1997 CEMC Pascal Problems/Problem 9

Problem

When a certain number is divided by $9$ , the quotient is $6$ and the remainder is $4$ . The number is

$\text{ (A) }\ 58 \qquad\text{ (B) }\ 42 \qquad\text{ (C) }\ 33 \qquad\text{ (D) }\ 67 \qquad\text{ (E) }\ 49$

Solution 1

Using the division algorithm, we can note that this means $qx + r = n$ , where $q$ is the quotient, $x$ is the number that we divided by, $r$ is the remainder of the division, and $n$ is the original number before division.

Plugging in the values of $q$ , $x$ , and $r$ given in the original problem, we have:

$qx + r = n$

$n = 6 \times 9 + 4 = 54 + 4 = \boxed {\textbf {(A) } 58}$

~anabel.disher

Solution 2 (answer choice elimination)

We can simply take the floor function of the division of the answer choices, and see if the quotient is too high, too low, or if it is correct:

Using answer choice E, we have:

$\lfloor \frac{49}{9} \rfloor = \lfloor 5.\overline{4} \rfloor = 5$

This is below the quotient, eliminating answer choices B, C, and E.

Using answer choice D, we have:

$\lfloor \frac{67}{9} \rfloor = \lfloor 7.\overline{4} \rfloor = 7$

This is above the quotient, eliminating answer choice D.

Since all of the other answer choices have been eliminated, the answer is $\boxed {\textbf {(A) } 58}$ .

~anabel.disher

Solution 3 (answer choices)

We can notice the answer choice must be between $9 \times 6 = 54$ and $9 \times 7 = 63$ because the quotient was $6$ , and had a remainder. The only answer choice that satisfies this is $\boxed {\textbf {(A) } 58}$ .

~anabel.disher

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1997 CEMC Pascal Problems/Problem 9

Contents

Problem

Solution 1

Solution 2 (answer choice elimination)

Solution 3 (answer choices)