For a number to be a multiple of 6 and 9, it should be a multiple of (LCM of 6 and 9).

LCM of 6 and 9 is 18.

Multiples of 18 between 1 and 1000 are:

18, 36, 54, .......990

This is an arithmetic progression with first term ft=18, common difference d = 18 and last term lt= 990

lt=ft+d(n-1)

990 = 18 + 18(n-1)

where n is the total number of terms in the series.

990-18 = 18(n-1)

972 =18(n-1)

n-1 = 972/18

n-1 = 54

n=54+1

n=55

LCM of 9 and 6 is 18 . 1000 divided by 18 gives us 55 as Quotient.

There are 55 multiples of 9 divisible by 6.

multiples of 9 are also multiples of 6, so they are multiples of 18 (least common multiple of 9 and 6)
18, 36, 54, ..., 990

990 = 18 + (n-1)b

990 = 18 + (n-1)18

990 = 18 + 18n - 18

990 = 18n

n = $\frac{990}{18}$

n = 55

so there are 55 numbers

the lowest number can be divided by 9 & 6 is " 18 "

so all multiples of 18 also can be divided by 9 & 6

so , # of 18 multiples within {1,1000} = 1000 / 18 = 55.56 multiple

neglect fraction, so # multiples = 55

Eu peguei os múltiplos comuns de 6 e 9.

9,18,27,36,..

6,12,18,24,30,...

e o múltiplo comum é 18

então eu dividi 1000/18 e deu o resultado de 55, removendo a parte decimal.

multiples of 9 are multiples of 6, so we need to find the least common multiplication of 9 & 6 that is 18. thus there are (1000/18 = 55,56) numbers. the answer is 55

Hello,this is one i did it arithmetically as multiples of 6 & 9 starting with series of 18,36,54,72.......990 as the range is [1,1000]....so by arithmetic, a=18,d=18,Tn=990

so Tn=a+(n-1)d

990=18+(n-1)18 18+18n-18=990 18n=990 n=990/18=55,therefore there are 55 multiples of 6 & 9, thanks...

The first multiple of 18 is 18 and last multiple of 18 below 1000 is 990. 18 X 1=18 and 18 X 55=990. So, 55 multiples are there.

In Multiples of 9 every alternate number is multiple of 6 as well. eg. 9,18,27,36,...... 18 then alternate number 36 are multiples of 6.

nos. divisible by 6 and 9 b/w 1 to 1000 are: AP: 18,36.........................990 therefore, a=18 , d=18 and An= 990 now, An=a+(n-1)d 990=18+(n-1)18 and by solving this we get n=55

The LCM of 6 and 9 = 18. So any multiple of 18 is multiple of both 6 and 9. Number of multiples of 18 between 1 and 1000 are 1000/18 => 55 (ignore decimal part)

In python programing language num=0; for i in range(1,1000): if(i%9==0 and i%6==0): num+=1; print num

Every other multiple of 9 is also a multiple of 6 due to every other multiple of 9 being in essence a multiple of 18, (which is (6*3) ).

999 = 9*111, therefore, (111/2) must equal the number of common multiples between 9 and 6 between the integers of 1 and 999.

However, as decimals would make no sense in this answer, 55.5 must be reduced to 55.

6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 - multiples of 6

0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 - multiple of 9? yes = 1, no = 0

9 18 27 36 45 54 63 72 81 90 99 - multiples of 9

0 1 0 1 0 1 0 1 0 1 0 - multiple of 6? yes = 1, no = 0

in n multiples of 9, only floor(n/2) are multiples of 6

e.g. there are 11 multiples of 9 but only 5 of it are multiples of 6 (see the list above)

then in integers from 1-1000, there are 111 multiples of 9, so the number of multiples of 6 in that range is floor(111/2) = 55

The integers which are multiples of both 9 and 6 are : 18,36,54,....

