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If three numbers 112, 232, and 400 are each divided by the number D, each of their quotients will have the same remainder R. Find R where R>1

How should I approach this?

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    Hint: consider the differences between pairs of those three numbers. – 2017-01-27
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    Find the possibilities for $D$ first. – 2017-01-27

4 Answers 4

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This is the question that when we subtract $x$ from $112, 232,400$, then we should get three numbers commonly divisible by some $D > x$. That is, $D | 112 - x, 232 - x, 400-x$ for $x < D$.

Before we think twice, I've got the answer: $x=16$ , and the numbers are $96,216,384$, commonly divisible by $24 = D$.

Note that the greatest common divisor of the numbers is $8$: $112 = 8*14, 232=8*29, 400=8*50$ . Hence, if $x$ is a multiple of $8$, then it is seen that the numbers $112-x, 232-x, 400-x$ are divisible by $8$ as well. However, if these all are divisible by $8$, so are their differences: $232-112 = 120, 400-232=168, 400-112 = 288$. These numbers, however, are also divisible by $24$. So we take the guess $24$ and it works out. So $R=16, D=24$ works.

Anything else that must work must be a divisor of $24$, as we have seen. We can rule out some of the possibilities:

1) $D=1,2,4,8$ can be ruled out, because on division by these, the remainder zero is left all the time.

2) On division by $3$, they all leave remainder $1$. So here is one answer, but then $R>1$ is given in the question, so this is not considered.

3) On division by $6$ and $12$, they leave the remainder $4$.

4) On division by $24$, they leave the remainder $16$.

The following is a complete description of the problem's answer.

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    I think there must be a flaw somewhere (or I've misinterpreted the problem). Don't all these numbers leave a remainder of $4$ when divided by $6$? – 2017-01-27
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    @BrianTung That is also true! Thank you for pointing this out. So, in that case, the $R=4$ value actually suffices! I should add this. – 2017-01-27
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    They leave a remainder of $4$ when divided by either $6$ or $12$, and they leave a remainder of $16$ when divided by $24$ (as you pointed out). I wonder if this is the answer the problem intended. – 2017-01-27
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    @BrianTung I am also confused, They way they talk about $R$, it seems unique. Anyway, I have answered the question the way it seems phrased. – 2017-01-27
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    Note that $R$ is supposed to be strictly greater than $1$ (in the OP), so that $D = 3$ is not a solution. – 2017-01-28
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    @BrianTung Ok, but then the rest are solutions, right? – 2017-01-28
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    Yes, as far as I can see. – 2017-01-28
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Hint: $$112 - R \equiv 0 \mod D$$

$$232 - R \equiv 0 \mod D$$

$$400 - R \equiv 0 \mod D$$

$$120 \equiv 0 \mod D$$ $$168 \equiv 0 \mod D$$

Hence $D$ is a common factor of both $120$ and $168$.

$$24 \equiv 0 \mod D$$

$D \in \{1,2,3,4,6,8,12,24 \}$

Use $R>1$ to identify which $D$ is possible.

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We have $112 \equiv 232 \equiv 400 \pmod{D}$. This is equivalent to $D$ dividing their pairwise differences: $D \mid 120,168,288$ and so $D \mid 24$. Therefore, $1,2,3,4,6,8,12,24$ are possibilities for $D$ but since $R > 1$ we have only $D = 6,12,24$.

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    In addition, $D \not= 3$, since $112 \div 3 = 37$ remainder $1$. – 2017-01-27
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    This is a special case of CRT - see my answer. – 2017-01-28
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We can solve it using general techniques, viz. CRT = Chinese Remainder Theorem. Recall

$$\begin{align} r \equiv a_1\!\!\!\pmod{d_1}\\ r \equiv a_2\!\!\!\pmod{d_2}\\ r \equiv a_3\!\!\!\pmod{d_3} \end{align}$$

is solvable $\!\iff\!$ the congruences are pairwise solvable $\!\iff\!$ $\,\gcd(d_i,d_j)\mid a_i-a_j\,$ for all $\,i\neq j$

We have $\,a_i = 112,232,400\,$ and $\,d_i = d\,$ $\Rightarrow$ $\,\gcd(d_i,d_j)=d,\,$ so the solvability condition is

$$ \begin{align}d\mid 232-112=120\\ d\mid 400-232=168\\ d\mid 400-112=288\end{align}\iff d\mid\gcd(120,168,288)=24$$

Now we need only check which divisors of $24$ yield remainder $\,r>1,\,$ which is easy.