Question

how many three digits Number are divisible by 7?​

Answer 1

Answer: 128

Step-by-step explanation:

Smallest 3 digit number 100 when divisible by 7 leaves a remainder of 2. So lowest 3 digit number divisible by 7 is 100+7-2=105

Likewise we can find that the greatest 3 digit number divisible by 7 is 994.

We have to find the number of terms of an AP whose first term a is 105, last term aₙ is 994, and common difference is 7

aₙ=994

⇒a+(n-1)d=994

⇒105 + (n-1)7=994

⇒7n = 994 - 105 + 7 =896

⇒ n =128

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

${\boxed{\sf\;{Arithemetic\;Progression}}}$

105, 112 , 119 , ...... , 994

Here we have :-

a = 105

$\rightarrow{d=t_{2}-t_{1}}$

= 112 - 105

= 7

${\boxed{\sf\:{Given\; Information}}}$

nth term = 994

★Formula of nth term :-

${\boxed{\sf\:{a_{n}=a+(n-1)d}}}$

105 + (n - 1) × 7 = 994

7(n - 1) = 994 - 105

7(n - 1) = 889

${\boxed{\sf\:{(n-1)=\dfrac{889}{7}}}}$

n - 1 = 127

n = 127 + 1

n = 128

★Therefore we get :-

$\fbox{128 numbers of 3 digits are divisible by 7}$