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When positive integer k is divided by 6 the remainder is 3. Which of t
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12 Jul 2021, 05:47
12 Jul 2021, 05:47
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When positive integer k is divided by 6 the remainder is 3. Which of the following CANNOT be an even integer?
a. k + 1
b. k -11
c. 4k + 2
d. (k-3)/3 +2
e. k/3
a. k + 1
b. k -11
c. 4k + 2
d. (k-3)/3 +2
e. k/3
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Re: When positive integer k is divided by 6 the remainder is 3. Which of t
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12 Jul 2021, 08:21
12 Jul 2021, 08:21
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Carcass wrote:
When positive integer k is divided by 6 the remainder is 3. Which of the following CANNOT be an even integer?
a. k + 1
b. k -11
c. 4k + 2
d. (k-3)/3 +2
e. k/3
a. k + 1
b. k -11
c. 4k + 2
d. (k-3)/3 +2
e. k/3
approach is to TEST values.
When positive integer k is divided by 6 the remainder is 3
So, k could equal 3, 9, 15, 21, etc
let's TEST k = 3
a. 3 + 1 = 4 (EVEN)
b. 3 -11 = -8 (EVEN)
c. 4(3) + 2 = 14 (EVEN)
d. (3-3)/3 +2 = 2 (EVEN)
At this point, we can already see the answer must be E.
Let's check E for "fun"
e. 3/3 = 1 (ODD)
Great!
Answer: E
Cheers,
Brent
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Re: When positive integer k is divided by 6 the remainder is 3. Which of t
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05 Sep 2022, 09:42
05 Sep 2022, 09:42
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When positive integer k is divided by 6 the remainder is 3
Theory: Dividend = Divisor*Quotient + Remainder
k -> Dividend
6 -> Divisor
a -> Quotient (Assume)
3 -> Remainders
=> k = 6*a + 3 = 6a + 3
Which of the following CANNOT be an even integer
Let's take each option and evaluate if it CANNOT be an even integer
a. k + 1
k + 1 = 6a + 3 + 1 = 6a + 4 => EVEN
b. k -11
k - 11 = 6a + 3 - 11 = 6a - 8 => EVEN
c. 4k + 2
4k + 2 = 4*(6a + 3) + 2 = 24a + 20 => EVEN
d. \(\frac{(k-3)}{3}\) + 2
\(\frac{(k-3)}{3}\) + 2 = \(\frac{(6a + 3 -3)}{3}\) + 2 = 2a + 2 => EVEN
e. \(\frac{k}{3}\)
\(\frac{k}{3}\) = \(\frac{6a + 3}{3}\) = 2a + 1 => ODD
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Remainders
Theory: Dividend = Divisor*Quotient + Remainder
k -> Dividend
6 -> Divisor
a -> Quotient (Assume)
3 -> Remainders
=> k = 6*a + 3 = 6a + 3
Which of the following CANNOT be an even integer
Let's take each option and evaluate if it CANNOT be an even integer
a. k + 1
k + 1 = 6a + 3 + 1 = 6a + 4 => EVEN
b. k -11
k - 11 = 6a + 3 - 11 = 6a - 8 => EVEN
c. 4k + 2
4k + 2 = 4*(6a + 3) + 2 = 24a + 20 => EVEN
d. \(\frac{(k-3)}{3}\) + 2
\(\frac{(k-3)}{3}\) + 2 = \(\frac{(6a + 3 -3)}{3}\) + 2 = 2a + 2 => EVEN
e. \(\frac{k}{3}\)
\(\frac{k}{3}\) = \(\frac{6a + 3}{3}\) = 2a + 1 => ODD
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Remainders
