What is the smallest number that leaves a remainder of 7 17 and 23 when divided by 12 22 and 28?

Question 14 Real Numbers Exercise 1B

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Answer:

Prime factors of 4, 7 and 13

4 = 2x2

7 and 13 are prime numbers.

LCM ( 4, 7, 13) = 364

We know that, the largest 4 digit number is 9999

Step 1: Divide 9999 by 364, we get

9999/364 = 171

Step 2: Subtract 171 from 9999

9999 - 171 = 9828

Step 3: Add 3 to 9828

9828 + 3 = 9831

Therefore 9831 is the number.

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Find the smallest number which when divided by 28 and 32 leaves remainder 8 and 12 respectively. Maths Q&A

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Here is a general way to solve.

It is equivalent to solving the system: $\;\begin{cases}x\equiv8&\bmod 28,\\x\equiv 12&\bmod32.\end{cases}$

There is a formula when the moduli are coprime. We'll reduce the problem to this case.

Any solution has to be divisible by $4$, so we'll set $x=4y$. The congruences can be written as $$\begin{cases}4y\equiv 8\mod 28\\4y\equiv 12\mod 32\end{cases}\iff \begin{cases}y\equiv 2\mod 7\\y\equiv 3\mod 8\end{cases}$$ Now a Bézout's relation between $7$ and $8$ is $8-7=1$, hence the solutions for $y$ are $$y\equiv 2\cdot 8-3\cdot 7=-5\mod 56,$$ whence $\;x=4y\equiv -20\mod 224$. So the smallest positive value is $\;\color{red}{x=204}$.

Added:

More generally, one shows a system of linear congruences $$x\equiv a_i\mod m_i\quad(i=1,\dots,r)$$ where the $m_i$ are not necessarily mutually coprime, has a solution if and only if $$\forall i\;\forall j,\enspace a_i\equiv a_j \mod\gcd(m_i,m_j)$$ and in this case, the solution is unique modulo $\operatorname{lcm}(m_1,\dots,m_r)$.