Difference between revisions of "2008 Mock ARML 1 Problems/Problem 7"

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== See also ==
 
== See also ==
{{Mock ARML box|year = 2008|n = 6|num-b=8|num-a=3|source = 206547}}
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[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 17:34, 29 May 2008

Problem

Compute the number of $3$-digit base-$5$ positive integer multiples of $7$ that are also divisible by $7$ when read in base $10$ instead of base $5$.

Solution

Let the number be $\overline{abc}$. Then $7 | 25a + 5b + c, 100a + 10b + c$, and so it must divide their difference, so $7 | 75a + 5b \Longrightarrow 7|-2(a+b)$, from which it follows that $7|a+b$. However, as $a,b < 5$, we have $\{a,b\} = \{3,4\}$, leading to $\boxed{2}$ solutions: $343, 434$.

See also

2008 Mock ARML 1 (Problems, Source)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8