Difference between revisions of "1990 USAMO Problems/Problem 1"
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Consider license plates of <math>n</math> digits, for some fixed <math>n</math>, issued with the same criteria. | Consider license plates of <math>n</math> digits, for some fixed <math>n</math>, issued with the same criteria. | ||
− | We first note that by the pigeonhole principle, we may have at most <math>10^{n-1}</math> distinct plates. Indeed, if we have more, then there must be two plates which agree on the first <math>n-1</math> | + | We first note that by the pigeonhole principle, we may have at most <math>10^{n-1}</math> distinct plates. Indeed, if we have more, then there must be two plates which agree on the first <math>n-1</math> digits; these plates thus differ only on one digit, the last one. |
We now show that it is possible to issue <math>10^{n-1}</math> distinct license plates which satisfy the problem's criteria. Indeed, we issue plates with all <math>10^{n-1}</math> possible combinations for the first <math>n-1</math> digit, and for each plate, we let the last digit be the sum of the preceding digits taken mod 10. This way, if two plates agree on the first <math>n-1</math> digits, they agree on the last digit and are thus the same plate, and if two plates differ in only one of the first <math>n-1</math> digits, they must differ as well in the last digit. | We now show that it is possible to issue <math>10^{n-1}</math> distinct license plates which satisfy the problem's criteria. Indeed, we issue plates with all <math>10^{n-1}</math> possible combinations for the first <math>n-1</math> digit, and for each plate, we let the last digit be the sum of the preceding digits taken mod 10. This way, if two plates agree on the first <math>n-1</math> digits, they agree on the last digit and are thus the same plate, and if two plates differ in only one of the first <math>n-1</math> digits, they must differ as well in the last digit. |
Latest revision as of 13:28, 20 November 2019
Problem
A certain state issues license plates consisting of six digits (from 0 through 9). The state requires that any two plates differ in at least two places. (Thus the plates and cannot both be used.) Determine, with proof, the maximum number of distinct license plates that the state can use.
Solutions
Consider license plates of digits, for some fixed , issued with the same criteria.
We first note that by the pigeonhole principle, we may have at most distinct plates. Indeed, if we have more, then there must be two plates which agree on the first digits; these plates thus differ only on one digit, the last one.
We now show that it is possible to issue distinct license plates which satisfy the problem's criteria. Indeed, we issue plates with all possible combinations for the first digit, and for each plate, we let the last digit be the sum of the preceding digits taken mod 10. This way, if two plates agree on the first digits, they agree on the last digit and are thus the same plate, and if two plates differ in only one of the first digits, they must differ as well in the last digit.
It then follows that is the greatest number of license plates the state can issue. For , as in the problem, this number is .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1990 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.