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Divisibility Problem from AMC 10A, 2003 | Problem 25

Try this beautiful problem from Number theory based on divisibility from AMC 10A, 2003.

Number theory in Divisibility - AMC-10A, 2003- Problem 25

Let $$n$$ be a $$5$$-digit number, and let $$q$$ and $$r$$ be the quotient and the remainder, respectively, when $$n$$ is divided by $$100$$. For how many values of $$n$$ is $$q+r$$ divisible by $$11$$?

• $$8180$$
• $$8181$$
• $$8182$$
• $$9190$$
• $$9000$$

Key Concepts

Number system

Probability

divisibility

Answer: $$8181$$

AMC-10A (2003) Problem 25

Pre College Mathematics

Try with Hints

Since $$11$$ divides $$q+r$$ so may say that $$11$$ divides $$100 q+r$$. Since $$n$$ is a $$5$$ digit number ...soTherefore, $$q$$ can be any integer from $$100$$ to $$999$$ inclusive, and $$r$$ can be any integer from $$0$$ to $$99$$ inclusive.

can you finish the problem........

Since $$n$$ is a five digit number then and $$11 | 100q+r$$ then $$n$$ must start from $$10010$$ and count up to $$99990$$

can you finish the problem........

Therefore, the number of possible values of $$n$$ such that $$900 \times 9 +81 \times 1=8181$$

Try this beautiful problem from Number theory based on divisibility from AMC 10A, 2003.

Number theory in Divisibility - AMC-10A, 2003- Problem 25

Let $$n$$ be a $$5$$-digit number, and let $$q$$ and $$r$$ be the quotient and the remainder, respectively, when $$n$$ is divided by $$100$$. For how many values of $$n$$ is $$q+r$$ divisible by $$11$$?

• $$8180$$
• $$8181$$
• $$8182$$
• $$9190$$
• $$9000$$

Key Concepts

Number system

Probability

divisibility

Answer: $$8181$$

AMC-10A (2003) Problem 25

Pre College Mathematics

Try with Hints

Since $$11$$ divides $$q+r$$ so may say that $$11$$ divides $$100 q+r$$. Since $$n$$ is a $$5$$ digit number ...soTherefore, $$q$$ can be any integer from $$100$$ to $$999$$ inclusive, and $$r$$ can be any integer from $$0$$ to $$99$$ inclusive.

can you finish the problem........

Since $$n$$ is a five digit number then and $$11 | 100q+r$$ then $$n$$ must start from $$10010$$ and count up to $$99990$$

can you finish the problem........

Therefore, the number of possible values of $$n$$ such that $$900 \times 9 +81 \times 1=8181$$

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