How many 3-digit numbers are a multiple of 7?

**
Details and assumptions
**

The number $12=012$ is a 2-digit number, not a 3-digit number.

The answer is 128.

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Very nice!

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
**

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What the ..... i must ask to my daddy in this level

Firman Aitken
- 7 years, 10 months ago

ok

Marylor Manje
- 7 years, 10 months ago

3=013 digit number,so 4 is not a digit number

Ismi Mursyid
- 7 years, 10 months ago

How many 3-digit numbers are a multiple of 7?

Details and assumptions The number 12=012 is a 2-digit number, not a 3-digit number.

Ismi Mursyid
- 7 years, 10 months ago

100<7n<1000 105≤7n≤998 15≤n≤142

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

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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.

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can you write it into English? I dont understanh your point.

Nguyễn Lê
- 7 years, 10 months ago

Log in to reply

an = a1 + (n - 1)*r

994 = 105 +(n -1)*7

7*(n-1) = 889

n-1 = 127

n = 128

Conclusion: There are 128 positive integers and three digit multiple of 7

I utilized an arithmetic progression to do this problem, if you will want to study. More questions, just ask. Bye.

Matheus Pinto
- 7 years, 10 months ago

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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

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good one, ou melhor você é brasileiro então boa resolução, não pensei em fazer por P.A

Victor Carnaúba
- 7 years, 10 months ago

????

Kush Mali
- 7 years, 10 months ago

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

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

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thnx

Syed Hossain
- 7 years, 10 months ago

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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
$\left \lfloor \dfrac{1000 - 100}{7} \right \rfloor = \color{#69047E}{\boxed{128}}$
.

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

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This is close, but should explain why it works. For example, from 100 to 200 with a range of 100, there are 14 multiples of 7. However, from 300 to 400 with a range of 7, there are 15 multiples of 7.

Hence, you will need a further explanation for why / when "divided by 7 and take the integer part" works. Alternatively, what is the idea?

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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);

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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 $142$ - $14$ = $128$

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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.

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1000- 104 = 896

896/7= 128

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

142th - 14th = 128

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

142-14=128

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*
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.

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×

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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.