777

The whole sequence of 3-digit numbers which are a multiple of 7 can be considered as an Arithmetic Progression (A.P) .

The first & the last 3-digit number divisible by 7 is 105 & 994.<br> The general term of an A.P is
$Tn = a + (n-1)d$

where $Tn$ is the $n$ th term,

$a$ is the 1st term,

$d$ is the common difference of the AP

Here, $Tn$ =994, $a$ =105 & $d$ =7

.: 994 = 105 + 7 $n$ - 7

Hence, $n$ = 128

Therefore, * _128_ * 3-digit numbers are multiple of 7.

There are 14 multiple from 1 to 100 [ $100/7$ ] and 142 multiple from 1 to 1000 [ $1000/7$ ]

And there we get $142-14 = 128$ 3 digit numbers multiple of 7.

And the answer is 128

105 is the smallest 3-digit number that is a multiple of 7 (15*7=105). 999-105=894 894/7=127.714 Therefor between 105 and 999 there are 127 numbers that are multiples of 7, add one for 105 and you have 128

100<7n<1000 105≤7n≤998 15≤n≤142

the numbers of n=142-15+1=128

an = a1 + (n - 1)*r

994 = 105 +(n -1)*7

7*(n-1) = 889

n-1 = 127

n = 128

Conclusão: Há 128 números inteiros positivos com três algarismos e múltiplos de 7.

no of 3 dig numbers div by 7 = no of numbers div by 7 till 999 - no of 2 dig numbers div by 7 999/7-99/7 (ignore remainders) 142 - 14 = 128 (ans)

a1 é 105, an é 994 , r é 7 an = a1 + (n-1) r

994=105+(n-1)7

994=105+7n-7

994=98+7n

994-98=7n

7n=896

n=896/7

n=128

128 múltiplos

$7 \times 15 = 105$ and $7 \times 142 = 994$

Hence, the total is $142-15+1 = \fbox {128}$

1000 / 7 = 142.85 = 142 integers ( 1 , 2 and 3 digits ) which are a multiple of 7 100 / 7 = 14.28 = 14 integers ( 1 and 2 digits ) which are a multiple of 7 3 digit integers which are a multiple of 7 = 142 - 14 = 128

100 <= 7x <=999 so x >= 100/7, because x is an integer so x>= 14 and x<=999/7, because x is an integer so x<=142 from 14 to 142 there are (142-14)+1=128 integer numbers

the little one is 105 the bigger one 994, the difference between then is 889, then you 889/7 it will be 127 but we count either 105 and 994 so the awnser is 128

The arithmetic progression is $105, 112 , 119 , \cdots 994$ . Let n terms be there in this sequence, then $994 = 105 + (n - 1)×7\\ 889 = (n - 1)×7\\ 127 = n - 1\\ n = \color{#69047E}{\boxed{128}}$ .

Alternative approach:-
The answer would be $\left \lfloor \dfrac{1000 - 100}{7} \right \rfloor = \color{#69047E}{\boxed{128}}$ .

999-99 gets us our total for three digit numbers; then I divided 900 by 7 and got my answer.

The smallest 3-digit multiple of 7 is 105 = 15x7 The largest 3-digit multiple of 7 is 994 = 142x7 So there are 142-14 = 128 ^^

Three digit numbers which are multiples of 7 are 105, 112, 119, .......994. So using the nth term of an arithmetic progression formula a+(n-1)d = 994 105 + (n - 1)7 = 994 On simplification, we get n = 128

first find out the three digit numbers which are divisible by 7 which is 999/7=142 then we calculate the 2 digit numbers which are divisible by 7 which is 99/7=14 now 3 digit numbers which are divisible by 7 is 142-14=128

I use this script (javascript) :

var count = 0; var i = 105; // the first 3-digit multiple of 7. while (i < 1000) { count++; i += 7; } alert(count);

Maximum possible for multiple of 7 under 1000 (3digit+2digit) is $7 \times 142$ = $994$ and maximum possible for multiple of 7 under 100 (2digit) is $7 \times 14$ = $98$

so many 3-digit numbers are a multiple of 7 is $142$ - $14$ = $128$

(1000-105)/7

The smallest 3-digit multiple of 7 is 105 = 15(7), The largest 3-digit multiple of 7 is 994 = 142(7), So as a result, 142-14 = 128. There are 128 3-digit multiples of 7, or 128 3-digit numbers that are divisible by 7.

All of the 3 digit numbers that are a multiple of 7 are in set Q such that if x∈Q, then x≤999, and x≥100, and x/7=y, where y is some integer. As such, there must exist the largest number that once multiplied by 7 is an element of set Q, and the smallest number that once multiplied by 7 is an element of set Q. By subtracting these numbers, you can determine the amount of elements in set Q. So, the first multiple of 7 that is in set Q is 105, and 105/7=15. The largest number in set Q is 994, and 994/7=142. 142-15=127, and adding the first element again as that has just been subtracted, 127+1=128. Therefore, there are 128 elements in set Q.

En español:

La tarea es encontrar todos los números de la forma $7x$ donde $99 < 7x < 1000$ . Despejando tenemos que $99/7 < x < 1000/7$

$14 1/7 < x < 142 6/7$ Es decir, todos los enteros pertenecientes al rango [15, 142]. La respuesta es 128.

1000- 104 = 896

896/7= 128

the smallest 3-digit number divisible by 7 = [15 x 7] = 105 the largest 3-digit number divisible by 7 = [142 x 7] = 994 so there are {142 - 14 = 128} that is 128,3-digit numbers divisible by 7.

994 = 142th * 7 && 98 = 14th * 7

142th - 14th = 128

There are 142 multiples of seven below 1000, and 14 are under 100.

142-14=128

The smallest 3-digit multiple of 7 is 105 = 15 7 The largest 3-digit multiple of 7 is 994 = 142 7 So there are 142-14 = 128 3-digit multiples of 7, i.e. 128 3-digit numbers that are divisible by 7.