Let the digits of the required number be x and y.
Now, the required number is 10x + y.
According to the question,
10x + y = 4[x + y]
So,
6x − 3y = 0
\[\Rightarrow\]2x − y = 0
\[x = \frac{y}{2}\] .....[1]
Also,
10x + y = 3xy .....[2]
From [1] and [2], we get
\[10\left[
\frac{y}{2} \right] + y = 3\left[ \frac{y}{2} \right]y\]
\[ \Rightarrow 5y + y = \frac{3}{2} y^2 \]
\[ \Rightarrow 6y = \frac{3}{2} y^2 \]
\[\Rightarrow y^2 - 4y = 0\]
\[ \Rightarrow y[y - 4] = 0\]
\[ \Rightarrow y = 0, 4\]
So, x = 0 for y = 0 and x = 2 for y = 4.
Hence, the required number is 24.
Contents
- 1 Problem 22
- 2 Video Solution
- 3 Video Solution for Problems 21-25
- 4 Solution
- 5 Solution 2
- 6 See Also
A
Video Solution
//youtu.be/7an5wU9Q5hk?t=2226
//www.youtube.com/watch?v=RX3BxKW_wTU
//youtu.be/AR3Ke23N1I8 ~savannahsolver
Video Solution for Problems 21-25
//www.youtube.com/watch?v=6S0u_fDjSxc
Solution
We can think of the number as
Solution 2
A two digit number is namely
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.