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Question 607897: six times the sum of the digits of a two digit number is one less than the number and eight more than the number obtained by reversing the digits of the two digit number. what is the product of the digits?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The two digits a & b
10a + b = "the number"
:
"six times the sum of the digits of a two digit number is one less than the number"
6(a + b) = 10a + b - 1
6a + 6b = 10a + b - 1
6b - b = 10a - 6a - 1
5b = 4a - 1
-4a + 5b = -1
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"and eight more than the number obtained by reversing the digits of the two digit number."
6(a + b) = 10b + a + 8
6a + 6b = 10b + a + 8
6a - a + 6b - 10b = 8
5a - 4b = 8
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multiply the 1st equation by 5, the 2nd equation by 4:
-20a + 25b = -5
+20a - 16b = 32
-----------------adding eliminates a, find b
0 + 9b = 27
b = 27/9
b = 3
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Find a
5a - 4b = 8
Replace b with 3
5a - 12 = 8
5a = 8 + 12
5a = 20
a = 4
:
43 is the number, the product of the digits is 12
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