Three-digit numbers divisible by 29

Query: is -123 a three digit number?

If yes, then the answer is zero.

The sum of the positive three digit numbers that are multiples of 29 is 17081, as explained in the other solutions.

The sum of the negative three digit numbers that are multiples of 29 is =-17081.

Thus the sum of all such three digit numbers, positive and negative, is 0.

The smallest three-digit number divisible by $29$ is $29 \times \left \lceil \dfrac {100}{29}\right \rceil = 29 \times 4$ and the largest three-digit number divisible by $29$ is $29 \times \left \lfloor \dfrac {999}{29} \right \rfloor = 29 \times 34$ . Therefore the sum of all three-digit multiples of $29$ is:

$\begin{aligned} S & = 29 \blue{(\underbrace{4+5+6+ \cdots + 34}_{\text{31 terms}})} & \small \blue{\text{Using the sum of AP }=\frac {n(a+l)}2} \\ & = 29 \times \blue{\frac {31(4+34)}2} \\ & = 29 \times 31 \times 19 \\ & = \boxed{17081} \end{aligned}$

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Notations:

• $\lceil \cdot \rceil$ denotes the ceiling function .
• $\lfloor \cdot \rfloor$ denotes the floor function .

The smallest three digit number divisible by $29$ is $29\times \lceil {\dfrac{100}{29}}\rceil=29\times 4$ , and the largest such number is $29\times \lfloor {\dfrac{1000}{29}}\rfloor=29\times 34$ . So the required sum is $29\times (4+5+6+...+34)=29\times (\dfrac{34\times 35}{2}-6)=\boxed {17081}$ .

Smallest 3 digit number divisible by 3 = 116

Largest 3 digit number divisible by 3 = 986, if n is the total number of 3 digit numbers which are divisible by 29

986 = 116 + (n-1)* 19 , solving this equation yields a value of n = 31

S = sum of n terms of an Arithmetic progression with initial term = a and common difference = d can be formulated as shown below

S = a + ( a + d) + ( a+2d)+....(a+(n-1)d) = n * a + d * (1+2+3+4....n-1) = n * a + (d * n * (n-1))/2 = n * (a+ d * (n-1)/2) = 31(116 + 15 * 29) = 17081