Questions on Algebra: Divisibility and Prime Numbers answered by real tutors!

Tutors Answer Your Questions about Divisibility and Prime Numbers (FREE)

Question 612341: what are the prime factors of 80,60,40?
Answer by radh(63)

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Algebra.com has solvers to find the factors of any number.

Solved by pluggable solver: Find factors of any number

80 is NOT a prime number: 80 = 2 * 2 * 2 * 2 * 5

Work Shown

80 is divisible by 2: 80 = 40 * 2.
40 is divisible by 2: 40 = 20 * 2.
20 is divisible by 2: 20 = 10 * 2.
10 is divisible by 2: 10 = 5 * 2.
5 is not divisible by anything.

This calculation used this Prime Factorization Algorithm.

You can see from the work shown, that obviously 5 is a prime factor, and so is 2 because 2 is not divisible by anything but itself and 1.

So, to find factors, you can use the solver (or factor it using logic), and look for the numbers that are not divisible by anything other than itself and 1. Try it for the other numbers.

Question 611475: i am a factor of 120, and a common multiple of 3, 4, and 10. The sum of my digits is 6.
Answer by KMST(893)

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120=12%2A10=3%2A4%2A10=3%2A2%2A2%2A10=3%2A2%2A2%2A2%2A5

So, the prime factorization of 120 is
120=2%2A2%2A2%2A3%2A5

while

and

A multiple of 4 needs to have 2%2A2

in its factorization.
A multiple of 10 needs to have 2%2A5

in its factorization.
A multiple of 3 needs to have

in its factorization.
The number 3%2A2%2A2%2A5=highlight%2860%29

is the smallest number that has all the factors needed, and is the answer to the problem.

60=3%2A2%2A2%2A5=%283%2A2%29%2A%282%2A5%29=6%2A10

so it is a multiple of 10.
60=3%2A2%2A2%2A5=3%2A%282%2A2%2A5%29=3%2A20

so it is a multiple of 3.
60=3%2A2%2A2%2A5=3%2A%282%2A2%29%2A5=3%2A4%2A5=%283%2A5%29%2A4=15%2A4

so it is a multiple of 4.

We could include more factors, but that would result in larger numbers, and most of them would be larger than 120, so they would not be factors of 120, and would not be a solution to the problem.
Adding factors, the smallest number we could get is when we include the smallest factor (another 2) as an extra factor. Then we get the only other factor of 120 that is also a common multiple of 3, 4, and 10.
3%2A2%2A2%2A5%2A2=%283%2A2%2A2%2A5%29%2A2=60%2A2=120

3%2A2%2A2%2A5%2A2=%283%2A2%2A2%2A5%29%2A2=60%2A2=120

, is a multiple of 120, but its digits do not add to 6.

Question 604114: find the prime
27,29,33,35
Answer by jim_thompson5910(23764)

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27 = 9*3 .... so 27 is NOT prime

33 = 3*11 ... so 33 is NOT prime

35 = 7*5 .... so 35 is NOT prime

By process of elimination and by the fact that 29 is indeed prime (the only factors are 29 and 1), the answer is 29.

Question 603923: i do not know how to use a step diagram to find the prime factorization of 196 please help

Answer by ewatrrr(7335)

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Hi,



             196

        4         49

       2 2       7  7

The prime factorization of 196 is 2·2·7·7  OR 2^2·7^2

Question 600447: 75=write the composite number as a product of prime numbers
Answer by lynnlo(331)   (Show Source):
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answer:exact solution: 3.5.5

Question 600110: 50/80000

Answer by Alan3354(23874)   (Show Source):
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duplicate

Question 600112: can you help me solve 50/80000 and can you show me a quicker way to solve it and can you show me how you done the working out. can you answer the question now. thnx....:)

Answer by Alan3354(23874)   (Show Source):
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can you help me solve 50/80000
------------
It's not solving, just reducing or simplifying.
50/80000
You do the same operation to the numerator and the denominator.
Divide both by 10
--> 5/8000
Divide both by 5
--> 1/1600
That's all you do as a fraction.
You can use a calculator (or other method) to get 0.000625

Question 599290: Try to find a number for x that when substituted in the formula X² - x + 23 yields a composite number.
Answer by jim_thompson5910(23764)   (Show Source):
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Plug in various values of x (start at x = 1 and increment by 1, so plug in x = 2, then x = 3, etc..) and make the following table.

x y ------------------------------------ 0 23 1 23 2 25 3 29 4 35 5 43 6 53 7 65 8 79 9 95 10 113

Looking at the table above, we can see that when x = 2, y = 25 (which is a composite number). So is composite when

Question 599148: How do you know when it is divisible by 2,9,5,8

Answer by rfer(10941)   (Show Source):
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By knowing your mutiplication tables.