This is an AP with a=18,d=18. So, now take 1000 as last term of AP and find n. The value of n will be 55.55555 showing that 1000 is not in AP but there are 55 terms before 1000 which are both multiples of 6 and 9. Thus, answer is 55

every third multiple of 6 is a multiple of 9, if we divide 1000 by 6 we get 166 and 4 as a remainder. now in every third multiple in these 166, is also a multiple of 9.(I hope you understand). now 166/3 = 55 complete and a remainder (the remainder does not matter).Thus our answer is 55.

E(1000/6) - E(1000/9) = 166 - 111 = 55 Where E is the integer part

For a number to be a multiple of 6 and 9, it should be a multiple of (LCM of 6 and 9).

LCM of 6 and 9 is 18.

Multiples of 18 between 1 and 1000 are:

18, 36, 54, .......990

This is an arithmetic progression with first term ft=18, common difference d = 18 and last term lt= 990

lt=ft+d(n-1)

990 = 18 + 18(n-1)

where n is the total number of terms in the series.

990-18 = 18(n-1)

972 =18(n-1)

n-1 = 972/18

n-1 = 54

n=54+1

n=55

from 1 to1000 , multiples of 6 and 9 make a A.P. series

18,36,54,.............................,990 so number of term in this series is given by

t= a+(n-1)d, where t=last digit , a=first of series n=total number of terms , d=difference between two terms,
so , 990=18+(n-1)*18,
n=55 answer

From 1 to 1000, there are 111 multiples of 9. in which every second multiple is a multiple of six. hence, 111/2=55.5. if v round 55.5 then the answer will be 56.=> that there are 112 multiples of 9. which is not the correct one so 55 is the correct answer.

The smallest multiple of 6 and 9 is 18

1000 / 18 = 55,55...

X = 55

M[6] = (9,18,27,36...)

M[9 ]=(6,12,18,27,...)

M[6,9] = (18,36,54,...,990)

R=18

a1 = 18

an = 990

an = a1+n*Q

990 = 18+n*18

18n = 972 -- > n = 54

54 + 1 = 55 termos

The L.C.M of 9 and 6 is 18. 1000/18= 55+ 10 remainder The answer is "55"

since , multiples of 6 and 9 between 1 to 1000 should be a a multiple of 6 and 9 i,e LCM of 6 and 9

LCM of 6 and 9 are 18

therefor the multiples of 18 between 1 to 1000 are

18,36,54........................................990

according to arithmetic progression , to find no of terms between any two numbers we should use this formula

an = a + ( n - 1 ) d

in which,

an= last term i,e 990 a = first term i,e 18 n= no. of terms d= common difference ( difference between two integer) i,e 36-18=18

an = a+ ( n - 1 )d

990 = 18 + ( n - 1 ) 18

990 = 18 + 18n - 18

990 = 18 n

n= 55

LCM OF 6,9=18

Divide 1 by 18 and get the quotient: 1/18=0 (WE ONLY WANT QUOTIENT)

Take the 0 and store it in your memory. Then divide 1000 by 18 and get the quotient again 1000/18=55

. Take the 55 and subtract the 0 we got beforehand.We get 55 HENCE ANSWER = 55

A number divisible both by 9 and 6 is divisible by 18.

Take the quotient of 1000 and 18 which is 55.56. This means that there are 55 integers between 1 to 1000 which are divisible by 18.

999/9 =111 ; 996/6 =166; 166-111=55

take LCM of 6.,9= 18. then divide the last number that is 1000 by 18, 1000/18= 55.5 clearly there are complete 55 multiples of 18 or rather 6 and 9 between 1 and 1000.

take LCM of 9&6.LCM=18.So,1000/18=55.55........i.e.Answer is 55

till 1000, 996 is the multiple of 6 & 999 is the multiple of 9 so, 996/6=166 and 999/9=111 therefore, 166-111=55 55 is the answer

1st no. to be multiple of 9 and also multiple of 6 is 18; let it be a

2nd no. is 36

whereas the last no. is 990

d= difference between 2nd no. and 1st no.

or, d= 36-18

or, d= 18

now, we know that last no. = a + (n-1)*d

therefore, 990 = 18 + (n-1)18

or, 990-18 = (n-1)18

or, 972 = (n-1)18

or, 972/18 = (n-1)

or, 54= n-1

or, 54 + 1 =n

or, 55 = n

where n= total no. of multiples.

