Think small

We can see that the number is odd by adding the last digits. By adding all the digits, the number is a multiple of $\boxed{3}$ .

Let me keep it very very simple!

Let the given expression be equal to $x$ .

By applying the rule of divisibility of $3$ , we know that a number is divisible by $3$ , if the sum of its digits is divisible by three. Applying this, we get that in the given expression, the $2^{nd}$ , $3^{rd}$ , $5^{th}$ , $6^{th}$ , $8^{th}$ , $9^{th}$ and $10^{th}$ terms are divisible by $3$ . And thus, the remaining expression, that is $1 + 1234 + 1234567$ will be left out as remainder.

So, we can write it as-

$x \equiv 3 \pmod{ [1 + 1234 + 1234567] }$

$x \equiv 3 \pmod{1235802}$

Again applying the rule of divisibility of $3$ , we get that $1235802$ is also divisible by $3$ . Therefore,

$x \equiv 3 \pmod{0}$

Hence, we get that $\fbox{3}$ is the smallest prime number that divides the given number $x$ .

cause 2 is first prime number, we must find are these terms is divisible by 2? so we just add the last digits of this terms:

$1+2+3+4+5+6+7+8+9+0$ = $45$ which not divisible by 2.

now we try 3 , second prime number , to make sure this term is divisible by 3 , we must add of ALL digits ,

$1+3+6+10+15+21+28+36+45+45$ = $210$ , which 210 is DIVISIBLE by 3 and it is the smallest possible so the answer is $3$

Take the prime factorization of every number in this sum and you'll notice that they are all multiples of $3$ except for $1$ , $1234$ , and $1234567$ . Adding these three numbers gives you a multiple of $3$ as well, making the entire sum divisible by $3$ , which is a prime number. Thus the answer is $3$ .

(This makes it the smallest because the sum, and odd number, is not divisible by the only smaller prime, 2.)

Upon adding the terms we get:

$\begin{matrix} 1+12+123+1234+12345+123456+\\1234567+ 12345678+123456789+1234567890 \end{matrix} = 1371742095$

Using the rules of divisibility we can start checking for prime numbers that divide 12371742095 . Right away we know that 5 divides it since the number ends in 5 .

The only prime smaller than 5 is 3 . The rule for divisibility by 3 is:

Add up all the digits of the number. If the sum is divisible by 3, then so is the number.

$1+3+7+1+7+4+2+0+9+5 = 39 \quad \Rightarrow \quad 39 = 13 \cdot 3$

So the smallest prime number that divides the expression is $\fbox{3}$ .

${1}+{12}+{123}+{1234}+{12345}+{123456}+{1234567}+{12345678}+{123456789}+{1234567890}={1371742095}$

Since the last digit of the sum is *5, this excludes 2 as a possible divisor.*

The next prime number after *2 is 3 and* $\frac{1371742095}{3}={457247365}$

Hence 3 is the smallest positive prime number that divides the above sum.

solution in python. gathered an ordered list of prime numbers

>>> bigNumber = 1 + 12 + 123 + 1234 + 12345 + 123456 + 1234567 + 12345678 + 123456789 + 1234567890
>>> 

>>> primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

>>> 

>>> for x in primes:
...   if bigNumber % x == 0:
...     print x
...     break
... 
3

same method!

the sums would be * (1+3 + 6 + 1+ 6 + 3 + 1 + 9 + 9 + 9) mod 9 = 3 mod 9 * hence it is divisible by 3 a prime number

if we sum all the integer the result is 1,371,742,095 then if we see the ones digit is end in digit "5" so it assume the answer is 5 because it is divisible by five but five is not the smallest prime number so the answer will be 2 or 3 only,, try trial and error and you'll get 3.... then the aswer is THREE (3),,

See the last digit and add all...

You get 5 as the last digit of the expression...

Put 5 in the answer box........Brilliant says wrong answer

Only 3 is left......cant be 2....

Just add them up, how difficult can it be?