How many 3-digit numbers leave remainder 1 when divided by 7?

Correct option is C

Given:

We need to find how many 3-digit numbers leave a remainder 1 when divided by 7.

Solution:
A number leaves remainder 1 when divided by 7 if it can be written in the form:

Let 7k + 1 be between 100 and 999:

​$$=100 \leq 7k + 1 \leq 999 \\ \ \\ = 99 \leq 7k \leq 998 \\ \ \\ =\left\lceil \frac{99}{7} \right\rceil \leq k \leq \left\lfloor \frac{998}{7} \right\rfloor \\ \ \\ =14.11 \leq k \leq 142.5$$

Number of values of k from 15 to 142 inclusive:

142 - 15 + 1 = 128

There are 128 such 3-digit numbers. 
Alternate Solution:
3 digits numbers divisible by 7 are: 105, 112, 119, ……, 994
In order to get one remainder, 106, 113, 129...........995
This is an arithmetic progression series in which first term, a = 106, difference, d = 7 and last term Tn = 995
=> Tn = a + (n – 1) × d
=> 995 = 106 + (n – 1) × 7
=> (n – 1) × 7 = 889
=> n – 1 = 127
∴ n = 128 three digits numbers are divisible by 7

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