Difference between revisions of "2021 Fall AMC 10A Problems/Problem 8"
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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4</math> | <math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4</math> | ||
| − | == Solution == | + | == Solution 1 == |
Note that the number <math>\underline{xy} = 10x + y.</math> By the problem statement, <cmath>10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1).</cmath> From this we see that <math>y(y-1)</math> must be divisible by <math>9.</math> This only happens when <math>y=9.</math> Then, <math>x=8.</math> Thus, there is only <math>\boxed{\textbf{(B) }1}</math> cuddly number, which is <math>89.</math> | Note that the number <math>\underline{xy} = 10x + y.</math> By the problem statement, <cmath>10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1).</cmath> From this we see that <math>y(y-1)</math> must be divisible by <math>9.</math> This only happens when <math>y=9.</math> Then, <math>x=8.</math> Thus, there is only <math>\boxed{\textbf{(B) }1}</math> cuddly number, which is <math>89.</math> | ||
~NH14 | ~NH14 | ||
| + | |||
| + | == Solution 2 == | ||
| + | Denote this number as <math>\overline{ab}</math>. | ||
| + | |||
| + | Hence, we have <math>\overline{ab} = a + b^2</math>. | ||
| + | |||
| + | This can be written as <math>10 a + b = a + b^2</math>. | ||
| + | |||
| + | Hence, <math>b \left( b - 1 \right) = 9 a</math>. | ||
| + | This implies <math>9 | b \left( b - 1 \right)</math>. | ||
| + | |||
| + | Hence, either <math>9 | b</math> or <math>9 | b - 1</math>. | ||
| + | Because <math>b \in \left\{ 0 , 1 , \cdots , 9 \right\}</math>, <math>b = 9</math> or 1. | ||
| + | |||
| + | For <math>b = 9</math>, we get <math>a = 8</math>. This is a solution. | ||
| + | |||
| + | For <math>b = 1</math>, we get <math>a = 0</math>. However, recall that <math>a \neq 0</math>. | ||
| + | Hence, this is not a solution. | ||
| + | |||
| + | Therefore, the answer is <math>\boxed{\textbf{(B) }1}</math>. | ||
| + | |||
| + | ~Steven Chen (www.professorchenedu.com) | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=A|num-b=7|num-a=9}} | {{AMC10 box|year=2021 Fall|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:02, 25 November 2021
Contents
Problem
A two-digit positive integer is said to be 
if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?
Solution 1
Note that the number 
By the problem statement, ![\[10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1).\]](http://latex.artofproblemsolving.com/7/9/9/79916ece27f39c825bfa425031e66cc209c3ee04.png)
From this we see that 
must be divisible by 
This only happens when 
Then, 
Thus, there is only 
cuddly number, which is 
~NH14
Solution 2
Denote this number as 
.
Hence, we have 
.
This can be written as 
.
Hence, 
.
This implies 
.
Hence, either 
or 
.
Because 
, 
or 1.
For , we get 
. This is a solution.
For 
, we get 
. However, recall that 
.
Hence, this is not a solution.
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
See Also
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
