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Counting - Case Work

SOURCE:COMPETITION
Number of Problems: 2.
FOR PRINT ::: (Book)

Problem Num : 1

Type:

Topic:Counting

Adjustment# : 0

Difficulty: 1

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According to the standard convention for exponentiation, $2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536.$
If the order in which the exponentiations are performed is changed, how many other values are possible?
$mathrm{(A) } 0qquad mathrm{(B) } 1qquad mathrm{(C) } 2qquad mathrm{(D) } 3qquad mathrm{(E) } 4$

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Category Case Work

Analysis

Solution/Hint
The best way to solve this problem is by simple brute force.
It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as $2uparrow 2uparrow 2uparrow 2$ , where $uparrow$ denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:

$2uparrow (2uparrow (2uparrow 2))$
$2uparrow ((2uparrow 2)uparrow 2)$
$((2uparrow 2)uparrow 2)uparrow 2$
$(2uparrow (2uparrow 2))uparrow 2$
$(2uparrow 2)uparrow (2uparrow 2)$

We can note that $2uparrow (2uparrow 2) = (2uparrow 2)uparrow 2 =16$ . Therefore options 1 and 2 are equal, and options 3 and 4 are equal. Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.
$((2uparrow 2)uparrow 2)uparrow 2 = 16uparrow 2 = 256$
$(2uparrow 2)uparrow (2uparrow 2) = 4 uparrow 4 = 256$
Thus the only other result is $256$ , and our answer is $oxed{ ext{(B)} 1}$ .

Problem Num : 2

Type:

Topic:Counting

Adjustment# : 0

Difficulty: 1

'
How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$mathrm{(A) } 41qquad mathrm{(B) } 42qquad mathrm{(C) } 43qquad mathrm{(D) } 44qquad mathrm{(E) } 45$
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Category Case Work

Analysis

Solution/Hint
If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.
Doing some casework:
If the middle digit is $1$ , possible numbers range from $111$ to $210$ . So there are $2$ numbers in this case.
If the middle digit is $2$ , possible numbers range from $123$ to $420$ . So there are $4$ numbers in this case.
If the middle digit is $3$ , possible numbers range from $135$ to $630$ . So there are $6$ numbers in this case.
If the middle digit is $4$ , possible numbers range from $147$ to $840$ . So there are $8$ numbers in this case.
If the middle digit is $5$ , possible numbers range from $159$ to $951$ . So there are $9$ numbers in this case.
If the middle digit is $6$ , possible numbers range from $369$ to $963$ . So there are $7$ numbers in this case.
If the middle digit is $7$ , possible numbers range from $579$ to $975$ . So there are $5$ numbers in this case.
If the middle digit is $8$ , possible numbers range from $789$ to $987$ . So there are $3$ numbers in this case.
If the middle digit is $9$ , the only possible number is $999$ . So there is $1$ number in this case.
So the total number of three-digit numbers that satisfy the property is $2+4+6+8+9+7+5+3+1=45Rightarrow E$

Solution 2 (much faster and slicker)

Alternatively, we could note that the middle digit is uniquely defined by the first and third digits, since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:
If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are $5cdot5=25$ numbers in this case.
If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are $4cdot5=20$ numbers here.
The total number, then, is $20+25=45Rightarrow E$