Consider the prime factorization of 20!.
The LCM of 1, 2, ..., 20 must contain all the primes less than 20 in its factorization, so
where is some integer not divisible by any of these primes.
Compare the factorizations of the remaining divisors of 20!, and check off any whose factorizations are already contained in the product of primes above.
- missing a factor of 2
- ✓
- missing a factor of 2²
- missing a factor of 3
- ✓
- missing a factor of 2
- ✓
- ✓
- missing a factor of 2³
- missing a factor of 3
- missing a factor of 2
From the divisors marked "missing", we add the necessary missing factors to the factorization of , so that
Then the LCM of 1, 2, 3, …, 20 is
Considering that the powers of 7 follow a pattern, it is found that the last two digits of are 43.
<h3>What is the powers of 7 pattern?</h3>
The last two digits of a power of 7 will always follow the following pattern: {07, 49, 43, 01}, which means that, for , we have to look at the remainder of the division by 4:
- If the remainder is of 1, the last two digits are 07.
- If the remainder is of 2, the last two digits are 49.
- If the remainder is of 3, the last two digits are 43.
- If the remainder is of 0, the last two digits are 01.
In this problem, we have that n = 1867, and the remainder of the division of 1867 by 4 is of 3, hence the last two digits of are 43.
More can be learned about the powers of 7 pattern at brainly.com/question/10598663
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Answer:
32°, 58°
Step-by-step explanation:
The second angle is (3x-38), and their sum is ...
x +(3x -38) = 90
4x = 128 . . . . . . . . . add 38 and simplify
x = 32 . . . . . first angle
3x -38 = 3·32 -38 = 58 . . . . . . second angle
The measures of the two angles are 32° and 58°.