Divisible By 13

The first 3-digit no. which is divisible by 13 is 104 and the series of the required 3-digit nos. divisible by 13 is in AP with first term(a)=104 and common difference(d)=13 and n= term no. We use here the formula of AP , $a_{n}=a+(n-1)d$

Let $a_{n}=1000$

$\implies a+(n-1)d=1000$

$\implies 104+(n-1)13=1000$

$\implies (n-1)13=896 \implies n-1=68.92 \implies n=\boxed{69.92}$

Since, 1000 is the last limit of the range for 3-digit nos. and is not in the series as n is fraction in case of $a_{n}=1000$ , the total no. of terms should be the value of n for the no. which is last of the series and less than and nearest to 1000 which is the 69th term as it is the nearest term to 1000 and is less than 1000.

Since $n=69.92$ in case of 1000, the total no. of terms $= \boxed{69}$

Divide 100 by 13 and get the quotient:

$\frac {100}{13}=7\frac{9}{13}$

Take the 7 and store it in your memory. Then divide 1000 by 13 and get the quotient again.

$\frac{1000}{13}=76\frac{2}{13}$

Take the 76 and subtract the 7 we got beforehand.We get 69.

$\boxed{Answer: 69}$

first term is 104

then last term is 988

therefore

988=104+(n-1)13

988=104+13n-13

988=91+13n

988-91=13n

897=13n

n=69*

Test for the lowest: (13 * 7 = 91, 13 * 8 = 104)

Test for the highest: (13 * 76 = 988, 13 * 77 = 1001)

Now, since 76 is the last number that, when multiplied by 13, yields a 3 digit number, it is our UPPER BOUND.

Next, since 8 is the lowest that yields a 3 digit number, it is the lower bound. BUT it is still in the range. Thus,

76 - 7 = 69 (number of 3-digit multiples of 13)

largest 3 digit number =1000 Smallest 3 digit number =100 the smallest 3 digit multiple of 13 is 13*8=104 The largest is 1000/13=76(equivalent) So the number of 3 digit numbers divisible by 13 is 76-8+1=69

here is a solution you can hang on 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,

Tn=a+(n-1)d 988=13+(n-1)13 988=13+13n -13 988=13n n=988/13 total n=76; 2degit divisible by 13 =7. so 76-7=69,,

Consider the sequence $100,101, ...,998 .999$

The smallest number greater than 100 and multiple of 13 is 104.

And the largest number less than 999 and multiple of 13 is 988.

Now, look at the sequence of three-digit numbers and multiples of 13

$104,117, ...,988$

This sequence is the same as

$\boxed{8} \cdot 13,\boxed{9} \cdot 13, ..., \boxed{76}\cdot 13$

We can associate this sequence the sequence

$8,9,...,76.$

Now, we just need to find the number of terms that sequence This number is $76 -8 +1 = 69$ .

Are $69$ numbers.

largest 3 digit no =999

smallest 3 digit no =100

999/13 -1000/13

76 - 7 (approx)

= 69

numbers between 0-1000 which divisible by 13 are 76. and numbers between 0-100 are 7. so 76 - 7 =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

For this problem the range of possible 3-digit number is between 100 till 999 which is permitted to find the divisibility of 13. Now, to solve for this problem all you have to do is to determine first the number or find the number (highest product) which is less than 100 and 999 respectively or greater than 100 and 999 respectively.

13x7= 91 (higest product less than 100) or 13x8=104 (Highest product greater than 100) 13x76=988 (highest product less than 999) or 13x77=1001 (Highest product greater than 999)

Let x be the number of 3 digit number divisible by 13 Equation: 13x=988-91 or 13x=1001-104 13x=897 13x=897 divide both side by 13 x=69 Answer. So finally there were 69 number of 3-digit number divisible by 13. Hope it helps....

First we need to find the largest multiple of 13 before 1000 because 1000 is not three digit number So we get (13)(76)=98 then we need to find the smallest multiple of 13 before 99 13(8)=104 hence 76-8=68

13 x 7 = 91, 13 x 8 = 104, 13 x 76 = 988, 13 x 77 = 1001. So, when 13 mutiple with each number from 8 to 76, it will give us a three-digit number <=> Numbers of Three-digit numbers are divisible by 13 is equal 76 - 8 + 1 = 69