I have an easy way,,.guyzz.. to find the multiples of 9 and 6 b/w 1 and 1000 ..we ll have 2 find the multiples of 18 b/w 1 and 1000...n we will get the answer,,, :):

common multiple of 6 and 9 are also multiple of 18

thus 1000/18=55.55(to take round finger)

Multiples of 9 and of 6 means: multiples of 3.3 and 3.2. Therefor is means multiples of 2.3.3 = 18. Now we divide 1000 by 18. 1000/18 = 55,... So the solution is 55.

For finding the multiples of 9 which are also multiples of 6.It is enough to find numbers which are multiple of 18.So we need to find all the multiples of 18 between 1 and 1000. 18,36,54,,,,990. This is an Arithmetic progression with first term 18 and last term 990.So we can get the number of terms by using the formula Last term(l)=First term +(number of terms-1) (common difference of AP) 990=18+(n-1) 18 solving we have number of terms as 55

This question can also be told as find the multiples of 18 between 1 and 1000. This is because 9 and 6 have their LCM. And so the answer comes 55(divide 1000 by 18..the answer will be in decimals, but atleast we can get the idea about how many multiples of 18 are there between 1 and 1000)

the number of integers divisible by nine are 111(1000/9=111+1/9). now we can see that every even multiple of nine is divisible by 2(9,18,27,36....). that means every even multiple of 9 is also a multiple of 6. hence (111-1)/2=55. 1 is been subtracted because the last multiple 999 is an odd multiple which we have to neglect.Thank you

Its Simple, Since we have to find the common multiples of 9 and 6, we take the LCM of 9 & 6 to get 18. Dividing 1000/18 gives you integer value - 55. which is our answer :)

c++ using ancient compiler XD

include<iostream.h>

include<conio.h>

main() { int num=0, total=0, ctr=0; cin>>num; for (int a=1;a<=num;a++) { if(a%9==0&&a%6==0) { ctr++ ; cout<<a<<"\n\n"; }

} cout<<"RESULT "<<ctr; getche(); }

HAHAHA

LCM of 6 and 9 is 18 and when you divide 1000 by 18 , you get it gets quotient as 55 and remainder as 10 and quotient is the answer.

integers divisible are multiples of 18 1000/18 gives 55 whole number so the solution is 55

Between 9 and 100 ,there are 11 multiples of 9(inluding 9 ),Multiplying this 10 times (for every 100) and adding 1 for 900,the total multiples are 111,of which every even multiple of 9 is a multiple of 6 as well,hence 111/2 and rounding to the lowest number as 9 being an odd number the anwser is arrived at as 55.

My solution goes like this

THe first integer that is multiple both of 6 and of 9 is 18...so 1000/18 is equal to 55.55. Take the whole number and that the number of integer which is multiples of both 9 and 6. So the answer is 55 integers which is both multiples of 6 and of 9.

• the least significant number less than 1000 that could be divided by 6 and 9 is 18.
• the most significant number less than 1000 that could be divided by 6 and 9 is 990.
• if we divide the most significant number by the least significant number, we will get total of number which can be divided by 9 and 6.
• therefore, the answer is 990/18 = 55

if we divide 999 by 9 we get 111 i.e,there are 111 multiples of 9 upto 1000. now the number which is divisible by 2 and 3 both is also divisible by 6 and every multiple of 9 is also a multiple of 3, now multiples of 9 has last digits as in order 9 8 7 ....so on.so numbers with last even digit are divisible by 2 so we simple divide 110 by 2 and got 55.

