Difference between revisions of "2014 AMC 8 Problems/Problem 22"
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==Solution== | ==Solution== | ||
| − | + | We can think of the number as <math>10a+b</math>, where a and b are digits. Since the number is equal to the product of the digits (<math>a\cdot b</math>) plus the sum of the digits (<math>a+b</math>), we can say that <math>10a+b=a\cdot b+a+b</math>. We can simplify this to <math>10a=a\cdot b+a</math>, and factor to <math>(10)a=(b+1)a</math>. Dividing by <math>a</math>, we have that <math>b+1=10</math>. Therefore, the units digit, <math>b</math>, is <math>\textbf{(E) }9</math> | |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=21|num-a=23}} | {{AMC8 box|year=2014|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:46, 26 November 2014
Problem
A 
-digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number?
Solution
We can think of the number as 
, where a and b are digits. Since the number is equal to the product of the digits (
) plus the sum of the digits (
), we can say that 
. We can simplify this to 
, and factor to 
. Dividing by 
, we have that 
. Therefore, the units digit, 
, is 
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
