Tutors Answer Your Questions about Divisibility and Prime Numbers (FREE)
Question 241454: what numbers are 6,925 divisible by?
Answer by Alan3354(6097)   (Show Source):
Question 241059: What is the sum of all the one-digit prime numbers?
A)15 B) 16 C) 17 D) 18
Answer by checkley77(7072)  (Show Source):
Question 240785: 11.2 divided by -2.8 = ____
Answer by Alan3354(6097) (Show Source):
Question 240488: What are the correct steps to answer this problem 52 divided by 323.18
Answer by rfer(2688) (Show Source):
Question 240490: A circular coffee table has a diameter of 5 ft. What will it cost to have the top refinished if the company chares $3 per square foot for the refinishing
Found 2 solutions by unlockmath, edjones:
Answer by unlockmath(121)  (Show Source):
You can put this solution on YOUR website!Hello,
Since we know the area of a circle is pi radius squared we can find the number of square feet by plugging in the numbers.
Area = 3.14 (2.5)^2 Do the math. (Note: It gives the diameter as 5 ft so the radius is 2.5 ft)
Area=19.625 square feet Now multiply $3 and we get:
Rounded off it comes to $58.88
RJ Toftness
www.math-unlock.com
Answer by edjones(3299)  (Show Source):
Question 240094: What is the smallest positive number divisible by four primes?
Answer by solver91311(5072)  (Show Source):
You can put this solution on YOUR website!
I presume you mean "what is the smallest positive number divisible by four different primes?"
If that is the case, you are looking for the product of the four smallest primes. The smallest prime is 2. 1 is not prime, by definition.
On the other hand, if your "four primes" can all be the same prime, then the answer is 
John
Question 239702: 1st penny = 1 gumballs
2nd penny = 4 gumballs
3rd penny = 10 gumballs
4th penny = 19 gumballs
5th penny = 31 gumballs
6th penny = ? gumballs
7th penny = ? gumballs
8th penny = ? gumballs
Answer by College Student(217)  (Show Source):
You can put this solution on YOUR website!The answers are: 46, 64, and 85.
.
When you paid the 2nd penny, you got 3 more gumballs.
When you paid the 3rd penny, you got 6 more gumballs.
When you paid the 4th penny, you got 9 more gumballs.
When you paid the 5th penny, you got 12 more gumballs.
So it is fair to expect that:
when you pay the 6th penny, you'll get 15 more than what you already have.
when you pay the 7th penny, you'll get 18 more than what you already have.
when you pay the 8th penny, you'll get 21 more than what you already have.
Question 238732: List the five (5) 2-digit primes each of whose digits is NOT prime
Answer by Alan3354(6097) (Show Source):
You can put this solution on YOUR website!89
----
The digits have to be 4, 6, 8 or 9.
------------
2, 3, 5 and 7 are prime numbers
-----------
Here are the first few prime numbers:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
------------
Only 89 fits. Who said there are 5?
Question 238469: 93xy is divisible by 72. then,
x=
y=
when this number is divided by 6,8,and 4, a there's always a reminder of 1?what is the least value of this number?
Answer by jsmallt9(594) (Show Source):
You can put this solution on YOUR website!There is an error in the problem. Since 72 is divisible by 6, 8 and 4 and since 93xy is divisible by 72, 93xy must be divisible by 6, 8 and 4. There will be no (or zero) remainders.
Please correct and repost.
Question 238461: The total number of prime divisors of the natural number 2002 is:
(A) 2 (B) 3 (C) 4 (D) 5 (E) None of the above.
Answer by jsmallt9(594) (Show Source):
You can put this solution on YOUR website!Since 2002 is even it is clearly divisible by 2:
2002 = 2 * 1001
Now we look for factors of 1001, if any. Since it is odd, 2 will not divide evenly. And since the digits add up to 2 (1+0+0+1 = 2), it is not divisible by 3. And since 1001 does not end in a 5 or 0, it is not divisible by 5. The next prime is 7. There is no rule for divisibility by 7 so we just have to divide by 7 to see if it works. And it does! Now we have:
2002 = 2 * 7 * 143
Now we look for prime factors of 143, if any. 7 does not work on this so we move on to the next prime which is 11. (If there is a rule for divisibility by 11 I am not aware of it.) Dividing 143 by 11 we find that it divides evenly giving us:
2002 = 2 * 7 * 11 * 13
The four factors are all prime so your answer is (C).
