Find all positive integers less than 100 whose 762nd power ends in the digits "41"

  21, 29, 71, 79

Step-by-step explanation:

Only integers ending in 1, 3, 7, or 9 have powers that end in 1. For the numbers so ending, their sequences of powers mod 100 have a repeat length that is a divisor of 20. That is, raising any of these numbers to the 762nd power mod 100 is equivalent to squaring them, mod 100.

For example, the powers of 29 mod 100 are, beginning with the first power, ...

  {29, 41, 89, 81, 49, 21, 9, 61, 69, 1}

This sequence is of length 10, typical of many of the numbers ending in 1, 3, 7, or 9.

The only positive integers less than 100 whose squares end in 41 are ...

  21, 29, 71, 79 . . . . . positive integers < 100 satisfying n^762 mod 100 = 41

the correct answer option is (2x-3)(2x+3).

step-by-step explanation:

we are given the following expression and we are to factorize it:

we are to find such factors that when multiplied they give a result of 36 (9 x 4 = 36) and when added they give a result of 0.

the two such factors of 36 are negative six and postive six.

factorizing the given expression to get:

therefore, the correct factorization of is (2x-3)(2x+3).

55 because 55 + 55 = 110 good luck babe