Difference between revisions of "2014 AMC 8 Problems/Problem 22"
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==Solution 2== | ==Solution 2== | ||
A two digit number is namely <math>10a+b</math>, where <math>a</math> and <math>b</math> are digits in which <math>0 < a \leq 9</math> and <math>0 \leq b \leq 9</math>. Therefore, we can make an equation with this information. We obtain <math>10a+b=(a \cdot b) + (a + b)</math>. This is just <math>10a+b=ab+a+b.</math> Moving <math>a</math> and <math>b</math> to the right side, we get <math>9a=ab.</math> Cancelling out the <math>a</math>s, we get <math>9=b</math> which is our desired answer as <math>b</math> is the second digit. Thus the answer is <math>\boxed{\textbf{(E)}9}</math>. | A two digit number is namely <math>10a+b</math>, where <math>a</math> and <math>b</math> are digits in which <math>0 < a \leq 9</math> and <math>0 \leq b \leq 9</math>. Therefore, we can make an equation with this information. We obtain <math>10a+b=(a \cdot b) + (a + b)</math>. This is just <math>10a+b=ab+a+b.</math> Moving <math>a</math> and <math>b</math> to the right side, we get <math>9a=ab.</math> Cancelling out the <math>a</math>s, we get <math>9=b</math> which is our desired answer as <math>b</math> is the second digit. Thus the answer is <math>\boxed{\textbf{(E)}9}</math>. | ||
| − | ~ | + | ~mathboy |
==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 13:27, 12 January 2022
Problem 22
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?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=2226
https://www.youtube.com/watch?v=RX3BxKW_wTU
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 
, which factors to 
. Dividing by 
, we have that 
. Therefore, the units digit, 
, is 
Solution 2
A two digit number is namely , where and are digits in which 
and 
. Therefore, we can make an equation with this information. We obtain 
. This is just 
Moving and to the right side, we get 
Cancelling out the s, we get 
which is our desired answer as is the second digit. Thus the answer is 
.
~mathboy
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. 
