Question

How many 3 digit numbers are there which leave reminder 2 on division by 7

Answer 1

Answer:

Since, all the numbers between 100 and 999 which are divisible by 7 and leaves the remainder 2 are in AP, where, Hence, there are 129 numbers between 100 and 999 which are exactly divisible by 7 and leaves the remainder 2.

Step-by-step explanation:

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Answer 2

Question:

How many 3 digit numbers are there which leave reminder 2 on division by 7?

Answer:

$\underline{ \: \bold{ 129} \: } \: is \: the \: number.$

Step-by-step explanation:

${1}^{st} \: three \: digit \: is \: 100$

If 100 is divided by 7 , remainder is 2.

We know that,

$101\: first \: number$

If 101 divided by 7 , remainder is 3.

$Common \: difference(d) = 7$

$\therefore \: series \: 101,108,115,.....$

Last three digit number is 999.

When 999 is divided by 7 remainder is 5.

Thus, 997 is the number when divides by 7 remaining 3

$\implies series \: 101,108,115,...,997$

$first \: term(a) = 101$

$Common \: difference(d) = 7$

$t _{n} = 997$

$no.of \: terms(n) = ?$

$\boxed{t_{n} = a + (n - 1)d}$

Now substituting the values,

$\implies 997 = 101 + (n - 1)7$

$7(n - 1) = 896$

$n - 1 = 128$

$\boxed{n = 129}$

$\therefore 129 \: is \: the \: number.$