Kiriti's number

Let $99*N=X$ . Since $X$ divisible by $9$ , then the digit sum of $X$ is also divisible by 9. So $X$ is the number that the total digit of 1 is a multiple of 9. If $X=111111111$ , then $X$ not divisible by $11$ , since $1-1+1-1+1-1+1-1+1=1$ and $1$ is not divisible by $11$ . So the possible value (and also the minimum) for $X$ is $X=111111111111111111$ Which total of digit $1$ is 18, and since $1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1=0$ is divisible by 11. So $N=\dfrac{111111111111111111}{99}=1122334455667789$ And the digit sum of $N$ is $2(1+2+3+4+5+6+7)+8+9=73$

Let M be the product of 99 and N.

99=9*11

Since M is divisible by 9, the sum of its digits must be divisible by 9, thus the number of digits of M must be a multiple of 9.

Since M is divisible by 11, the difference between alternating digits must be divisible by 11, thus the number of digits should be divisible by 2.

Hence the smallest value that M can take consists of 18 digits of 1.

Thus the value of N is 1122334455667789, hence the sum of its digit is 73

The product of N and 99 should be divisible by 99, which means it should be divisible by 9 as well as 11.

The product contains only ones. For it to be divisible by 9, the sum of its digits should be a multiple of 9, so the the no. of ones in the product must be a multiple of 9. Since we need to find the minimum value, we will first assume that the product is 9 ones. - 111,111,111

But this product is not divisible 11, so it cant be a multiple of 99. So next we will assume that the product is 18 ones. - 111,111,111,111,111,111

This time the product is divisible by both 9 and 11. So it must be the required product.

N*99=product

So, the value of N will be simply the product (18 ones) divided by 99.

N=1122334455667789

The sum of digits of N is 73.

This problem is simple if you know the divisibility tests for 9 and 11, since a number that is divisible by 99 must be divisible by these two numbers as well. The divisibility test for 9 states that in order to be divisible by 9, the digit sum of a number must be divisible by 9. This means that the number of digits in our large number must be a multiple of 9. The divisibility test for 11, a strange one, states that in order to be divisible by 11 the difference between the digit sum of every other digit and the digit sum of the skipped over digits is 0. Since every digit in our number is 1, that means there must be an even number of digits in our large number. Therefore the smallest number of digits we can have is $9 \times 2 = 18$ . Our large number, $111111111111111111$ divided by 99 gives $1122334455667789$ , which results in a digit sum of $73$ .

The resultant number must be a multiple of $9$ and $11$ . The smallest such number is a number with $18 1's$ . So next, we look to find $\frac {111111111111111111}{99}$ . Dividing be 11 we get $\frac {10101010101010101}{9}$ . For those who aren't familiar with this pattern, this divides down into $1 122 334 455 667 789$ . So the digit sum is $73$ .

let 99xN=x this implies that X is divisible by 9 and 11 since all the digits are 1, then the number of one's should be a multiple of 9. using the divisibility rule of 11 with all digits the same, the number of digits should be an even number. so the smallest number possible is when x has 18 one's so N=x/99=11223344556677889

digit sum = 73

To begin with, note that the number whose digits are all $1$ 's has to be divisible by $99$ (due to the fact that it is $99N$ ). If we find what that number is, we will easily be able to find $N$ as well.

Therefore, let us derive a divisibility rule with $99$ .

Since $99=11\cdot 9$ , it is obvious that the number has to be divisible with both $11$ and $9$ .

Divisibility by 9

The divisibility rule for $9$ states that the a number is divisible by $9$ only if the sum of its digits is also divisible by $9$ . This means that the number we are looking for has $9k$ digits, for any positive integer $k$ .

Divisibility by 11

The divisibility rule for $11$ states that a number is divisible by $11$ only if the alternating sum of the digits is divisible by $11$ (difference between the sum of the digits put on odd spots in the number and the sum of the digits put on the even spots in the is divisible by $11$ ).

More details on how this particular rule works may be found here .

In any case, from this we can conclude that since the digits of the number are only $1$ 's, the possible results of the alternating sum of the digits are $1$ or $0$ . Since $1$ is not divisible by $11$ and we know that $0$ is divisible by any number, it is obvious that the alternating sum of the digits equals $0$ .

This is only achievable if the number has an even number of digits (because all of the even "1's" will negate the odd "1's").

Therefore, from both of the divisibility rules, the sum of the digits of the number we are looking for has got to be an even number that is divisible by $9$ . Since we are looking for the smallest possible number with this property, it is obvious that $99N$ has exactly $18$ digits, which means that

$99N=111111111111111111 \Rightarrow N=1122334455667789$ .

The digit sum equals of $N$ equals $(1+2+3+4+5+6+7)\cdot 2 +8+9=73$ .

Thus, the correct answer is $73$ .

99xN = 11x9xN

We get: 111111111111111111 Becase digit sum 111111111111111111 is 18 (divisible by 9 and 11)

Consider: 99xN = 111111111111111111 or 9xN = 10101010101010101

Then N = 1122334455667789

The Answer = 73

All multiples of 99 have a digit sum of 18

i.e. 99 * 1=99 (9+9=18); 99 * 2=198 (1+9+8=18); 99 * 3=297 (2+9+7=18)

Therefore, since the product of 99N must have every digit as a one, the product is

111111111111111111.

111111111111111111/99=1122334455667789 (Divide this by hand because on a calculator the answer is in scientific notation.)

1+1+2+2+3+3+4+4+5+5+6+6+7+7+8+9=73

The answer is 73

(99)(N) = 111...11, (100-1)(N) = 111...11, (100)(N)-N = 111...11, The smallest N = 1122334455667789. So the sum of its digit = 2+4+6+8+10+12+14+17=73

The product of 99 and N clearly must be a multiple of 99. Since 99 = 9 x 11 and 9 and 11 are coprime, that means a number which 99 divides must be divisible by both 9 and 11. Our task is now to find the smallest positive integer which is composed entirely of the digit 1 and is divisible by both 9 and 11 (and thus is divisible by 99).

Let us begin by searching for positive integers composed entirely of the digit 1 which are divisible by 9. It is a well-known fact that an integer is only divisible by 9 if the sum of its digits is divisible by 9; thus, our smallest potential value for 99N would be 111,111,111 (nine ones), because its digits sum to 9. However, this number fails the test for divisibility by 11 (the test that I used involves determining whether or not the difference between the sums of the alternating digits of the number is divisible by 11).

Then, our next smallest potential value for 99N would be 111,111,111,111,111,111 (eighteen ones), because its digits sum to 18. Indeed, this number is divisible by 11 as well. After performing the tedious long division, N comes out to be 1,122,334,455,667,789, the sum of whose digits is 73.

choose the number with all digits as1's that is divisible by 99. It meanss it should be divisible by both 9 and 11 exactly. By checking 10101010101010101 is exactly divisible by 9 and the number 111111111111111111(number with 18 1's is smallest number that is divisible by 99)which when divided by 99 gives 1122334455667789 =N ,the required number and whose digit sum is 73