Question

how many three digits number is divisibal by 7?​

Answer 1

Answer:

mark me as brainlest

Step-by-step explanation:

105/7=15

Answer 2

Answer:

128 is three digit number which is divisible by 7

Step-by-step explanation:

Number divisible by 7 are

• 7, 14, 21, 28....

$\bf \: lowest \: 3 \: digit \: number : - \\ \\ \sf \longmapsto \: \frac{100}{7} = 14 \frac{3}{7} \\ \\ \sf \longmapsto \frac{101}{7} = 14 \frac{3}{7} \\ \\ \sf \longmapsto \frac{102}{7} = 14 \frac{4}{7} \\ \\ \sf \longmapsto \frac{105}{7} = 15 \\ \\ \large\frak{ \blue{so \: lowest \: number \: is \: 105}}$

$\bf \: highest \: 3 \: digit \: number : - \\ \\ \sf \longmapsto \frac{999}{7} = 142 \frac{5}{7} \\ \\ \sf \longmapsto \frac{998}{7} = 142 \frac{4}{7} \\ \\ \sf \longmapsto \frac{997}{7} = 142 \frac{3}{2} \\ \\ \sf \longmapsto \frac{994}{7} = 142 \\ \\ \red{ \large{ \frak{ \: so \: height \: number \: is \: 142}}}$

$\bf \: thus \: our \: series \: will \: be \\ \\ \color{red} \tt \large105, 112, 119 , 126...994$

$\large \sf\: Since, \: difference \: is \: same \: it \: is \: a \: A.P$

$\sf \: Last \: term \: of \: series \: is \: 994$

$\sf \: We \: need \: to \: find \: number \: terms \: i.e \: \bf{n}$

$\sf \: Hence,$

$\qquad \qquad\bf \: a_{n} = 994 , \bf a = 105, \bf \: d = 7$

Putting values in formals

$\qquad \: \bf \: a_{n} = a + (n -1 )d \\ \\\longrightarrow \sf 994 = 105 + (n - 1) \times 7 \\ \\ \longrightarrow \sf994 = 105 + 7n - 7 \\ \\\longrightarrow \sf 994 - 105 + 7 = 7n \\ \\ \longrightarrow \sf889 - 7 = 7n \\ \\ \longrightarrow \sf896 = 7n \\ \\ \longrightarrow \sf \cancel \frac{896}{7} = n \\ \\ \longrightarrow \sf128 = n \\ \\ \qquad \boxed{ \blue{ \frak{n = 128}}}$

Therefore, 128 3-digit number are divisible by 7