Triple-Digit Digit Product

Let's call our number $\overline{abc}$ . We want $abc \geq 100$ . Remember $0 \leq a,b,c \leq 9$ , this is important for the inequalities.

Start by finding the lowest possible value of a. We know $bc \leq 81$ . If a=1, $abc \leq 81$ , never yielding a 3-digit product. When $a=2, abc \leq 162$ . Only the smallest solution is important, so we can ignore values for a other than 2 (obviously, a solution in the 200s is less than a solution in the 300s or higher).

From $abc \geq 100$ and a=2, $bc \geq 50$ . Now let's look for the lowest possible value for b. If $b \leq 5$ , the previous condition for bc cannot be met. Therefore, 6 is the lowest possible value for b. Again, we can ignore higher values of this digit. We now find that c=9 to meet the initial conditions. $\overline{abc}=269$ . Sure enough, the product $2*6*9=108$ is a 3-digit number.

Clearly, if we have 3 single digit numbers $(x,y,z)$ that fulfill $x \times y \times z ≥ 100$ , we would select $(a,b,c)$ from amongst them such that $a≤ b ≤ c$ so that the 3-digit number formed will be the smallest possible for such a combination of 3 digits.

Thus, we may consider only 3-digit numbers such that $a ≤ b ≤ c$ . If $a = 1$ , then $b \times c ≤ 9 \times 9 = 81 < 100$ , hence this will not fulfill the condition.

Trying $a = 2$ , we need $b \times c ≥ 50$ , and it is easy to see that if $b ≤ 5$ , then $b \times c ≤ 5 \times 9 = 45 < 50$ , so we must have $b > 5$ . Trying $b = 6$ , we obtain that $c$ has to be $9$ in order for $b \times c ≥ 50$ to hold.

Hence the smallest possible such 3-digit number is $269$ .

The basic idea is that out of the 3 digits, the first has to be 2, since 1xy can never give a 3 digit number by multiplying the digits. It could maximum be 199 resulting in 81. This, the starting digit has to be 2. Now, 2 * 50 gives 100, i.e. a 3 digit no. So, think of 2 digits whose product is 50 or more: smallest would be 69 as the product is 54 and 2*54 = 108. Hence, the smallest no. is 269.

Note that the maximum product of two digits is 81, which is less than 100. So, all the digits must be greater than 1. We then set the first digit equal to 2. The product of the remaining two digits must then by greater than or equal to 50. The only pairs of digits that satisfy this are (6,9), (7,9), (8,9), (9,9), (7,8), (8,8). It is then easy to see that 269 is the lowest possible number.

You go through all the one hundreds, but realize that if the first number is 1, and the other two are 9, it is still only 81. Therefore, it can't start with a 1. So you go through the two hundreds. If the first number is 2, the second number is 2, then the third number needs to be 25, which is not a single digit. So, you check the second number with 3, which means the last number has to be about 17, which is still too high. Then you try the second number with 4, and the third number needs to be 12.5, still too high. The fourth number being 5, you would need a 10 as the last number which is still not a single digit. Finally, if you try a 2 and a 6, you get 8.33333, and if you choose the next-highest integer, it will be 9 (a single-digit). 12x9=144 and is the lowest possibility of a 3-digit number.

We want to minimize $N=\overline{abc}$ for which $a\times b\times c\ge 100$ .

If $a=1$ , $N$ is at most $199$ , but $1\times 9\times 9=81<100$ , so for all $N=\overline{1bc}$ , $1\times b\times c$ is not a 3-digit number. Hence, $a$ cannot be $1$ .

If $a=2$ , $N$ is at most $299$ , and $2\times 9\times 9=162\ge 100$ . So we can find some $N=\overline{2bc}$ such that $2\times b\times c\ge 100$ , or $b\times c\ge 50$ .

If $1\le b\le 5$ , $\overline{bc}$ is at most $59$ , but $5\times 9=45<50$ .

For $b=6$ , $c=9$ will satisfy the requirement, and the answer is $N=269$ .

Let \overline{abc} be the 3-digit number, where a,b,c are nonnegative 1-digit integers and a is not 0.

Since we want the 3-digit number to be as small as possible, we start with a=1.

Note that if a=1, then the maximum value of b c is 9 9 = 81, making a b c only a 2-digit number. Thus, a cannot be 1.

If a=2, then b*c>=100/2=50.

If b<=5, then c>=10, but c should be a 1-digit integer. Thus, b>5.

If b=6, then c>=25/3. Since c is an integer, then c=9.

Testing this, we get that a b c=2 6 9=108, which is indeed a 3-digit number and 269 is the smallest possible since those smaller than it were shown to be not possible.

Therefore, the 3-digit number we are looking for is 269.

Let the three digit number have digits a, b, c as the hundreds, tens, and ones.

a cannot be 1 since the maximum product would be 81. When a is 2, after trying b in ascending order, you find that 2 6 9 obtains a three digit number while 2 6 8 doesn't so 269 is the answer.

If 3-digit number is denoted as 'abc', For a=1, the product of the individual digits is always less than 3-digit number because the maximum value of the product from given situation is 1 9 9 = 81. For a=2, the product of the individual digits that is always more than 2-digit number should be equal or more than 100. Thus, a b c >= 100 2 b c >= 100 b c >= 50
In order to obtain the smallest 3-digit number, b value must be - as minimal as possible and - satisfy this inequalities : b c >= 50 so c value should be as high as possible, which is 9. Hence, b value that closest but more than 50 is 6. So, the smallest 3-digit positive number is 269

To let the 3-digit positive number smaller, we can assume that the number is 1 , but the largest number 199 is 1 9 9 =81, is wrong. So the number must be 2 . The last two digits' product must higher than 50 and we should let the second digit to be smallest. Then we find 6 9=54>50, while 6 8=48<50,5*9=45<50. So the 3-digit number is 269.

Let $N = \overline{abc} = 100a + 10b + c$ . To get the smallest possible $N$ , we set $1 \leq a \leq b \leq c \leq 9$ .

Since $a \times b \times c \geq 100$ and $b \leq c \leq 9$ we have $a \geq \frac {100}{81}$ , so $a = 2$ is the smallest possible value.

Likewise, $2 \times b \times c \geq 100$ with $c \leq 9$ thus $b \geq \frac {100}{18}$ , so $b=6$ is the smallest possible value.

Now $2 \times 6 \times c \geq 100 \Rightarrow c \geq \frac{100} {12}$ , so $c = 9$ is the smallest possible value.

Therefore $N=269$ is the smallest such number.

9 is the largest digit so mostly unit place 100/9=11.111 so product of smallest digit with any digit>11.111 1 multiplied will not give so 2 is one digit fixed to hundreds place as we want small number so 2 multiplied by smallest number which gives greater than 11.111 is 6 so it is fixed as tenth place and then if unit digit is 8 it gives product of 96 but if 9 gives 108>100 so the number is 269!!!