Difference between revisions of "1986 AIME Problems/Problem 10"
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This reduces <math>m</math> to one of 136, 358, 580, 802. But also <math>a+b+c=\frac{N+m}{222}>\frac{N}{222}>14</math> so <math>a+b+c\geq 15</math>. Only one of the values of <math>m</math> satisfies this, namely <math>358</math>. | This reduces <math>m</math> to one of 136, 358, 580, 802. But also <math>a+b+c=\frac{N+m}{222}>\frac{N}{222}>14</math> so <math>a+b+c\geq 15</math>. Only one of the values of <math>m</math> satisfies this, namely <math>358</math>. | ||
== See also == | == See also == | ||
| − | |||
| − | |||
{{AIME box|year=1986|num-b=9|num-a=11}} | {{AIME box|year=1986|num-b=9|num-a=11}} | ||
| + | * [[AIME Problems and Solutions]] | ||
| + | * [[American Invitational Mathematics Examination]] | ||
| + | * [[Mathematics competition resources]] | ||
Revision as of 13:52, 6 May 2007
Problem
In a parlor game, the magician asks one of the participants to think of a three digit number (abc) where a, b, and c represent digits in base 10 in the order indicated. The magician then asks this person to form the numbers (acb), (bca), (bac), (cab), and (cba), to add these five numbers, and to reveal their sum, 
. If told the value of , the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if 
.
Solution
Let 
be the number 
. Observe that 
so
This reduces to one of 136, 358, 580, 802. But also 
so 
. Only one of the values of satisfies this, namely 
.
See also
| 1986 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
