Difference between revisions of "2014 AMC 8 Problems/Problem 22"
(→Solution 2) |
|||
| (33 intermediate revisions by 16 users not shown) | |||
| Line 1: | Line 1: | ||
| − | + | ==Problem 22== | |
| + | A <math>2</math>-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? | ||
| + | |||
| + | <math>\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad \textbf{(E) }9</math> | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/7an5wU9Q5hk?t=2226 | ||
| + | |||
| + | ==Video Solution 3== | ||
| + | https://youtu.be/AR3Ke23N1I8 ~savannahsolver | ||
| + | |||
| + | ==Solution== | ||
| + | We can think of the number as <math>10a+b</math>, where a is the tens digit and b is the unit digit. Since the number is equal to the product of the digits (<math>ab</math>) plus the sum of the digits (<math>a+b</math>), we can say that <math>10a+b=ab+a+b</math>. We can simplify this to <math>10a=ab+a</math>, which factors 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>\boxed{\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}} | ||
Latest revision as of 09:51, 23 July 2024
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 by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=2226
Video Solution 3
https://youtu.be/AR3Ke23N1I8 ~savannahsolver
Solution
We can think of the number as 
, where a is the tens digit and b is the unit digit. 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 
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. 
