anonymous
  • anonymous
25th root of 10 minus 7 is divisible by what number?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathstudent55
  • mathstudent55
Is this \( \sqrt[25]{10} - 7\) or \(\sqrt[25]{10 - 7} \)? Also the root part gives an irrational answer. How do you use divisibility with irrational numbers?
anonymous
  • anonymous
this is the equation 10^25 -7 is divisible by what number?
mathstudent55
  • mathstudent55
Oh, you mean 10 to the 25th power minus 7, not root. \(10^{25} - 7 \)

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anonymous
  • anonymous
yup
mathstudent55
  • mathstudent55
Look at the following pattern: \(10^1 - 7 = 10 - 7 = 3\) divisible by 3 \(10^2 - 7 = 100 - 7 = 93\) divisible by 3 \(10^3 - 7 = 1000 - 7 = 993\) divisible by 3 \(10^4 - 7 = 10000 - 7 = 9993\) divisible by 3 \(10^5 - 7 = 100000 - 7 = 99993\) divisible by 3 Can you predict what will happen when the exponent is 25?
anonymous
  • anonymous
i tried it,its beyond billion,but that is in my questionaire
anonymous
  • anonymous
got it,thank u so much :)
mathstudent55
  • mathstudent55
Notice that 10^n, where n is a natural number (1, 2, 3, ...) is simply the number made up of a 1 followed by n zeros. For example, 10^1 = 10 (1 followed by 1 zero). 10^4 = 10000 (1 followed by 4 zeros) 10^25 = 10000000000000000000000000
mathstudent55
  • mathstudent55
If you subtract 7 from such a number, you get either 3 (for 10 - 7), or one or more 9's followed by a 3, as I showed above. For 10^25, you will have many, many 9's followed by a single 3. That number is divisible by 3.
mathstudent55
  • mathstudent55
If you subtract 7 from such a number, you get either 3 (for 10 - 7), or one or more 9's followed by a 3, as I showed above. For 10^25, you will have many, many 9's followed by a single 3. That number is divisible by 3.
mathstudent55
  • mathstudent55
Great. You're welcome.
mathstudent55
  • mathstudent55
Great. You're welcome.
mathstudent55
  • mathstudent55
Great. You're welcome.
mathstudent55
  • mathstudent55
Great. You're welcome.
mathstudent55
  • mathstudent55
Great. You're welcome.
anonymous
  • anonymous
i still have some equations here can you please give me your email add
mathstudent55
  • mathstudent55
Just post them. Someone will help you.

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