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castortr0y [4]
3 years ago
7

A teacher that likes two different numbers p and q and states that p plus q equals 0 which statement could be true about these t

wo numbers
(A)both numbers are positive
(B)both numbers are negative
(C)one number is zero in the other is positive
(D)one number is positive and the other is negative
​
Mathematics
1 answer:
pychu [463]3 years ago
7 0
D, one number is positive and the other is negative. The only way it could equal 0 is to cross the other out. An example is, -5+5=0
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In base $10$, the number $2013$ ends in the digit $3$. in base $9$, on the other hand, the same number is written as $(2676)_{9}
skelet666 [1.2K]
Note that

2013_{10}=2(10)^3+1(10)^1+3(10)^0

so that

\dfrac{2013}{10}=200+1+\dfrac3{10}

i.e. the remainder upon dividing 2013 (in base 10) by 3 is 3. The point is that any base-b representation of 2013_{10} will end in the digit 3 whenever division of 2013 by b leaves a remainder of 3.

First, we require that b\ge4, because any smaller base simply won't have 3 as a possible digit.

Second, we clearly can't have b\ge2014, because any value of b beyond that point will have 2013 as its first digit. That is, in base 2014, for instance, if we separate the digits of numbers by colons, we would simply have 2013_{10}=0:0:0:2013_{2014} (the 0s aren't necessary here, but only used for emphasizing that 2013 would be its own digit, regardless of how we represent digits outside of 0-9).

Third, we can't have b=2013 because 2013 divides itself and has no remainder. That is, 2013_{10}=10_{2013}.

So we've reduced the possible domain of solutions from all positive integers to just those lying within 4\le b\le2012.

Now, in terms of modular arithmetic, we're essentially solving the following equivalence for b:

2013\equiv3\pmod b

which, if you're not familiar with the notation or notion of modular arithmetic, means exactly "2013 gives a remainder of 3 when divided by b". It's equivalent to saying "there exists an integer n such that 2013=bn+3.

From this equation, we have 2010=bn. Now, we know both b,n must be integers, so we need only find the factors of 2010. These are

1, 2, 3, 5, 6, 10, 15, 30, 67, 134, 201, 335, 402, 670, 1005, 2010

So we have 16 total possible choices for b,n. But we're omitting 1, 2, and 3 from the solution set, which means there are 13 base-b representations of the base-10 number 2013_{10} such that the last digit 2013_b is 3. They are

2013_{10}=31,023_5
2013_{10}=13,153_6
2013_{10}=2013_{10}
2013_{10}=8:14:3_{15}
2013_{10}=2:7:3_{30}
2013_{10}=30:3_{67}
2013_{10}=15:3_{134}
2013_{10}=10:3_{201}
2013_{10}=6:3_{335}
2013_{10}=5:3_{402}
2013_{10}=3:3_{670}
2013_{10}=2:3_{1005}
2013_{10}=1:3_{2010}

- - -

Added a link to some code that verifies this solution. In case you're not familiar with Mathematica or the Wolfram Language, in pseudocode we are doing:

1. Start with a placeholder list called "numbers", consisting of all 0s, of length 2010. We know 4\le b\le2013, which contains 2010 possible integers.
2. Iterate over b:
2a. Start with b=4.
2b. Ensure b doesn't exceed 2013.
2c. After each iteration, increment b by 1; in other words, after each step, increase b by 1 and use this as a new value of b.
2d. Check if the last digit of the base-b representation of 2013 ends is a 3. In the WL, IntegerDigits[x, n] gives a list of the digits of x in base-n. We only want the last digit, so we take Last of that. Then we check the criterion that this value is exactly 3 (===).
2d1. If the last digit is 3, set the (b-3)-th element of "numbers" to be a list consisting of the base b and the list of digits of 2013 in that base.
2d2. Otherwise, do nothing, so that the (b-3)-th element of "numbers" remains a 0.
3. When we're done with that, remove all cases of 0 from "numbers" (DeleteCases). Call this cleaned list, "results".
4. Display "results" in Grid form.
5. Check the length of "results".
7 0
3 years ago
Albert is traveling 60 mph in his car. After 3 hours, how far will he have traveled? 
Dahasolnce [82]
He will have traveled 180 miles, though with real situations the distance may vary.
6 0
3 years ago
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Solve for x<br><br> 2(x-3)+5=4-7(2x-1)<br><br> A. -1/10<br> B. 3/4<br> C. 1/2<br> D. 1/16
Stolb23 [73]
Hello 
<span>2(x-3)+5=4-7(2x-1)
2x-6+5 = 4 -14x+7
2x+14x =6-5+4+7
16x=12
x=12/16
x=(4×3)/(4×4)
x=3/4 (simplified by 4)
answer B</span>
6 0
3 years ago
Read 2 more answers
7. Which of the following is an example of a
Klio2033 [76]
By looking at the answer a is eliminated by default since it is not between 6 and 7 which also eliminates e. now you’re left with b and c and you have to remember the definition of an irrational number (in short, it’s a number that never ends) which would leave you with c
3 0
3 years ago
Read 2 more answers
4(5f + 9)
Marina86 [1]

Answer: hopes this helps i spent a lot of time on it !!!:)lol

2 0 + 3 6

− 8 − 1 2

2 0 + 1 0

1 0 ℎ − 1 5

− 1 2 − 1 4

5 6 − 6 3

8 − 1 8

− 8 + 1 2

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
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