Question 238468: 93xy is divisible by 72. then,
x=
y=
Answer by vleith(1977) (Show Source):
You can put this solution on YOUR website!Find the prime factors of both constant terms
Factors of 93 --> 3,31
Factors of 72 --> 3,3,2,2,2
For the left side to be divisible with a remainder of 0, the left side must include all the factors of 72. Since 93 already has a factor of 3, you need x and y to include the other factors of 72 (the second 3 and all of the 2s)
So the product of x and y must include 3, 2, 2, 2
There are an infinite number of x and y that can do that.
One pair would be x = 3 y=8
Another is x=6 y=4
Basically ANY combination of xy that is a multiple of 24 will do the trick.
Make sense?
Question 238190: If a three digit number is divided by 5 or 6,the remainder is 1 in each case. What is the least such three digit number?
Answer by solver91311(5072) (Show Source):
You can put this solution on YOUR website!
First find the smallest 3-digit number that is evenly divisible by both 5 and 6. Then add 1.
Since 5 is prime and not a factor of 6, the smallest positive integer evenly divisible by both 5 and 6 must be 5 times 6 or 30. So the question becomes what is the smallest 3-digit number evenly divisible by 30? 1, 2, and 3 times 30 each produce a 2-digit number. Hence the smallest integer multiplier of 30 that produces a 3-digit product is 4, and the smallest 3-digit integer divisible by 30 is 120. Hence, 120 is the smallest 3-digit integer evenly divisible by both 5 and 6. And finally, 121 is the smallest 3-digit integer when divided by either 5 or 6 leaves a remainder of 1 in each case.
John
Question 238106: how many 4 digit prime numbers are there?
Answer by solver91311(5072) (Show Source):
You can put this solution on YOUR website!
Here are two lists:
Prime numbers less than 1000: http://www.factmonster.com/math/numbers/prime.html
Prime numbers less than 10000: http://aleph0.clarku.edu/~djoyce/numbers/primes.html
The answer to your question is the number of items in the second list minus the number of items in the first list. Fortunately, both of these lists are arranged in rows and columns, so it should be fairly simple to count the number of elements in each.
Good luck.
John
Question 237034: what is the equation for36 and 48 prime fatorarization
Answer by solver91311(5072) (Show Source):
You can put this solution on YOUR website!
I don't know what a "fatorarization" is. Be that as it may, there is no "equation" for performing the process of determining a prime factorization.
Step 1: Use your calculator to determine the square root of the number you want to factor. You only need to know the integer part, so just jot that down somewhere.
Step 2:
Divide by 2 as many times as you can. That is, divide your original number by 2, then divide the quotient by 2, then divide that quotient by 2, etc. Every time you can divide by 2 means there is a prime factor of 2.
Once you get to an odd quotient, check to see if you can divide by 3. You can divide by 3 if the sum of the digits is divisible by 3. For example 111 is divisible by 3, while 113 is not. If you can divide by 3, do it, and keep doing it as many times as possible. Every time you can divide by 3, you have a prime factor of 3 in your original number.
Once you can no longer divide by 3, try 5, then 7, then 11, and so on using every successive prime number as a divisor. The only question is how high in the list of prime numbers to you have to go with your trial divisors. The answer to that is found back at step 1. Once the next prime number in your list is larger than the square root of the number to be factored, you are done.
Rules for divisibility: Ends in 0 or 5, divisible by 5, but you won't be at this step unless your intermediate quotient ends in some odd number, so look for ends in 5.
http://math.about.com/library/bldivide.htm has a couple of tests for divisibility by 7, but they are so convoluted in my mind that just trying to divide by 7 is faster and easier.
Three digit numbers that are divisible by 11 have the property that the center digit is the sum of the two outer digits. (watch for a possible carry to the hundreds place)
Larger divisors are trial and error. If you have larger numbers to struggle with, I suggest you use one of several prime factorization calculators on the web -- mathwarehouse.com has one
Let's try this process with one of your numbers. 7 times 7 is 49, so you know that the square root of 48 is a little smaller than 7 and that the largest trial divisor you would ever have to use is 5. If you get that far, that is.
