Difference between revisions of "1984 AIME Problems/Problem 2"
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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>. | 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>. | ||
| − | == Solution == | + | == Solution 1== |
Any multiple of 15 is a multiple of 5 and a multiple of 3. | Any multiple of 15 is a multiple of 5 and a multiple of 3. | ||
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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>. | 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== | ||
| + | |||
| + | 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 the answer is <math>80+800+8000=\boxed{8880}.</math> | ||
== See also == | == See also == | ||
Revision as of 18:23, 6 June 2024
Contents
Problem
The integer 
is the smallest positive multiple of 
such that every digit of is either 
or 
. Compute 
.
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 only contains the digits 0 and 8, the units digit of must be 0.
The sum of the digits of any multiple of 3 must be divisible by 3. If has 
digits equal to 8, the sum of the digits of is 
. For this number to be divisible by 3, must be divisible by 3. We also know that 
since is positive. Thus must have at least three copies of the digit 8.
The smallest number which meets these two requirements is 8880. Thus the answer is 
.
Solution 2
Notice how 
for all integers 
. Since we are restricted to only the digits 
, because 
we can't have an in the optimal smallest number. We can just 'add' fives to quickly get 
to get our answer. Thus the answer is 
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 | ||
