Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If the remainder when $x$ is divided by 5 equals to the remainder when x is divided by 4 then x could be any of the following a)20 , b)21 , c)22 , d)23 e)24 (Ans=(e)24)

Now I could only tell till $x-5q=x-4p$.

Using equations how can I predict the values of x ?

share|improve this question
5  
Any of (a), (b), (c), and (d) work: leave remainders of 0, 1, 2, and 3 respectively. 24 is not a valid answer, because the remainder when divided by 4 is $0$, but the remainder when divided by 5 is $4$, and $0\neq 4$. –  Arturo Magidin Jul 17 '12 at 4:30
add comment

3 Answers 3

You are better off using congruences, not equations. In any case, the solution is only well-defined up to a multiple of $20$: given any valid solution, add $20$ to it ($20$ is the lcm of $4$ and $5$), and the result is still a solution. Subtract $20$ and the same is true.

Write $x = 5k+r$, with $0\leq r\lt 5$. Then $x = 5k+r = 4k + (k+r)$.

What we know is that the remainder of dividing $x$ by $4$ must be $r$ as well. That means that the remainder of dividing $k+r$ by $4$ must be $r$; this means that $k$ itself must be a multiple of $4$ and $r$ must be smaller than $4$. That is, $k=4\ell$ for some $\ell$, and $0\leq r\lt 4$. So $x = 5k+r = 5(4\ell)+r = 20\ell + r$, with $0\leq r\lt 4$. Thus, $20$, $21$, $22$, and $23$ all work (as do $0$, $1$, $2$, and $3$, $40$, $41$, $42$, $43$, $160$, $161$, $162$, $163$, etc). But 24 does not work (remainder when dividing by $5$ is too large).

This is much easier to do with congruences, though.

share|improve this answer
add comment

Hint $\ $ The common remainder $\rm\,r \le 3,\,$ being a remainder mod $4.\,$ Since both $4$ and $5$ divide $\rm\:x-r\:$ so too does their lcm $20.\:$ Thus $\rm\:x = 20\,k + r,\:$ for $\rm\:0\le r\le 3,\:$ hence $\rm\:x\ne 24.$

Remark $\ $ This is a trivial constant case of CRT = Chinese Remainder. If $\rm\:x\equiv r\pmod{n_k}$ has a constant residue $\rm\,r\,$ for all moduli, then, as above, we infer $\rm\:x \equiv r\pmod n,\,$ for $\rm\:n = lcm\,\{n_k\}.\,$ Further, if $\rm\,r\,$ denotes a remainder (vs. congruence class), $ $ so $\rm\, 0\le r < n_k,\,$ then $\rm\: x = j\,m + r,\:$ where $\rm\:0\le r < n_0,\,$ for $\rm\:n_0 = min\, \{n_k\},\,$ and some $\rm\,j\in\Bbb Z.$

So that's how to generally "predict the values". However, for test problems like this, it is usually quicker to brute-force search for the non-solution, since each test takes at most a few seconds.

share|improve this answer
add comment

If $x=5a+r$ with $0\le r\le4$ and $x=4b+r$ with $0\le r\le3$ then $5a=4b$ so $a=4c$ and $b=5c$ for some $c$, and we have $$x=20c+r{\rm\quad with\quad}0\le r\le3$$ and that equation tells you which values of $x$ work and which ones don't.

share|improve this answer
    
I lost you at $a=4c$ How did you get c here and after that ? –  Rajeshwar Jul 17 '12 at 4:49
    
$5a=4b$, so $5a$ is a multiple of 4, but 5 is relatively prime to 4, so $a$ is a multiple of 4, so $a=4c$ for some $c$. Similarly for $b=5d$ for some $d$. But $5a=4b$, so $20c=20d$, so $c=d$. –  Gerry Myerson Jul 17 '12 at 5:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.