A division problem given in school resulted in a quotient of 57 and a
remainder of 52. One student who had solved the problem wasn't quite sure of his figuring efficiency and so he made a test by multiplying the quotient with unfortunately was in no way identical with the original dividend. The reason was that the student when doing the multiplication had mistaken a six, the second figure of the divisor from the right, for a zero. What was the original division problem?
Answers
Step-by-step explanation:
A division problem given in school resulted in a quotient of 57 and a
remainder of 52. One student who had solved the problem wasn't quite sure of his figuring efficiency and so he made a test by multiplying the quotient with unfortunately was in no way identical with the original dividend. The reason was that the student when doing the multiplication had mistaken a six, the second figure of the divisor from the right, for a zero. What was the original division problem?
Answer:
A division problem given in school resulted in a quotient of 57 and a
remainder of 52. One student who had solved the problem wasn't quite sure of his figuring efficiency and so he made a test by multiplying the quotient with unfortunately was in no way identical with the original dividend. The reason was that the student when doing the multiplication had mistaken a six, the second figure of the divisor from the right, for a zero. What was the original division problem?