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Zina [86]
2 years ago
12

205%29%20-%20%20%20%7B%284%20%2B%202%29%7D%5E%7B2%7D%20" id="TexFormula1" title=" {3}^{4} \times {3}^{5 } +2(5 - 2 \times 5) - {(4 + 2)}^{2} " alt=" {3}^{4} \times {3}^{5 } +2(5 - 2 \times 5) - {(4 + 2)}^{2} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
neonofarm [45]2 years ago
7 0
3 {}^{4} \times {3}^{5} + 2(5 - 2 \times 5) - (4 + 2) \\ = 3 {}^{4 + 5} + 2(5 - 10) - (6) \\ = 3 {}^{9} + 2( - 5) - 6 \\ = 3 {}^{9} - 10 - 6 \\ = 19667

3^4 × 3^5 Is the same as 3^9.
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Mashutka [201]

Answer: \frac{5}{35}

Step-by-step explanation:

Since, the total number of contestant = 7,

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Answer:

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Step-by-step explanation:

23 is a prime number. That fact informs the factorization of 23 and 230.

The sums of digits of the other two numbers are multiples of 9, so each is divisible by 9 = 3^2. Dividing 9 from each number puts the result in the range where your familiarity with multiplication tables comes into play.

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<em>Comment on divisibility rules</em>

Perhaps the easiest divisibility rule to remember is that a number is divisible by 9 if the sum of its digits is divisible by 9. That is also true for 3: if the sum of digits is divisible by 3, the number is divisible by 3. Another divisibility rule fall out from these: if an even number is divisible by 3, it is also divisible by 6. Of course any number ending in 0 or 5 is divisible by 5, and any number ending in 0 is divisible by 10.

Since 2, 3, and 5 are the first three primes, these rules can go a ways toward prime factorization if any of these primes are factors. That is, it can be helpful to remember these divisibility rules.

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