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Stels [109]
2 years ago
12

A palindromic number is a number that reads the same forwards as it does backwards (example: 282 or 51715). How many 5-digit pal

indromic numbers have digits that add up to 14?
Mathematics
1 answer:
Anuta_ua [19.1K]2 years ago
8 0

Palindromes are numbers whose reverse is the same as the original number

There are twenty-five 5-digit palindromic numbers, that have digits that add up to 14

<h3>How to determine the count of 4-digit palindromes</h3>

The digit of the palindromic number is 5.

This means that:

  • The first digit of the palindrome can be any of 1 to 7 (because 7 is half of 14)
  • The second digit can be any of 0 - 6 (i.e. 7 digits)
  • The third digit can be any of 9 digits
  • The fourth digit must be the second digit (i.e. 1 digit)
  • The last digit must be the first digit (i.e. 1 digit)

So, the total number of digits is:

Total = 7 + 7 + 9 +  1 +1

Total = 25

Hence, there are twenty-five 5-digit palindromic numbers, that have digits that add up to 14

Read more about palindromes at:

brainly.com/question/26527765

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<h2>Answer:</h2>

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In the question,

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On putting the points in the circle we get,

x^{2}+y^{2}+ax+by+c=0\\(a_{1})^{2}+(b_{1})^{2}+a(a_{1})+b(b_{1})+c=0\\and,\\(a_{2})^{2}+(b_{2})^{2}+a(a_{2})+b(b_{2})+c=0\\and,\\(a_{3})^{3}+(b_{3})^{3}+a(a_{3})+b(b_{3})+c=0\\

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(a_{1})^{2}+(b_{1})^{2}+a(a_{1})+b(b_{1})+c=0\\a(a_{1})+b(b_{1})+c=-(a_{1})^{2}-(b_{1})^{2}\,.........(1)\\and,\\a(a_{2})+b(b_{2})+c=-(a_{2})^{2}-(b_{2})^{2}\,.........(2)\\and,\\a(a_{3})+b(b_{3})+c=-(a_{3})^{3}-(b_{3})^{3}\,.........(3)\\

On solving these equation using, <u>Matrix method we can get the required equation of the circle,</u>

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Step-by-step explanation:

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