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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 0r<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
2726=bx+r(1)4472=ax+r(2)5054=cx+r(3)6412=dx+r(4)
Now we should subtract equation (1) from (2),(2) from (3), (3) from (4) and (4) from (1).
Upon subtracting, we get the following equations.
1746=(ab)x582=(ca)x1358=(dc)x3686=(da)x
We can express these equations numerically in the following way-
1746=2×3×3×97582=2×3×971358=2×7×973686=2×19×97
So from the above expansion, we can observe that 97is the only two digit number which is in common for all four numbers.
The value of mis 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.