2005 PMWC Problems/Problem I9

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Problem

There are four consecutive positive integers (natural numbers) less than $2005$ such that the first (smallest) number is a multiple of $5$ , the second number is a multiple of $7$ , the third number is a multiple of $9$ and the last number is a multiple of $11$ . What is the first of these four numbers?

Solution

Let $n$ be the first number. Then

$\begin{eqnarray*}n &\equiv& 0 \pmod{5}\\n \equiv -1 &\equiv& 6 \pmod{7}\\n \equiv -2 &\equiv& 7 \pmod...$

Applying the Chinese Remainder Theorem, we can eventually find that $n \equiv 1735 \pmod{3465}$ . (We can do this by solving the first two to get $n \equiv 20 \pmod {35}$ and the second two to get $n \equiv 52 \pmod{99}$ , and then list out all of the numbers that fit the second condition until we hit one that fits the first condition. This is not as tedious as it may seem, since we know that the result must be divisible by 5.) Since $0 < n < 2005$ , our answer is $n = 1735$ .

2005 PMWC (Problems)
Preceded by Problem I8	Followed by Problem I10
I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10

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2005 PMWC Problems/Problem I9

From AoPSWiki

Problem

Solution

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