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Difference between revisions of "1984 AIME Problems/Problem 2"

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== Problem ==
 
== Problem ==
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The [[integer]] <math>n</math> is the smallest [[positive]] [[multiple]] of <math>15</math> such that every [[digit]] of <math>n</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>.
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== Solution ==
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{{solution}}
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== Solution 1==
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Any multiple of 15 is a multiple of 5 and a multiple of 3.
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Any multiple of 5 ends in 0 or 5; since <math>n</math> only contains the digits 0 and 8, the units [[digit]] of <math>n</math> must be 0. 
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The sum of the digits of any multiple of 3 must be [[divisible]] by 3.  If <math>n</math> has <math>a</math> digits equal to 8, the sum of the digits of <math>n</math> is <math>8a</math>.  For this number to be divisible by 3, <math>a</math> must be divisible by 3. We also know that <math>a>0</math> since <math>n</math> is positive. Thus <math>n</math> must have at least three copies of the digit 8. 
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The smallest number which meets these two requirements is 8880.  Thus the answer is <math>\frac{8880}{15} = \boxed{592}</math>.
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== Solution 2==
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Notice how <math>8 \cdot 10^k \equiv 8 \cdot (-5)^k \equiv 5 \pmod{15}</math> for all integers <math>k \geq 2</math>. Since we are restricted to only the digits <math>8,0</math>, because <math>8\equiv -7 \pmod{15}</math> we can't have an <math>8</math> in the optimal smallest number. We can just 'add' fives to quickly get <math>15 \equiv 0 \pmod{15}</math> to get our answer. Thus n is <math>80+800+8000=8880</math> and <math>n/15=8880/15=\boxed{592}</math>
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== See also ==
 
== See also ==
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* [[1984 AIME Problems/Problem 1 | Previous problem]]
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{{AIME box|year=1984|num-b=1|num-a=3}}
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* [[1984 AIME Problems/Problem 3 | Next problem]]
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* [[AIME Problems and Solutions]]
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* [[1984 AIME Problems]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
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[[Category:Intermediate Number Theory Problems]]

Latest revision as of 00:24, 18 July 2024

Problem

The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$. Compute $\frac{n}{15}$.

Solution 1

Any multiple of 15 is a multiple of 5 and a multiple of 3.

Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0.



The sum of the digits of any multiple of 3 must be divisible by 3. If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$. For this number to be divisible by 3, $a$ must be divisible by 3. We also know that $a>0$ since $n$ is positive. Thus $n$ must have at least three copies of the digit 8.

The smallest number which meets these two requirements is 8880. Thus the answer is $\frac{8880}{15} = \boxed{592}$.

Solution 2

Notice how $8 \cdot 10^k \equiv 8 \cdot (-5)^k \equiv 5 \pmod{15}$ for all integers $k \geq 2$. Since we are restricted to only the digits $8,0$, because $8\equiv -7 \pmod{15}$ we can't have an $8$ in the optimal smallest number. We can just 'add' fives to quickly get $15 \equiv 0 \pmod{15}$ to get our answer. Thus n is $80+800+8000=8880$ and $n/15=8880/15=\boxed{592}$

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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