The smallest three-digit number divisible by 13 is 13 x 8 = 104, so there are seven two-digit multiples of 13. The greatest three-digit number divisible by 13 is 13 x 76 = 988. Therefore, there are 76 - 7 = 69 three-digit numbers divisible by 13.

Let k be any positive integer so that 13k is a multiple of 13. For the smallest three-digit number, 13k>100 and k>100/13=(approx)7.7. For the greatest three-digit number, 13k<999 and k<999/13 =(approx) 76.8. The number k can range from 8 to 76 so there are 69 three digit numbers.

I've solved it using a simple c code. Here it is:

include <stdio.h>

int main() { int i,counter=0; for(i=100;i<=999;i++) { if(i%13==0) counter++; if(i>=999) break; } printf("%d",counter); }

and it returns 69 :D

Simple:

$\frac { 1000 }{ 13 } -\frac { 100 }{ 13 } =\frac { 900 }{ 13 } =69,23076923076923...$

You need only the integer part. Therefore the answer is $\boxed{69}$

Doing simple multiplication, find the highest product of 13 less than 100 (the lowest possible 3-digit number). In this case, the closest you can get is 13*7=91

Now find the highest product of 13 greater than 999 (the highest possible 3-digit number). In this case it's 13*77=1001

So we can now easily see that the products of 13 that are 3 digit numbers are exclusively given as: {8, 9, 10...75,76,77} or {8 through 77}

To find the actual amount of products in that range, just subtract 8 from 77

77-8=69

There are 69 3-digit numbers divisible by 13, so the answer is C!!! :)

Tn=a+(n-1)d, here a=104(bcz it is the 1st 3 digit multiple of 13) , d=13,(bcz we need the multiples of 13), Tn=988(bcz it is the highest 3 digit multiple of 13). Therefore find n to get the number of # digit multiples of 13. NOTE:In Tn n is the subscript . It is NOT TXn

Hello all,

as for 3 digits that are divisible by 13,the range is [104,988],i made it as an arithmetic, therefore the number of terms,n = total numbers 3 digits that are divisible by 13 in the range [104,988],

as a=104(1st term that is divisible by 13), d=13 , Tn =988(last term that is divisible by 13)

by applying Tn = a + (n-1)d

988 = 104 + (n-1) 13

13n - 13 = 988 - 104

n = (988 - 104 + 13) / 13 = 69,

Therefore there are 69 (3 digit numbers).....

thanks...

999/13=76,84...

99/13=7,61...

76-7=69

forms an A.P from 104 to 988

The smallest three-digit number divisible by 13 is 13 x 8 = 104, so there are seven two-digit multiples of 13. The greatest three-digit number divisible by 13 is 13 x 76 = 988. Therefore, there are 76 - 7 = 69 three-digit numbers divisible by 13

HERE First term i.e., a=104 , common difference , d=13 and last term(nth term)= 988 Therefore, an = a+ (n-1) d ;
988= 104 + (n-1)
13 ;
988-104 = 13 n - 13;
884 +13 =13
n ;
n= 69

starting 3 digit number divisible by 13 is 104 and last number is 988.so, 104+(n-1)13=988 (last term in an AP)..by solving u get n =69