Difference between revisions of "2007 Alabama ARML TST Problems/Problem 5"
(New page: ==Problem== How many positive 5-digit odd integers are palindromes? ==Solution== The units digit must be odd, and the ten-thousands digit must be equal to the units digit. So there are 5 ...) |
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==Problem== | ==Problem== | ||
− | How many positive 5-digit odd integers are | + | How many positive 5-digit odd integers are [[palindrome]]s? |
==Solution== | ==Solution== | ||
− | The units digit must be odd, and the ten-thousands digit must be equal to the units digit. So there are 5 possible choices there. Then there are ten choices for the tens and hundreds digits, but the thousands digit is fixed. Therefore there are 5 | + | The units digit must be odd, and the ten-thousands digit must be equal to the units digit. So there are 5 possible choices there. Then there are ten choices for the tens and hundreds digits, but the thousands digit is fixed. Therefore there are <math>5 \cdot 10 \cdot 10=\boxed{500}</math> 5-digit odd palindromes. |
==See also== | ==See also== | ||
+ | {{ARML box|year=2007|state=Alabama|num-b=4|num-a=6}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] |
Latest revision as of 08:16, 18 June 2008
Problem
How many positive 5-digit odd integers are palindromes?
Solution
The units digit must be odd, and the ten-thousands digit must be equal to the units digit. So there are 5 possible choices there. Then there are ten choices for the tens and hundreds digits, but the thousands digit is fixed. Therefore there are 5-digit odd palindromes.
See also
2007 Alabama ARML TST (Problems) | ||
Preceded by: Problem 4 |
Followed by: Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |