Math, asked by pinturathod15926, 10 months ago

How many 7's are present in the following sequence of numbers which are both preceded and followed by perfect square numbers ? 8 6 3 7 9 7 4 3 6 9 7 2 1 7 4 9 5 2 5 4 7 9 5 2 9 7 1 4 7 1 "

Answers

Answered by saisagar6129
3

Answer:

Step-by-step explanation:

five 7's are there which succeded and preceeded by perfect squares

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Answered by sourasghotekar123
0

Answer:

There are 5 seven's from the given sequence- 8 6 3 7 9 7 4 3 6 9 7 2 1 7 4 9 5 2 5 4 7 9 5 2 9 7 1 4 7 1 were the seven comes in between the the perfect square.

Step-by-step explanation:

  • What is perfect square? When you multiply an integer (a “whole” number, positive, negative or zero) times itself, the resulting product is called a square number, or a perfect square or simply “a square.” So, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, and so on, are all square numbers.
  • So in the given series- 1,4,9 are the perfect square.
  • Finding the seven (7) which occurs or which is followed by perfect square that is preceding and succeeding by perfect square.
  • The first seven is in between 3 and 9, 9 is perfect square but three is not. Second square comes between 9 and 4 which follows the given conditions. So in such as way we will find other 7 which comes in between perfect squares.
  • So there are 5 such 7s which comes in between perfect square.

Hence, there are five sevens which comes in between the perfect square.

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