Divide by 2, equals 24. First factor of 2
Divide by 2, equals 12. Second factor of 2
Divide by 2, equals 6. Third factor of 2
Divide by 2, equals 3. Fourth factor of 2
Divide by 2, not an integer, therefore no more factors of 2.
3 is prime, One factor of 3, and you are done.
Hence the prime factorization of 48 is: 
Oh, and yes, spelling counts. It is all about you having the courtesy to take enough time and care in your written communications with someone who is trying to help you -- for free.
John
Question 237000: 3x^2 - 8x + 5 = 0.
Answer by thrasherlax74(11) (Show Source):
Question 234957: The greatest common factor of 72 and X is 3. The least common multiple of 72 and X is 1800. What must be true?
a) X is less than 72
b) X is multiple of 5.
c) X is a multiple of 72.
d) X is a factor of 3.
e) X is even
Answer by jsmallt9(594) (Show Source):
You can put this solution on YOUR website!Probably the easiest way to solve this is to understand the relationship that exists between the LCM and GCF of two numbers. (Important: This only applies to two numbers, not 3, 4, etc.)
If our two numbers are a and b:
LCM = (a*b)/GCF
Putting the numbers (and x) from your problem we get:
which simplifies to:
Solving this (by dividing both sides by 24):
From this we can see that (a), (c), (d) and (e) are not true. Only (b) is true.
Question 234483: what is product of the divisors of 1000 (including 1 and 1000)?
a) 10^24 b) 10^16 c) 10^8 d) 10^30 e) 10^32
Answer by Edwin McCravy(2922) (Show Source):
Question 233431: What is the largest prime factor of ( 30^2 - 7^2 ) ?
Answer by Alan3354(6097) (Show Source):
Question 233402: If n is a whole number then the largest number that n(n+1)(2n+1) is divisible by for all n is
a) 2 b) 6 c) 10 d) 3 e) none
Answer by jsmallt9(594) (Show Source):
You can put this solution on YOUR website!Assuming n must be an integer (or a whole number or a natural number), here the keys to this problem: - All even numbers are divisible by two.
- Whenever integers are multiplied, if one or more of the numbers is even, then product will be even.
- If n is even, n+1 will be odd. Or if n is odd, then n+1 will be even. The point is that either n or n+1 must be an even number.
- So n(n+1)(2n+1) must be even.
- So n(n+1)(2n+1) must be divisible by 2.
The answer is definitely not (e). I cannot find any way to show that n(n+1)(2n+1) is always divisible by anything but 2 so I believe the answer is (a).
Question 232799: What two-digit positive integer is one more than a multiple of 2, 3,4, 5 and 6 ?
Answer by ankor@dixie-net.com(6693)  (Show Source):
You can put this solution on YOUR website!What two-digit positive integer is one more than a multiple of 2, 3,4, 5 and 6 ?
:
Primes representing all these numbers: 2, 2, 3, 5,
:
2 * 2 * 3 * 5 = 60 + 1 = 61
Question 232459: which of the following groups of numbers are all prime numbers?
a)2,3,5,9
b)3,11,23,31
c)2,5,15,19
d)7,17,29,49
Answer by stanbon(26297) (Show Source):
Question 232460: which of the following groups of numbers are all prime numbers?
a)2,3,5,9
b)3,11,23,31
c)2,5,15,19
d)7,17,29,49
Answer by rfer(2688) (Show Source):
Question 232017: Use the following list of numbers for exercises 13 and 15.
0, 1, 15, 19, 23, 31, 49, 55, 59, 87, 91, 97, 103, 105
13. Which of the given numbers are prime?
15. List all the prime numbers between 30 and 50.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Use the following list of numbers for exercises 13 and 15.
0, 1, 15, 19, 23, 31, 49, 55, 59, 87, 91, 97, 103, 105
13. Which of the given numbers are prime?
Ans: 19, 31, 59, 87, 91, 97, 103
----------------------------------------------------------
15. List all the prime numbers between 30 and 50.
Ans: 31, 37, 41, 43, 47
=================================
Cheers,
Stan H.
Question 228978: -5/8 divided by 3/4
Found 2 solutions by rfer, CassieLynn:
Answer by rfer(2688) (Show Source):
Answer by CassieLynn(3) (Show Source):
Question 228338: Word problems over prime factorization or factoring the monomial.