Question 597407: Write 360 as a product of prime factor
Answer by rapaljer(4601)   (Show Source):
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First, find any number that divides evenly into 360. Perhaps the number that comes to mind, since it ends in a 0, is 10. Since 10 divides into 360 and the result is 36, you can write 360 = 10*36.

Next, break down the number into 2*5, and 36 into 6*6 (or some other product will do just as well!)

360=10*36
360=2*5*6*6

Keep breaking numbers down until you get PRIME NUMBERS, like 2, 3, 5, 7, etc.

Next, you need to break down the 6 into 2*3, so this is what you have:
360=2*5*2*3*2*3

Write the numbers in order from smallest to largest:
360=2*2*2*3*3*5

Finally, write this product using exponents. Notice that you have 2*2*2, which would be and .

I have a section explaining FACTORING NUMBERS on my own website, using "Factor Trees." The easiest way to find the website is to use the easy to spell and easy to remember link www.mathinlivingcolor.com. On this single page website, there is a link at the bottom of the page that takes you to my Homepage.

Near the top of my homepage, look for the link "Basic, Intermediate, and College Algebra: One Step at a Time", and choose Basic Algebra (Chapter 2). Look for Section 2.02 FACTORING NUMBERS.

There you will find explanations that may be easier to understand than your traditional textbooks. If you like my website, please tell your friends and family. It's all FREE. You can contact me by Email at rapaljer@seminolestate.edu.

Dr. Robert J. Rapalje, Retired Professor of Mathematics
Seminole State College of Florida

Question 597402: Write 250 as a product of prime numbers
Answer by edjones(7481)   (Show Source):
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250/2=125
125/5=25
25/5=5
.
2*5*5*5
.
Ed

Question 597022: Hi. Can someone please teach me how to solve this question? It's quite easy but I want to know how. It would mean a lot. Thanks!
How many integers between 197 and 303 are divisible by 4 or 10?
Found 2 solutions by AnlytcPhil, math-vortex:
Answer by AnlytcPhil(1177)   (Show Source):
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The other tutor counted the multiples of both 4 and 10 twice. Here's the way I do it. First we find the number of multiple of 4 there are between those two numbers. Multiples of 4 come every fourth number, so to find out how many we do this: There are 303 numbers from 1 to 303. We do not use the 196 numbers from 1 throught 196. So to find out have many numbers there are between 197 and 303, we subtract 303-196 and get 107. Multiples of 4 come every fourth number, So can find out how many there are by dividing 107 by 4 and get 26.75. That doesn't make it to 27, so we use the largest whole number that doesn't exceed that and so we round 26.75 down to 26. That's the number of multiples of 4 between 197 and 303. Now some of those 26 multiples of 4 are also divisible by 10, so the only ones not counted among the 26 are the ones that are multiples of 10 that are not multiples of 4. Since any number of 100's is divisible by 4 we only need to look at the last two digits to find the numbers which are multiples of 10 that are not multiples of 4. They are these five numbers: 10, 30, 50, 70, and 90. So that means 210, 230, 250, 270, and 290 are the only ones between 197 and 303 that are mutiples of 10 that are not multiples of 4. So that's a total of 26+5 or 31. Edwin

Answer by math-vortex(216)   (Show Source):
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Find the first number divisible by 4 starting with 197.
200 is divisible by 4 because 4*50=200. After that, every fourth number if divisible by 4
.
200, 204, 208, 212,..., 300
.
300 is the last number that is divisible by 4 because 4*75=300, and the next integer would be 304.
.
There are 26 integers in this range; (300-200)/4 is 25, plus 1 extra because we include the numbers at both end of the range.
.
To find the multiples of 10, we use a similar process.
200 is the first (10*20=200), and 300 is the last (10*30).
.
200, 210, 220, 230,..., 300
.
There are 11 integers in this range; (300-200)/10 is 10, plus 1 extra because we include the integers at both ends of the range.
.
That's it. Feel free to email via gmail if this explanation is unclear.
.
Ms.Figgy
math.in.the.vortex@gmail

Question 596369: How to find the smallest prime number when there are more than one prime factors ?
Answer by richard1234(5244)   (Show Source):
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There aren't any.

