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Question 261686: Can you please help me with this License plate number problem
the license plate was in two parts, a two digit number and a three digit number. The two digit number was a prime and the sum of the digits was a two digit prime. The tens digit was larger than the units digit.
the last 3 digits of his license plate were all odd and different. The sum of the three digits was palindromic. The sum of the first and third digits was one half of the sum of the 1st and 2nd digits.
what were the numbers on the license plate?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! the license plate was in two parts, a two digit number and a three digit number. The two digit number was a prime and the sum of the digits was a two digit prime. The tens digit was larger than the units digit.
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Of the prime numbers between 10 and 100 only 6 of them is their sum a two digit prime.
Of those six, only 83 has a two digit sum where the 10's digit is greater than the units.
1st two numbers are 83
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Let the last 3 digits be: x, y, z
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the last 3 digits of his license plate were all odd and different.
That would be 1, 3, 5, 7, or 9
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The sum of the three digits was palindromic.
The only palindromic sum of 3 odd different digits is 11! therefore:
x + y + z = 11
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This would be 1, 3, 7, but we don't know which is which:
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Equation for the statement:
" The sum of the first and third digits was one half of the sum of the 1st and 2nd digits."
x + z = .5(x + y)
x + z = .5x + .5y
z = .5x - x + .5y
z = .5y - .5x
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Look as this equation, the only values that would work using 1, 3, 7 would be
3 = .5(7) - .5(1)
3 = 3.5 - .5
:
x = 1; y = 7; z = 3
:
The lic number 83-173
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