Tutors Answer Your Questions about Divisibility and Prime Numbers (FREE)
Question 84158: The Prime numbers between successive gaps of ten numbers are:
2, 3, 5, and 7.
11, 13, 17, and 19.
23 and 29.
31 and 37.
41, 43, 47, and (not 49 - square of 7)
53 and 59.
61 and 67.
71, 73, and 79 and (not 77)
83 and 89.
91 and 97.
Can you see a pattern forming? Despite of this I was not able to formulate a simple and general expression so that I could tell whether or not a number is prime by simply looking at the number and applying the above sequence. Can you help? Please! I am very excited about this project. It is my own thought and has not been assigned to me as a homework.
Thankyou.
Click here to see answer by rapaljer(4671)  |
Question 86742: I need help with the following problem.
Someone spilled ink on a bill for 36 sweatshirts. If only the first and last digits were covered and the other three digits were, in order, 8, 3, 9 as in
?83.9?, how much did each cost?
I do not have a clue how to do this problem. Please help.
Thank you for your time, it is appreciated.
Sincerely,
R. Franke
Click here to see answer by Edwin McCravy(20054)  |
Question 86742: I need help with the following problem.
Someone spilled ink on a bill for 36 sweatshirts. If only the first and last digits were covered and the other three digits were, in order, 8, 3, 9 as in
?83.9?, how much did each cost?
I do not have a clue how to do this problem. Please help.
Thank you for your time, it is appreciated.
Sincerely,
R. Franke
Click here to see answer by ankor@dixie-net.com(22740)  |
Question 86869: I need help with this problem. The problem states:
Two digits of this number were erased: 273*49*5. However we know that 9 and 11 divide the number. What is it?
What I have done so far:
2+7+3+*+4+9+*+5 = 30+*+*
The only number divisible by both 9 and 11 is 99. However, to make the sum of this number equal 99 the two missing digits would have to be 35 and 34. Would these numbers be correct? I have tried other single digit numbers and haven't found a sum divided by both 9 and 11.
Your help would be appreciated. Thanks.
Sincerely
R. Franke
Click here to see answer by ankor@dixie-net.com(22740)  |
Question 86740: I need help with this problem. The problem states:
Two digits of this number were erased: 273*49*5. However we know that 9 and 11 divide the number. What is it?
What I have done so far:
2+7+3+*+4+9+*+5 = 30+*+*
The only number divisible by both 9 and 11 is 99. However, to make the sum of this number equal 99 the two digits would have to be 35 and 34. I am not sure if these numbers qualify? I tried other single numbers and haven't found a sum divided by both 9 and 11.
Your help would be most appreciated. Thanks.
Sincerely,
R. Franke
Click here to see answer by scott8148(6628)  |
Question 88232: Hi there,
I have to express my answer in simplest form after adding this problem. The problem reads like this: 6/3x-2 + 3/2x.
I'm not even really sure what to do, so any help you can give me would be greatly appreciated!
Thank you so much!!
Click here to see answer by longjonsilver(2297)  |
Question 95663: When using prime factorization method for finding the least common multiples, why are only certain prime number used instead of all prime numbers? How will I know what prime numbers to use?
6= 2.3
8= 2.2.2=2
LCM=2.2.2.3=24
What tells me to only use three of the four 2s?
Click here to see answer by stanbon(75887) |
Question 97845: Can you take me step by step in helping me with the prime factorization for the numbers 34, 37, 67, 123, and 71? I would love to learn how to do theses so that I may understand when testing time comes around.
Thank you.
Click here to see answer by coyote(32) |
Question 97923: Prove that n^3 + (n+1)^3 + (n+2)^3 is divisible by 9 for all n in Natural numbers. I have tried, let mathematical induction, let P(n) be the statement, then then n^9 - n is divisible by 9. I'm stuck...
Thanks
Click here to see answer by aaaaaaaa(138) |
Question 97923: Prove that n^3 + (n+1)^3 + (n+2)^3 is divisible by 9 for all n in Natural numbers. I have tried, let mathematical induction, let P(n) be the statement, then then n^9 - n is divisible by 9. I'm stuck...
Thanks
Click here to see answer by mathslover(157)  |
Question 103551: Ok, I'm stumped. Divide the rational expressions.
x^4-x^3+x^2+x / x^3-4x^2+x-4 (these are the numerators)
2x^3+2x^2+x+1 / 2x^3-8x^2+x-4 (these are the denominators)
Here's what I have.
X(x^3-x^2+x)* 2x^2(x-4)+(x-4) nuerators
2x^2(x+1)(x+1) * x^2 (x-4)(x-4) denominators
now I'm stumped on factoring the first numerator
Click here to see answer by jim_thompson5910(35256) |
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Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295
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