+0

+22
182
7

How many positive integers less than or equal to 100 have a prime factor that is greater than 4?

Dec 18, 2022

#1
+3

prime numbers greater than 4 are 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

you can count from there

Dec 19, 2022
#5
-8

Proyaop, you usually present competent math solutions; however, this is NOT one of them....

Your presentation does not even attempt to answer the question. You have listed the prime numbers greater than (4). So what? ... What’s the student supposed to do with them?  ...

[Usually I’d add some snarky comments on what the student is supposed to do with them, like “Fold them neatly and shove them up her ass!” But I saved them for Mr. BB. Also I don’t want to risk pissing-you-off too much because you are one of the Forum’s better student mathematicians – a rare species in the best of times and very rare presently.]

GA

--. .-

GingerAle  Dec 21, 2022
edited by GingerAle  Dec 21, 2022
#2
0

(5, 7, 10, 11, 13, 14, 15, 17, 19, 20, 21, 22, 23, 25, 26, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 97, 98, 99, 100)==80 such integers.

Dec 19, 2022
#6
-7

Mr. BB, you never present competent math solutions. Your answer (via your computerized algorithm) is correct, but how does a student use this to SOLVE the problem?

Let me guess: sprinkle Magic Pixie Dust on it, and click her heals three times while saying, “Abracadabra Bibbidi-Bobbidi-Boo!"

GA

--. .-

GingerAle  Dec 21, 2022
#3
+23

Aww was hoping for a more clever less tedious answer... Thank you anyways!

Dec 20, 2022
#4
+2

Well there is. Just find any integer less than 100 that is a 1, 2, 3 or have their prime factorization be 2^a * 3^b.

You can do easy counting from there. No need to brute force.

proyaop  Dec 20, 2022
#7
-5

Here’s a more clever, less tedious answer ...with a solution:

Start the process by assembling multiples of (2) such that the product does not exceed 100.

Do the same with factors of (3). Then assemble the multiples of (2) and (3) .

Count these, add one (1) to the count because (1) doesn’t have a prime greater than (4).

Subtract this count from 100.

$$\left[ {\begin{array}{ccc} 2^1 = 2 \;\;\; | & 3^1 = 3 \; \;\;\; | & 2^1 * 3^1 = 6 \;\;\;\;\;| & 2^1 * 3^2 = 18 \;\;\; | & 2^1 * 3^3 = 54\\ 2^2 = 4 \;\;\; | & 3^2 = 9 \; \;\;\; | & 2^2 * 3^1 = 12 \;\;\;| & 2^2 * 3^2 = 36 \;\;\; | \\ 2^3 = 8 \;\;\; | & 3^3 = 27 \;\;\; | & 2^3 * 3^1 = 24 \;\;\;|& 2^3 * 3^2 = 72 \;\;\;|\\ 2^4 = 16 \;\; | & 3^4 = 81 \;\;\; | & 2^4 * 3^1 = 48\;\;\;|\\ 2^5 = 32 \;\; |& \hspace{4em} | & 2^5 * 3^1 = 96 \;\;\; |\\ 2^6 = 64 \;\; |& \end{array}} \right]\underbrace{_\text {There are nineteen (19) integers ... plus one (1) }} _\text {...don't forget the (1)}$$

$$100 - 20 = 80 \leftarrow \small \text{ The number of positive integers less than or equal to 100 that have (at least) one prime factor that is greater than 4. }\\$$

GA

--. .-

GingerAle  Dec 21, 2022