Problem 1:::: You are arranging 70 plants in a rectangular garden with the same number of plants in each row. How many ways can you arrange the garden?
Problem 2:::: A dog kennel groups the dogs in order to determine at what time they will be given a treat. Each group should have the same number of dogs. There are 120 dogs in the kennel. How many groups are possible?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Problem 1:::: You are arranging 70 plants in a rectangular garden with the same number of plants in each row. How many ways can you arrange the garden?
---
70 = 2*5*7
Ans:
1 row of 70
2 rows of 35
5 rows of 14
7 rows of 10
10 rows of 7
14 rows of 5
35 rows of 2
70 rows of 1
---------------------
Problem 2:::: A dog kennel groups the dogs in order to determine at what time they will be given a treat. Each group should have the same number of dogs. There are 120 dogs in the kennel. How many groups are possible?
---
120 = 2*2*2*3*5
1 row of 120
2 rows of 60
4 rows of 30
etc.
=============
Cheers,
Stan H.
Question 228805: What is the first three digit prime number?
Answer by nerdybill(2448)  (Show Source):
Question 227613: write the provided fraction in simplest form.
60a2b/36ab3
Answer by user_dude2008(716) (Show Source):
Question 227612: write the prime factorization of the provided number
504
Answer by user_dude2008(716) (Show Source):
Question 227608: write the provided improper fraction as a mixed number or a whole number
149/143
Answer by user_dude2008(716) (Show Source):
Question 227267: write a 4 digit number that is divisible by both 6 and 9. Explain how you know without dividing that it is divisible by both 6 and 9.
Found 2 solutions by solver91311, Theo:
Answer by solver91311(5072) (Show Source):
You can put this solution on YOUR website!
If the sum of the digits is divisible by 3, then the number is divisible by 3. If the number is even it is divisible by 2. If it is divisible by 2 AND 3, it is divisible by 6. If the sum of the digits is divisible by 9, then the number is divisible by 9 (and also divisible by 3). So you need to make yourself an EVEN 4 digit number where the sum of the digits is divisible by 9.
A number that ends in 2 is even, so let's start with that -- the one's digit is 2. The other three digits must then add up to either 7, 16, or 25.
3222, 1242, 2232, 3132, and lots of others all work.
John
Answer by Theo(675)  (Show Source):
You can put this solution on YOUR website!multiply 60 * 90 to get 3600
this is divisible by 6 and 9 because I created it using factors that contained 6 and 9
3600 / 6 = 600
3600 / 9 = 400
Question 227124: how many two digit prime numbers have a remainder of 2 when divided by 7?
Answer by solver91311(5072) (Show Source):
You can put this solution on YOUR website!
The only way I know how to do this one is to get a list of all the two-digit prime numbers (go to: http://www.math.utah.edu/~pa/math/primelist.html -- they are all in the first row of the data presented). Then divide each of them by 7 and count the ones that have a remainder of 2. If you are using your calculator, you will get a decimal fraction when you divide by 7 (you are only dividing into primes, so none of them will come out even). If the decimal fraction begins .285714, then you know you have a remainder of 2. This isn't as onerous as it seems, since there are only 21 two-digit primes.
John
Question 226910: is 450 divisible by 5, 9, 10
Answer by Alan3354(6097) (Show Source):
Question 226645: 5= n dived by 6
Answer by rfer(2688) (Show Source):
Question 226491: dividing polynomials
Answer by solver91311(5072) (Show Source):
You can put this solution on YOUR website!
...is a truly fascinating subject. Was there a specific problem with which you need help? If you just want to review the subject in general, PurpleMath has a very good treatise on the subject at:
http://www.purplemath.com/modules/polydiv2.htm
Wikipedia has a good discussion that contains both polynomial long division and synthetic division (you need to know both)
http://en.wikipedia.org/wiki/Polynomial_long_division
John
Question 226428: What is the least number greater than 1,000 divisible by 10?
Answer by jim_thompson5910(13794) (Show Source):
You can put this solution on YOUR website!We know that 1,000 is divisible by 10 since the last digit is 0. So just add 10 to this number to get 1,010. So 1,010 is the smallest number divisible by 10 that happens to be greater than 1,000 (since the last digit is also 0).
