Question

Find the least number which is divided by 5, 6, 8, 9, 12 leaves a remainder 1 but when divided by 13 leaves no remainder.
A.3601
B.1469
C.2091
D.4879

Answer 1

Hint: Here, we will first find the LCM of 5, 6, 8, 9 and 12. We will assume that the required number is 1 greater than a multiple of the LCM. Then we will find the required number using Euclid’s Division Lemma.

Formulas used:
 We will use the formula of Euclid’s Division lemma which is given by $$a=bq+r$$.

Complete step-by-step answer:
Let the required number be $$x$$. We know that when $$x$$ is divided by 13, it does not leave any remainder. This means that $$x$$ is divisible by 13.
Any number that leaves a remainder of 1 on being divided by 5, 6, 8, 9 and 12 will also leave a remainder of 1 on being divided by the LCM of 5, 6, 8, 9 and 12. We will find the L.C.M. of 5, 6, 8, 9 and 12:

Now, the L.C.M. is given by:
$$\begin{array}{l}\mathrm{L}\mathrm{C}\mathrm{M}=2\times 2\times 2\times 3\times 3\times 5\\ \mathrm{L}\mathrm{C}\mathrm{M}=360\end{array}$$
As the number will leave a remainder 1 on being divided by 360, we will substitute 1 for $$r$$, $$x$$ for $$a$$ and 360 for $$b$$ in Euclid’s division lemma, $$a=bq+r$$. Therefore, we get
$$x=360q+1$$……………………$$\left(1\right)$$
We will substitute different values for $$q$$ in the above equation and check whether the $$x$$ we obtain is divisible by 13. We will do this till we find a number divisible by 13. This method is called the method of trial and error.
Substituting 1 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360+1\\ \Rightarrow x=361\end{array}$$
361 is not divisible by 13.
Substituting 2 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360\left(2\right)+1\\ \Rightarrow x=721\end{array}$$
721 is not divisible by 13.
Substituting 3 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360\left(3\right)+1\\ \Rightarrow x=1081\end{array}$$
1081 is not divisible by 13.
Substituting 4 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360\left(4\right)+1\\ \Rightarrow x=1441\end{array}$$
1441 is not divisible by 13.
Substituting 5 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360\left(5\right)+1\\ \Rightarrow x=1801\end{array}$$
1801 is not divisible by 13.
Substituting 6 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360\left(6\right)+1\\ \Rightarrow x=2161\end{array}$$
2161 is not divisible by 13.
Substituting 7 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360\left(7\right)+1\\ \Rightarrow x=2521\end{array}$$
2521 is not divisible by 13.
Substituting 8 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360\left(8\right)+1\\ \Rightarrow x=2881\end{array}$$
2881 is not divisible by 13.
Substituting 9 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360\left(9\right)+1\\ \Rightarrow x=3241\end{array}$$
3241 is not divisible by 13.
Substituting 10 for $$q$$ in equation $$\left(1\right)$$, we get
$$\begin{array}{l}x=360\left(10\right)+1\\ \Rightarrow x=3601\end{array}$$
3601 is divisible by 13.
$\therefore$ The least number which is divided by 5, 6, 8, 9, 12 leaves a remainder 1 but when divided by 13 leaves no remainder is 3601.
Option A is the correct option.

Note: The above method is a lengthy and tedious method. We can also find the answer by choosing the least number from the options and checking whether it is divisible by 13 and leaves a remainder of 1 on being divided by 360. Option B is divisible by 13 but doesn’t leave a remainder of 1 on being divided by 360. Option C is not divisible by 13. Option fulfils both the criteria, so it’s the correct option.
If a number (say $$a$$) on being divided by another number (say b) leaves a remainder $$r$$ and has a divisor $$q$$, then according to Euclid’s Division lemma: $$a=bq+r$$
where $$0\le r<b$$ and $$a$$ and $$b$$ are integers.