The nearest number that is multiple of 6,9 is 18 and the second is 36 so the sequence will be (18,36,54,......,990) as 990 is the last number that is multiple of 6 and 9 in the 1000 integer. a=18.....d=18......L=990 L=a+(n-1)d 990=18+(n-1)18 972=(n-1)18 (divide by 18) n-1=54 n=55

I noticed that for every three multiples of 6 there is one which is also a multiple of 9. For example, 6 12 18 24 30 36 ...... 18 and 36 are multiples of 9 as well as multiples of 6. Then I calculated the number of the multiples of 6 which exist between 1 and 1000. Which is 166: 100 until 600 , 50 more until 900, 10 more until 960 and 6 more until 196. 100+50+10+6=166. Then I divided it by three ( which is the ratio multiples of 9/ multiples of 6). 166= 55.xxxxxx . So, the integer closest to 55.xxxx which is 55 is the number of common multiples of 6 and 9 from 1 to 1000.

9=3x3 and 6=2x3 Therefore, multiples of 9 that are also multiples of 6 are multiples of 18 (18 = 2x3x3). So we're looking for the number of multiples of 18 from 1 to 1000 ; How many times 18 can enter in 1000? 55 times. Which means that there's 55 ''package'' of 18 numbers, and each of these ''package'' contains 1 multiple of 18 (because after 18 the next multiple is 18+18...etc.). Which explains why the answer is 55

first consider the multiple of 9 is

9

18

27

36

45

here u see that one number after one number is multiple of 6. between 1 to 1000 there are 111 numbers which are multiple of 9. and first 9 and last 999 this two are only multiple of 9. we can discard one. so we have now 110. so multiple of six is 110/2=55.

9x1 =9 9x2 =18(Multiple of 6) 9x3 =27 9x4 =36(Multiple of 6) ... For this pattern, we can see every second pair of the multiples of 9 are also multiples of 6. Total multiples of 9 is 111x9 = 999 <1000 So there's a total 110/2 of the second pairs, 55pairs of multiples of 6. Answer = 55.

First, we need to find the number of multiples of 9 between 1 and 1000. To do so we divide 1000 by 9 and take only the quotient while excluding the remainder. On solving , we find that there are 111 multiples of 9 between 1 and 1000. Now, every even multiple of 9 is a multiple of 6 (9 2=18 is a multiple of 6, 9 4=36 is a multiple of 6 and so on). So we can say that every alternate multiple of 9 is a multiple of 6. Hence, to find the common multiples, we divide 111 by 2 and consider only the quotient. The answer, on solving, is 55

The smallest common multiple of 9 and 6 is 18. We need to find the multiple of 18 which is between 1 and 1000. Let 1000/18 = 55,5(6) ==> 55 is the answer!

Starting with one simple factorization with the number 6 and 9 , we get 2x3x3=18 , now split for 1000 following the result of factorization, so we have the values ​​of 55 quotient and 10 rest , then we can say that before the unit thousand have 55 thousand multiples of 18 or 55 multiples of 6 and 9 simultaneously, and the last digit would be the value of the dividend subtracted from the rest. Bem desculpem-me pelo meu inglês , realmente esse não é meu forte , e isso realmente é um aspecto fraco meu , e se alguém não entender muito bem inglês vou deixar em português a resolução logo abaixo. Começamos fatorando o 6 e o 9 , e tiramos o MMC ( que resulta em 18), depois partimos para uma divisão utilizando o MMC como divisor e o numero proposto como dividendo , essa divisão resultará em 55 e em um resto 10 ,o valor 55 representa a quantidade de algarismos que são múltiplos de 18 antes do numero dividido, já o resto se subtraído do dividendo será o ultimo múltiplo do divisor.

6 = 6,12,18,24,30,36,42,48,54,60,66,72,78,84,90 9 = 9,18,27,36,45,54,63,72,81,90 So, I got 5 that multiples of 9 that also multiples of 6. 15 : 5 = 3 Because 1000 ; 6 - 166,666... that circles to down to be 166 166 : 3 = 55,333... circled down be 55. THAT'S THE ANSWER! I can't explained it but that's how I solved this question. I'm sorry for bad english.