Prime numbers are only divisible by 1 and themselves. 1 is defined as not a prime, so primes only have one prime factor.

Question 595147: Formulas that yield prime numbers, for example: x squared - x + 41
Select 5 numbers for X,- 0 , two even, two odd- substitute them in the formula see if prime numbers occur. try to find a number for X that when substituted in the formula yields a composite number.
This might as well be in Greek to me...
Can you please help?

Answer by Alan3354(23874)   (Show Source):
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Formulas that yield prime numbers, for example: x squared - x + 41
Select 5 numbers for X,- 0 , two even, two odd- substitute them in the formula see if prime numbers occur. try to find a number for X that when substituted in the formula yields a composite number.
----------------
Prime numbers have no factors other than 1 and itself. eg, 7 = 1*7
Composite numbers have other factors, eg, 15 = 3*5
------------------

Sub numbers for x and get the result
For zero --> 0^2 - 0 + 41 = 41
That's called f(0), the value of the function at zero.
----------
f(1) = 1^2 - 1 + 41 = 41
f(2) = 2^2 - 2 + 41 = 43
f(3) = 3^2 - 3 + 41 = 47
f(4) = 4^2 - 4 + 41 = 53
f(5) = 5^2 - 5 + 41 = 61
-------
That's 6 of them, and they're all prime numbers.
------
f(41) = 41^2 - 41 + 41 = 1681 which is not prime, it's 41*41

Question 593323: What is the sum of all odd one digit prime positive numbers
Answer by Edwin McCravy(7325)   (Show Source):
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3 + 5 + 7 = 15

Question 592541: Find a value of n that makes the expression divisible by 2,3, and 5.
My problems are:
1. 9n+3
2. n-5
3. 5n
4. 2n-4
Answer by solver91311(13355)   (Show Source):
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30 is the lowest number divisible by 2, 3, and 5. Set your expressions equal to 30 and solve for

John

My calculator said it, I believe it, that settles it

Question 591789: What is the smallest odd number you can obtain from the product
of four different prime numbers? Show or explain how you got
your answer.

Found 2 solutions by richard1234, solver91311:
Answer by richard1234(5244)   (Show Source):
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Cannot use 2, because 2 is even and the product of an even number with any set of integers is even.

Therefore, use the next four smallest primes, 3,5,7,11 and multiply them.

Answer by solver91311(13355)   (Show Source):
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The product of any two odd numbers is odd:

.

Since

is even, hence

is odd.

Any product that has a factor of 2 is even, so in order to get 4 prime factors of an odd number, 2 must be excluded. The smallest product is therefore obtained from the next four prime numbers.

John

My calculator said it, I believe it, that settles it

Question 589144: i have to know the definition of groups and EXAMPLES

Answer by richard1234(5244)   (Show Source):
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A set that is defined under some binary operation. That is, any two elements in the group can be combined with the operation to produce another element in the group (closure). For example, the set of integers forms a group under addition, because any two integers add up to another integer, which is also in the group.

Question 588799: Name a number that is >200 that is divisible by both 3 and 5.
What are the steps you take to solve this question?
Answer by solver91311(13355)   (Show Source):
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The only numbers that are divisible by 5 end in either 0 or 5. Numbers where the sum of the digits is divisible by 3 are divisible by 3.

Since the desired number must be strictly greater than 200, then next number that ends in either 0 or 5 is 205. But the sum of the digits of 205 is 7 which is not divisible by 3. The next number that is divisible by 5 is 210. The sum of the digits of 210 is 3 which is divisible by 3, hence 210 is divisible by 5 because it ends in 0 and 210 is divisible by 3 because its sum of digits is divisible by 3. Not only did we find a number >200 divisible by both 3 and 5, but we found the smallest number >200 that is divisible by both 3 and 5.

John

My calculator said it, I believe it, that settles it

Question 588005: Prove or disprove the following: The expression n2 – n +41 produces a prime number for every positive integer n. (Hint: try the following numbers 1, 10, and 41).
Answer by richard1234(5244)   (Show Source):
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n = 41 fails because 41^2 - 41 + 41 = 41^2, obviously not a prime number.

Question 587074: N to the second power=49 n=?

