6412=dx+r\to \left( 4 \right) \\
\end{align}\]
Now we should subtract equation \[\left( 1 \right)\] from \[\left( 2 \right)\],\[\left( 2 \right)\] from \[\left( 3 \right)\], \[\left( 3 \right)\] from \[\left( 4 \right)\] and \[\left( 4 \right)\] from \[\left( 1 \right)\].
Upon subtracting, we get the following equations.
\[\begin{align}
& 1746=\left( a-b \right)x \\
& 582=\left( c-a \right)x \\
& 1358=\left( d-c \right)x \\
& 3686=\left( d-a \right)x \\
\end{align}\]
We can express these equations numerically in the following way-
\[\begin{align}
& \Rightarrow 1746=2\times 3\times 3\times 97 \\
& \Rightarrow 582=2\times 3\times 97 \\
& \Rightarrow 1358=2\times 7\times 97 \\
& \Rightarrow 3686=2\times 19\times 97 \\
\end{align}\]
So from the above expansion, we can observe that \[97\]is the only two digit number which is in common for all four numbers.
\[\therefore \] The value of \[m\]is \[97\].
Note: Using the Euclid division algorithm we can also find the HCF of the numbers. We must have a point to note that the numbers must be positive in order to apply the Euclid division algorithm in order to obtain a unique quotient and remainder.
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