Question

The least number which when divided by \[5,\ 6\ ,\ 7\] and \[8\] leaves a remainder \[3\], but when divided by \[9\] leaves no remainder is
A. \[1677\]
B. \[1683\]
C. \[2523\]
D. \[336\]

Answer 1

Hint: In this question, we need to find the least number which when divided by \[5,\ 6\ ,\ 7\] and \[8\] leaves a remainder \[3\], but when divided by \[9\] leaves no remainder. First, we need to find the LCM of \[5,6,7\] and \[8\] . Then we need to assume that the required number is \[3\] greater than a multiple of the LCM. Then by using the formula of Euclid’s Division Lemma , we can find the number .

Complete answer:
Given, the least number is divided by \[5,\ 6\ ,\ 7\] and \[8\] .
Let us consider the number as \[x\].
Also given that \[x\] is divisible by \[9\] . Any number that leaves a remainder of \[3\] on being divided by \[5,\ 6,\ 7\] and \[8\] will also leave a remainder of \[4\] on being divided by the LCM of \[5,\ 6,\ 7\] and \[8\].
Now we need to find the LCM of \[5,\ 6,\ 7\] and \[8\]
Now we can list the prime factors of the given number,
Factors of \[5\] ,
\[\Rightarrow \ 5 \times 1\]
Factors of \[6\] ,
\[\Rightarrow \ 3 \times 2\]
Factors of \[7\] ,
\[\Rightarrow \ 7 \times 1\]
Factors of \[8\] ,
\[\Rightarrow \ 2 \times 2 \times 2\]
Therefore the LCM is \[2 \times 2 \times 2 \times 3 \times 5 \times 7\]
On multiplying,
We get,
The LCM as \[840\] .
As the number will leave a remainder \[3\] on being divided by \[840\], we will substitute \[3\] for \[r\] , \[x\] for \[a\] and \[840\] for \[b\] in Euclid’s division lemma.
Euclid’s division lemma, \[a = bq + r\]
Now on substituting the values,
We get,
\[\Rightarrow \ x = 840q + 3\]
Now , we need to substitute different values for \[q\] in the above equation (1) and check whether the \[x\] is divisible by \[9\]. We will do this till we find a number divisible by 9. This method is known as the method of trial and error.
Now let us substituting \[1\] for \[q\] in equation (1) ,
We get,
\[\Rightarrow \ x = 840 \times 1 + 3\]
On simplifying,
We get,
\[x = 843\] which is not divisible by \[9\] .
Then let us substitute \[q = 2\] ,
On substituting,
We get,
\[\Rightarrow \ x = 840 \times 2 + 3\]
On simplifying,
We get,
\[x = 1683\] which is divisible by \[9\].
Thus , the least number which when divided by \[5,\ 6\ ,\ 7\] and \[8\] leaves a remainder \[3\], but when divided by \[9\] leaves no remainder is \[1683\] .
Final answer :
The least number which when divided by \[5,\ 6\ ,\ 7\] and \[8\] leaves a remainder \[3\], but when divided by \[9\] leaves no remainder is \[1683\] .
Option B). \[1683\] is the correct answer.

Therefore, the correct option is B

Note: In order to solve these types of questions, we should have a strong grip over Euclid’s division lemma. The concept of Euclid’s division lemma is if a number \[x\] on being divided by another number \[y\] leaves a remainder \[r\] and has a divisor \[q\] , then according to Euclid’s division lemma: \[x = yq + r\] where \[0 \leq r < y\] and \[x\] and \[y\] are integers. We should be careful while taking the LCM of the given number.