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==AOPS Contributions== | ==AOPS Contributions== | ||
| + | Observe that if <math>1 \le n \le 60</math> such that n is a solution to the desired equation, so is <math>n + 60\cdot m</math>, where m is an integer, <math>0 \le m \le 9</math>. | ||
| + | \\So we only need to consider n from 1 to 60. | ||
| + | As shown in Solution 2, there are 4 cases which we will split into 2 main cases: | ||
| + | \\ | ||
| + | \\Case 1: <math>4 \mid n</math> or <math>6 \mid n</math>, <math>5 \mid (n+1)</math> | ||
| + | \\Case 2: <math>4 \mid (n+1)</math> or <math>6 \mid (n+1)</math>, <math>5 \mid n</math> | ||
| + | \\There are 4 values of n where <math>1 \le n \le 12</math> satisfying <math>4 \mid n</math> or <math>6 \mid n</math>. | ||
| + | \\ | ||
| + | \\I claim that there are 4 values of <math>n \le 60</math> satisfying Case 1. Suppose x is one value of n satisfying <math>4 \mid n</math> or <math>6 \mid n</math>, and <math>n \le 12</math>. | ||
| + | \\Hence the solutions satisfying <math>4 \mid n</math> or <math>6 \mid n</math>, <math>n \le 60</math> are of the form <math>x + 12m</math>, so the values of <math>n + 1</math> are <math>x + 12m + 1 \equiv x + 2m + 1 \equiv 0</math> (mod 5), so <math>2m \equiv 4 + 4x</math> (mod 5) and hence the value of m is unique since <math>0 \le m \le 4</math> to satisfy <math>1 \le n \le 60</math> and 2 and 5 are relatively prime. | ||
| + | \\ | ||
| + | \\A similar approach can be used to show the same for Case 2, that there are 4 values of <math>n \le 60</math>. | ||
| + | \\ | ||
| + | \\Hence our answer is <math>(4+4)*10 = \fbox{080}</math>. | ||
Revision as of 09:44, 22 February 2022
AOPS Contributions
Observe that if 
such that n is a solution to the desired equation, so is 
, where m is an integer, 
.
\\So we only need to consider n from 1 to 60.
As shown in Solution 2, there are 4 cases which we will split into 2 main cases:
\\
\\Case 1: 
or 
, 
\\Case 2: 
or 
, 
\\There are 4 values of n where 
satisfying or .
\\
\\I claim that there are 4 values of 
satisfying Case 1. Suppose x is one value of n satisfying or , and 
.
\\Hence the solutions satisfying or , are of the form 
, so the values of 
are 
(mod 5), so 
(mod 5) and hence the value of m is unique since 
to satisfy and 2 and 5 are relatively prime.
\\
\\A similar approach can be used to show the same for Case 2, that there are 4 values of .
\\
\\Hence our answer is 
.