Answer by algebrahouse.com(960)   (Show Source):
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n² = 49
n = ±√49 {took the square root of both sides}
n = 7 or -7 {evaluated the square root of 49}
www.algebrahouse.com

Question 584220: simplify
Answer by JBarnum(1961)   (Show Source):
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use cross cancelation reduction

Question 581552: 6z-6=12-9z
Answer by dfrazzetto(205)   (Show Source):
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6z-6=12-9z
15z = 18
z = 18/15 = 6/5

Question 581545: how do you solve 4(5x-2)=92

Answer by dfrazzetto(205)   (Show Source):
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4(5x-2)=92

distribute the 4:

20x - 8 = 92

add 8 to both sides:

20x = 100

divide by 20:

x = 5

CHECK:
4(5(5) - 2) = 92
4(25-2) = 92
4(23) = 92 √

Question 580624: Prime factors for Mathematics have been likened to atoms in the study of matter. By comparison, discuss the importance of this.
Answer by dfrazzetto(205)   (Show Source):
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Just as atoms are the "fundamental" building blocks of matter, so too are prime factors the building blocks of all real numbers
They are the smallest integer factors that any given number can be reduced to.

Question 578828: I have three factors. the sum of my factors is 7. What am I?
Answer by edjones(7481)   (Show Source):
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12
factors are 2, 2, 3

Question 573634: What is a 4 digit natural number divisible by 4 but not by 8?
Found 2 solutions by Alan3354, richard1234:
Answer by Alan3354(23874)   (Show Source):
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What is a 4 digit natural number divisible by 4 but not by 8?
----------
9996, if you mean evenly divisible.
9996/8 = 1249.5

Answer by richard1234(5244)   (Show Source):
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1004

Question 572038: 406/6
Answer by bulls400(11)   (Show Source):
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67 2/3

Question 572591: How to calculate the Multiplicative inverse modulo.
like how to calculate 7inverse mod (20). It will be helpful, if the solution or the method is calculative
Answer by stanbon(50467)   (Show Source):
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How to calculate the Multiplicative inverse modulo.
like how to calculate 7 inverse mod (20).
---
1/7 = (1+20x)/7 = 3 when x = 1
----
1/7 = 3 (mod 20)
====================
Cheers,
Stan H.
============

Question 567583: The positive integer N has exactly eight different positive integral factors. Two of these factors are 15 and 21. What is the value of N?
Answer by richard1234(5244)   (Show Source):
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Note that 15 = 3*5 and 21 = 3*7. The LCM of 15 and 21 is 3*5*7 = 105, however this already has eight factors (use the algorithm for finding the # of factors of a number, add 1 to each exponent, multiply). Therefore N = 105.

Question 566460: Hey I am having problems with this just Prime Numbers what are all the prime numbers?
Answer by richard1234(5244)   (Show Source):
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Any integer greater than 1 that is only divisible by 1 and itself.

E.g. 2, 3, 5, 7, 11, 13, 17, ..., infinitely many primes.

Question 565188: i am a multiple of 5. two is not one of my factors. i am not prime and i am not square. i am less than 30. what number am i?

Answer by jim_thompson5910(23764)   (Show Source):
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Multiples of 5 less than 30: 0, 5, 10, 15, 20, 25

Since 2 is NOT a factor, we cross out 0, 10, and 20

So we're left with: 5, 15, 25

The number is NOT prime, so 5 is out. The number is NOT a square, so 25 is also out.

So the answer is 15.

Question 563311: Your task is to find two different numbers that sum to 1000 . One number must be a multiple of 19 and the number must be a multiple of 47 . Show or explain how you got your answer .
Answer by richard1234(5244)   (Show Source):
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19a + 47b = 1000

Solving for a,

We want 1000-47b to be congruent to 0 mod 19. Since 1000 is 12 mod 19, 47b must be 12 mod 19. Also, 47 is congruent to 9 (mod 19), so we have

If b = 14, then 9b = 126 which is 12 mod 19, so this works. Hence 47b = 47*14 = 658, 19*18 = 342. If you know Diophantine equations you can use this solution to generate infinitely many solutions.

Question 560364: A square has area 18225 square feet. What is the perimeter of the square. Could you please how me a step by step how to complete this problem
Answer by Earlsdon(6215)   (Show Source):
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If the area of a square is 18225,the length of one side (s) is:

The perimeter p is 4(s), so...

