TheXSquaredFactor

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UsernameTheXSquaredFactor
Score2446
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Questions 3
Answers 956

 #1
avatar+2446 
+2

Let's define a few variables to start.

 

Let A = one digit of the three-digit number

Let B = one digit of the three-digit number

Let C = one digit of the three-digit number

 

For problems like these, it is generally helpful to find an expression that relates everything together.  \(ABC, ACB, BAC, BCA, CAB, CBA\) represent all the six different 3-digit numbers. We can then represent the sum of all the unique 3-digit numbers with the expression \(222(A+B+C)\) because each digit is added twice in the units, tens, and hundreds digit. 

 

Also, it is generally helpful to find a bound of some kind that restricts the possibilities. This way, it is not necessary to check every single possibility mindlessly. We can achieve this by recognizing that we can represent the sum in a different way. The sum must be at least \(3231+123=3354\) and must not exceed \(3231+987=4218\).

 

With this information, we can solve an inequality that restricts the possibilities for \(A+B+C\).

\(3354\leq 222(A+B+C)\leq4218\\ 15.108...\leq A+B+C \leq 19\)

 

Of course, in the context of this problem, \(A+B+C\) is restricted to the set of whole numbers. Therefore, \(A+B+C\) equals 16,17,18, or 19. After this point, we will just test all the possibilities since we only have 4 to check anyway.

 

We can represent the 6th number as \(222(A+B+C)-3231\). Below, I have included a table for each possibility for \(A+B+C\):

 

\(A+B+C\) 6th Number
\(16\) \(321\)
\(17\) \(543\)
\(18\) \(765\)
\(19\) \(987\)

 

From this table, only one of possibilities for the 6th number reigns supreme, 765, because 765 is the only number where its digits add to its corresponding \(A+B+C\) value.

May 15, 2020
 #2
avatar+2446 
+1

I did this same exercise, and I got 387!

 

If you are curious about the code I wrote, then here you go:

public static void main(String[] args)
    {
        int numPassed = 0;
        int numToTest = 0;
        double temp = 0.0;
        for (int hundredDigit = 0; hundredDigit<=9; hundredDigit++)
        {
            for (int tenDigit = 0; tenDigit<=9; tenDigit++)
            {
                for (int oneDigit = 0; oneDigit<=9; oneDigit++)
                {
                    numToTest = hundredDigit*100+tenDigit*10+oneDigit;
                    temp = Math.sqrt((double) numToTest);
                    if (temp != Math.floor(temp) && numToTest%2==1 && numToTest%5!=0)
                    {
                        System.out.println(numToTest);
                        numPassed++;
                    }
                }
            }
        }
        System.out.println(numPassed);
    }

 

Here is the list, if you are curious: 3,7,11,13,17,19,21,23,27,29,31,33,37,39,41,43,47,51,53,57,59,61,63,67,69,71,73,77,79,83,87,89,91,93,97,99,101,103,107,109,111,113,117,119,123,127,129,131,133,137,139,141,143,147,149,151,153,157,159,161,163,167,171,173,177,179,181,183,187,189,191,193,197,199,201,203,207,209,211,213,217,219,221,223,227,229,231,233,237,239,241,243,247,249,251,253,257,259,261,263,267,269,271,273,277,279,281,283,287,291,293,297,299,301,303,307,309,311,313,317,319,321,323,327,329,331,333,337,339,341,343,347,349,351,353,357,359,363,367,369,371,373,377,379,381,383,387,389,391,393,397,399,401,403,407,409,411,413,417,419,421,423,427,429,431,433,437,439,443,447,449,451,453,457,459,461,463,467,469,471,473,477,479,481,483,487,489,491,493,497,499,501,503,507,509,511,513,517,519,521,523,527,531,533,537,539,541,543,547,549,551,553,557,559,561,563,567,569,571,573,577,579,581,583,587,589,591,593,597,599,601,603,607,609,611,613,617,619,621,623,627,629,631,633,637,639,641,643,647,649,651,653,657,659,661,663,667,669,671,673,677,679,681,683,687,689,691,693,697,699,701,703,707,709,711,713,717,719,721,723,727,731,733,737,739,741,743,747,749,751,753,757,759,761,763,767,769,771,773,777,779,781,783,787,789,791,793,797,799,801,803,807,809,811,813,817,819,821,823,827,829,831,833,837,839,843,847,849,851,853,857,859,861,863,867,869,871,873,877,879,881,883,887,889,891,893,897,899,901,903,907,909,911,913,917,919,921,923,927,929,931,933,937,939,941,943,947,949,951,953,957,959,963,967,969,971,973,977,979,981,983,987,989,991,993,997,999

 

I assume the discrepancy occurred because Guest included 1, which should not be included because it is a square number.

May 15, 2020
 #1
avatar+2446 
+2

Quincy suspects that the two triangles above are similar, but there is not enough information to prove this. This is because \(\triangle JKL\) only has two defined side lengths, which is not enough to determine the other unknown angles or sides. 

 

2a) The SAS Similarity Theorem proves the similarity of two triangles by showing that the following is true:

 

  • Both side lengths of one triangle are proportional to the side lengths of the other triangle.
  • The included angle in one triangle is congruent to the corresponding angle of the other triangle.

Quincy would need the measure of \(\angle K\). This is the included angle of \(\overline{JK}\text{ and }\overline{KL}\) . We can determine the measure of the corresponding angle, \(\angle Q\) , by using the Law of Cosines. 

 

2b) The SSS Similarity Theorem proves the similarity of two triangles by showing that the following is true:

 

  • All three side lengths of one triangle are proportional to the corresponding sides of another triangle.

Quincy would need the length of \(\overline{JL}\) . We already know the lengths of the other two side lengths of \(\triangle JKL\) and all the side lengths of \(\triangle PQR\), so this is the only extra bit of information needed to prove similarity. 

 

2c) The AA Postulate proves the similarity of two triangles by showing that the following is true:

 

  • Two corresponding angles of the two triangles are congruent. 

Of course, Quincy would need to know the measure of two angles before proving their congruency, so it is an unlikely candidate to prove their similarity. Yes, it is possible to do, but it would be quite unnecessary.

Feb 16, 2019