Question

How many three digit numbers are divisible by 7.
A.128
B.64
C.124
D.None of these.

Answer 1

Hint:Here, first, we have to find the lowest and highest 3 digit numbers which are divisible by 7, that we will get as 105 and 994. Then, we have to find the number of terms between 105 and 994 which are divisible by 7 using the formula of AP, where $a_n=a_1+(n-1)d$, we know the first term, last term and the common difference. We have to find the value of $n$.

Complete step-by-step answer:
First consider the smallest 3 digit number 100. We have to check whether it is divisible by 7 or not.
Now, dividing 100 by 7 we get,
${\displaystyle \frac{100}{7}}=14{\displaystyle \frac{2}{7}}$, where 2 is the remainder,
Therefore 100 is not divisible by 7.
Similarly check for 101. i.e. by dividing 101 by 7 we get:
${\displaystyle \frac{101}{7}}=14{\displaystyle \frac{3}{7}}$, where 3 is the remainder.
Therefore, 101 is not divisible by 7.
Now consider 102, i.e. by dividing 102 by 7 we obtain:
${\displaystyle \frac{102}{7}}=14{\displaystyle \frac{4}{7}}$, where 4 is the remainder.
Hence, 102 is not divisible by7.
Now consider 103. By dividing 103 by 7 we get:
${\displaystyle \frac{103}{7}}=14{\displaystyle \frac{5}{7}}$, where 5 is the remainder.
Therefore, 103 is also not divisible by 7.
Next, consider 104, by dividing it by 7 we get:
${\displaystyle \frac{104}{7}}=14{\displaystyle \frac{6}{7}}$, Here 6 is the remainder.
Hence 104 is also not divisible by 7.
Now, consider 105, by dividing it by 7 we get:
${\displaystyle \frac{105}{7}}=15$, where the remainder is zero.
Therefore, we can say that 105 is divisible by 7.
Hence 105 is the smallest 3 digit number that is divisible by 7.
Next we have to find the largest 3 digit number which is divisible by 7.
First, consider 999 the largest 3 digit number, by dividing it by 7 we get:
${\displaystyle \frac{999}{7}}=142{\displaystyle \frac{5}{7}}$, where the remainder is 5.
Therefore, 999 is not divisible by 7.
Next, consider 998, by dividing it by 7 we obtain:
${\displaystyle \frac{998}{7}}=142{\displaystyle \frac{4}{7}}$, where the remainder is 4.
Hence, we can say that 998 is not divisible by 7.
Now, consider 997, by dividing it by 7 we get:
${\displaystyle \frac{997}{7}}=142{\displaystyle \frac{3}{7}}$, where 3 is the remainder.
Therefore, 997 is not divisible by 7.
Next, take the number 996, by dividing it by 7 we obtain:
${\displaystyle \frac{996}{7}}=142{\displaystyle \frac{2}{7}}$, where 2 is the remainder.
Therefore, 996 is not divisible by 7.
Now, consider 995, i.e. by dividing it by 7 we get:
${\displaystyle \frac{995}{7}}=142{\displaystyle \frac{1}{7}}$, where the remainder is 1.
Hence, 995 is also not divisible by 7.
Now take the number 994, by dividing it by 7 we get:
${\displaystyle \frac{994}{7}}=142$, where the remainder is 0.
Therefore, we can say that 994 is divisible by 7.
Hence, 994 is the largest 3 digit number which is divisible by 7.
So, the numbers that are divisible by 7 are:
105, 112, 119, ….., 987, 994.
Since the common difference of the numbers in the above sequence is the same which is 7. Therefore, we can say that the above sequences are in arithmetic progression. i.e. AP.
Next, we have to find the number of terms in the sequence which are divisible by 7.
If the first term and common difference is given, we have a formula to find the nth term. i.e.
$a_n=a_1+(n-1)d....\text{ (1)}$
In the above formula $a_1$ is the first term, $a_n$ is the nth term, $d$ is the common difference and $n$ is the number of terms.
Here, in the sequence we have:
$\begin{array}{rl}& a_1=105\\ & a_n=994\\ & d=a_2-a_1\end{array}$
$d=112-105$, where the second term $a_2=112$
$d=7$
Next we have to find n.
By substituting all the above values in equation (1), we obtain:
$994=105+(n-1)7$
By multiplying $n-1$ by 7 we get:
$\begin{array}{rl}& 994=105+7n-7\\ & 994=98+7n\end{array}$
In the next step take 98 to the left side, then 98 becomes -98. Hence, we get:
$\begin{array}{rl}& 994-98=7n\\ & 896=7n\end{array}$
 Now by cross multiplication we get:
$n={\displaystyle \frac{896}{7}}$
By cancellation, we obtain:
$n=128$
Hence, we can say that there are 128 three digit numbers that are divisible by 7.
Therefore, the correct answer for this question is option (a).

Note: Here, to find the smallest 3 digit number that is divisible by 7 you have to start from 100 and for what number you get the remainder zero that is the smallest number. To find the largest you have to start from 999 to backwards. Similarly for what number you get the remainder zero that is the largest. In this way you will get the first and last term which then substitute in the formula and find the answer.
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