Question 558849: REMOVED FOR LANGUAGE
Answer by KMST(893)   (Show Source):
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DO NOT POST SUCH FILTH IN HERE.

Question 557800: What is the smallest number that can be divided by all the numbers 1 to 10?
Answer by Edwin McCravy(7325)   (Show Source):
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Let the number be N 1 = 1 so N must have factor 1 (every integer has factor 1, so we can ignore 1) 2 = 2 so N must have factor 2 3 = 3 so N must have factor 3 4 = 2*2 so N must have factor 2*2 5 = 5 so N must have factor 5 6 = 2*3 so N must have factor 2*3 7 = 7 so N must have factor 7 8 = 2*2*2 so N must have factor 2*2*2 9 = 3*3 so N must have factor 3*3 10 = 2*5 so N must have factor 2*5 So the smallest N can be and have at least as many of every factor above is the product of three 2's, two 3's, one 5, and one 7, which means N = 2*2*2*3*3*5*7 = 2520 Edwin

Question 553709: How to solve this problem 5\9of450
Answer by prateekagrawal(49)   (Show Source):
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just we need to multiply 5/9 to 450.
so,
5/9 of 450=
so, 250 is the answer.

Question 553497: Find the greatest common divisor of 96 and 54.
Find the least common multiple of 30 and 42.
Answer by nyc_function(2705)   (Show Source):
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Greatest Common Factor: The highest number that divides exactly into two or more numbers. It is the "greatest" thing for simplifying fractions!
What is the BIGGEST number that evenly divides 54 and 96?

How about 6?

54/6 = 9

96/6 = 16

Yes, that's correct.

The least common multiple (LCM) of two numbers is the smallest number (not including zero) that is a multiple of both.

Multiples of 3O: 30, 60, 90, 120, 150, 180, 210

Multiples of 42: 42, 84, 126, 168, 210

On the list provided, what number is found in both?

The number 210 is found in both.

Then 210 is the LCM.

Question 513810: FIND AL NUMBERS LESS THAN 50THAT ARE PRIMES AND HAVE A REMAINDER OF 3 WHEN THEY ARE DIVIDED BY 7.
Found 2 solutions by MathTherapy, JBarnum:
Answer by MathTherapy(647)   (Show Source):
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FIND AL NUMBERS LESS THAN 50THAT ARE PRIMES AND HAVE A REMAINDER OF 3 WHEN THEY ARE DIVIDED BY 7.

Multiply all numbers, from 1 to 6, by 7, and than add 3 to see if the result is a prime number.

N.B. Multiplying 7 by 7, and adding 3 would yield 52 (7*7 + 3), which would be too large

_7 * 1 = 7 + 3 = 10 (NOT PRIME)
7 * 2 = 14 + 3 = 17 (PRIME)
7 * 3 = 21 + 3 = 24 (NOT PRIME)
7 * 4 = 28 + 3 = 31 (PRIME)
7 * 5 = 35 + 3 = 38 (NOT PRIME)
7 * 6 = 42 + 3 = 45 (NOT PRIME)
7 * 7 = 49 + 3 = 52 (NOT PRIME, and as stated above, 52 > 50)

Therefore, the numbers being sought are and

Send comments and “thank-yous” to “D” at MathMadEzy@aol.com

Answer by JBarnum(1961)   (Show Source):
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list all primes up to 50
1 2 3 5 7 11 23 29 31 37 41 43 47
1-7 easily cancel out
11 23 29 31 37 41 43 47
./.../.../...(31).../.../.../.../
7*4=28 28+3=31

Question 493228: What are the two smallest numbers with only 10 divisors?
Answer by JBarnum(1961)   (Show Source):
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100
1 100
2 50
4 25
5 20
10 10
__________
90
1 90
2 45
3 30
5 18
9 10
_____________
80
1 80
2 40
4 20
5 16
8 10

____________
80,90 the 2 lowest i found

Question 539852: How many integers between 1 and 200 are multiples of both 3 and 5 but not of either 4 or 7?
Answer by JBarnum(1961)   (Show Source):
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use the 3 trick when counting with 5s starting at 15 1+5=6 so its divisible by 3
15,30,45,60,75,90,105,120,135,150,165,180,195
above are the possibilities for 3 n 5 but some are divisible by 4 and 7 so go through and knock those out
15,30,45,,75,90,,,135,150,165,,195