Note: this is like saying that 4 is divisible by 2 and 4+2=6 is also divisible by 2.
Question 226094: find the prime factorization of 125
Answer by checkley77(7072) (Show Source):
Question 225540: I am having a hard time in Math class, can someone please help me..
X2-X=41
Select some numbers for X, substiute them in formula, and see if prime number occur. Try to find a number for X that when subsituted in the formula yields a composite number. I must substitute at least 5 numbers; 0 (zero), 2 even and 2 odd into the equation for x. I have to show all stepsThe symbols ^ and * should not be used I am so lost can someone please help me..
Answer by Alan3354(6097) (Show Source):
Question 225297: What is the greatest prime you must consider to test whether 503 is prime?
I tried the smaller numbers, 2, 3, 5, 7 and then looked at a prime chart and it said it was prime, but how do I test it and choose the greatest prime?
Answer by ankor@dixie-net.com(6693) (Show Source):
You can put this solution on YOUR website!What is the greatest prime you must consider to test whether 503 is prime?
:
Find the square root of 503 = 22.47....
You know the greatest prime (if it has one) has to be less than 22
that would be 19, you can check it with all primes less than 19
Question 223547: What is the prime factorization of 999 to the 10th power?
Please help! :)
Answer by jsmallt9(594) (Show Source):
You can put this solution on YOUR website!It would be a huge waste of time and effort and make the problem extremely difficult to actually try to raise 999 to the 10th power. What we will do is find the prime factorization of 999.
3 will divide into 999:
3 will divide into 333:
3 will divide into 111:
37 is prime so we have all our prime factors. We just need to simplify:
Question 223539: What is the greatest prime you must consider to test whether 503 is prime? Why?
Please help! :)
Answer by jsmallt9(594) (Show Source):
You can put this solution on YOUR website!To make sense of the answer, let's look at the factors of 24:
1 24
2 12
3 8
4 6
6 4
8 3
12 2
24 1
As you look down this list of pairs of factors you will notice that the last half of the list is the same as the first half only in reverse order. There will be a "halfway" point for any list of factors for any number. If you are checking for divisibility there is no need to check beyond this "halfway" point because any factor found in the upper "half" is paired with a number in the lower "half" (which you would have already found).
So how do you determine the "halfway" point beyond which there is no point searching for more factors? Answer: The square root of the number you are trying to factor. Factors above the square root, if any, will always be paired with a factor below the square root. So there will be no factors above the square root that you would not have already found when checking the factors below the square root.
So for determining the primeness of 503, there is no point in checking prime factors above  (which is approximately 22.4). The last prime before 22.4 is 19. So the only factors to check for 503 are: 2, 3, 5, 7, 11, 13, 17, 19. If none of these divide into 503, then 503 is prime.
Question 224166: If a man buys two products by the pound and pays $0.26 for one and $0.19 for the other. His total amount payed is $10,530.03 and his total Weight is 42,110 pounds. What is the weight of product bought at $0.26 and what is the weight of product bought at $0.19 ??? Please help!
Found 2 solutions by checkley77, stanbon:
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!.26x+.19(42,110-x)=10,530.03
.26x+8,000.9-.19x=10,530.03
.07x=10,530.03-8,000.9
07x=2,529.13
x=2,529.13/.07
x=36,130.43 pounds of $.26 product.
42,110-36,130.43=5,979.57 pounds of $.19 product.
Proof:
.26*36,130.43+.19*5.979.57=10,530.03
9,393.91+1.136.12=10,530.03
10,530.03=10,530.03
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!If a man buys two products by the pound and pays $0.26 for one and $0.19 for the other. His total amount payed is $10,530.03 and his total Weight is 42,110 pounds. What is the weight of product bought at $0.26 and what is the weight of product bought at $0.19 ???
------------------------------
Weight Equation: x + y = 42110
Value equation: 0.26x + 0.19y = 10530.03
------------------------------
Multiply thru the Value Eq. by 100
Multiply thru the Weight Eq. by 26
---
26x + 19y = 1053003
26x + 26y = 26*42110
---------------------
Subtract the 1st from the 2nd and solve for "y":
7y = 41857
y = 5979.5714 lbs (Weight of product bought at $0.19)
==========================================================
Cheers,
Stan H.
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