9+9=18

6+6+6=18

now we know every double of 9 is aqual to triple of 6 so we need to find the highest multiple of 18 hence 1000/18=55.55 but we assume as 55

//include iostream statement goes here

using namespace std;

typedef unsigned short int USHORT;

int main (int argc, char* argv []) {

int counter (0);

for (int i = 1; i <= 1000; i++) {

if (i % 6 == 0 && i % 9 == 0) {

  counter++;

}

}

cout << counter << endl;

return (0);

}

Well, I didn't even use any known functions on this as far as I know I just made up a general equation.
1- My first approach: every other multiple of 9 is also a multiple of 6.
For example, multiples of 9 are 9X1=9, 9X2=(18), 9X3=27, 9X4=(36), 9X5=45, 9X6=(54), 9X7=63, 9X8=(72), 9X9=81,(9Xa=b)
all the numbers between parentheses are also multiples of 6.
2- My second approach: making a general function ...... (9a)/18 =c where a is an even number between 1 and 1000 and c is the number of multiples of 9 that are also multiples of 6. You can plug in values to check that out.
3- My third approach: the highest value of a (must be even) that can be chosen to get a multiple of 9 that is less than 1000 is 110. So, use the equation for a=110 ------> 9X110/18=55............so there are 55 multiples of 9 that are also multiples of 6.

Se um inteiro é múltiplo de dois ou mais números simultaneamente, então ele representa os múltiplos do menor múltiplo comum entre eles. Por exemplo, M.M.C(9,6) = 18. Os múltiplos inteiros de 18 entre 1 e 1000 são todos os números da forma 18k, com k inteiro positivo. Dividindo 1000 por 18 encontramos 55 como quociente que representa o número de múltiplos inteiros positivos de 18 menores que 1000.

The given situation forms an A.P. a=18 , d=18 990 is the last integer which is divisible by 9 as well as 6 and hence 990 is the last term of the A.P. Let $a_{n}$ be the last term of the A.P. Therefore,

$a_{n}$ = a + (n - 1)d

990 = 18 + (n - 1)(18)

990 - 18 = (n - 1)(18)

972/18 = (n - 1)

54 = (n - 1)

n = 54 + 1

n = 55

Hence, there are 55 integers from 1 to 1000, which are multiples of 9 and are also multiples of 6.

LCM 9 and 6 is 18
the 1000 divade 18 = 55 ( take result without decimal)
the answer is 55

lcm of 6,9 is 18 hence the solution is [1000/18] = 55...

when we check the multiples of 9 we find that every alternate multiples of 9 are a multiple of 6 .there are total 110 multiples of 9 therefore 110%2=55.are the multiples of 6.

use c++

here is my source code http://pastebin.com/F07fjtYY

To b a multiple of both 9 and 6 it must b the LCM of 9 &6 ...or a multiple of the LCM of 6&9.

So the total number of multiples will the sum of multiples from 1 to 1ooo i.e the numbers which are divided by the LCM.

As we can see 18 is the lcm of 6&9

So , ... we can find the total number of multiples ranging from 1 to 1000 , dividing 1000 by 18

1000/18= 55.55

taking the integer value we get the answer 55.

first we should know that which first no. is multiple of both 6 & 9 in 1-1000 i.e. = 18 similarly 2nd multiple of both i.e=36 now we have to know the last multipl of both i.e=990 these makes a pattern of series i.e. 18,36,............ 990 that is arithmatic series using nth term formulae we count no. of multiples of both.. i.e. 18+(n-1)18=990 from here n=55

the multiples of 9 (=3x3) and 6 (=2x3) is the multiplies of 18 (=2x3x3)

so the number of multiplies of 18 from 1 to 1000= ⌊1000/18⌋-⌊1/18⌋=55-0=55