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Please help me out with this. Not getting any approach to solve this.

Let 'k' be a +ve integer such that k+4 is divisible by 7. Then the smallest +ve integer 'n' , greater than 2, such that k+2n is divisible by 7 equals :

PS: I am confused b/w "perfectly divisible" and "divisible" rather in this case because I did a question from my same text book which is previous to this one being asked. It said "perfectly divisible". Although the question was different from this one yet I got this feeling that in "perfectly divisible" you get no remainder so is this case a different one or both the statements are used "interchangeably"

May be I am leaving out some mere details or some basic maths but its okay...often we do mistakes...so please solve this with a better explanation of the concept and especially the PS section. Thanks in advance...:)

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  • $\begingroup$ What's a "+ve integer"? $\endgroup$ – fleablood Mar 3 '16 at 23:15
  • $\begingroup$ @fleablood Lol, that stands for "positive". It comes as "positive-integer". $\endgroup$ – Shivam Tripathi Mar 3 '16 at 23:16
  • $\begingroup$ Anyway, the answer is the same as asking what is the smallest $n>2$ such that $2(n-2)$ is divisible by $7$. And this is $n=9$. $\endgroup$ – Crostul Mar 3 '16 at 23:19
  • $\begingroup$ @Crostul Yes bro, your answer is correct. Can you elaborate it, I mean how did you break it? $\endgroup$ – Shivam Tripathi Mar 3 '16 at 23:23
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$k + 4 = m7$ for some integer $m$.

$k + 2n = j7$ for some integer $j$.

Solve for $n$.

$2n = j7 - k = j7 - m7+4 = (j-m)7 + 4$

$n = \frac{j-m}{2}7 + 2$

what is the smallest such positive number greater than 2. Clearly $n =9$

Then $k + 2n = k+4$ is divisible by 7.

If you are familiar with modular notation.

$k + 4 \equiv 0 \mod 7 \iff k \equiv -4 \equiv 3 \mod 7$.

$k + 2n \equiv 2n + 3 \mod 7 \equiv 0$

$2n \equiv -3 \equiv 4 \mod 7$

$n \equiv 2 \mod 7 \equiv 2 +7 = 9 \mod 7$.

Smallest such positive $n>2$ is $9$.

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"perfectly divisible" and "divisible" probably doesn't mean anything different. "divisible" implies divisible by an integer with an integer result with no remainder. "perfectly divisible" is probably simply used to rules out non-integer results.

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  • $\begingroup$ Correct answer is '9' buddy. $\endgroup$ – Shivam Tripathi Mar 3 '16 at 23:29
  • $\begingroup$ Oh, you're right. I didn't see the n > 2 requirement. $\endgroup$ – fleablood Mar 3 '16 at 23:30
  • $\begingroup$ Same thing though. n = 2 is smallest positive integer. Just add 7 to get the next one. $\endgroup$ – fleablood Mar 3 '16 at 23:31
  • $\begingroup$ Yo..thanx bro..I really appreciate it. I think there's no need for modular notation though...Well thanx once again..:) $\endgroup$ – Shivam Tripathi Mar 3 '16 at 23:38
  • $\begingroup$ Well, modular notation, once you know it, is usually considered easier and more basic than $k + 4 = 7m$. We don't give a flip about m or j or (m-j) or the restriction that (m-j) is even so why should we bother writing them? Modular arithmetic is much cleaner. $k + 4 = 0 (mod 7)$. End of story. $\endgroup$ – fleablood Mar 3 '16 at 23:50

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