Brilli counting

36 (base 10) is 100 in base 6, likewise 216 (base 10) is 1000 in base 6. So we need to count how many multiples of 7 fall between 36 and 216.

$7 \times 6 = 42$ is the least multiple between the bounds.

$7 \times 30 = 210$ is the greatest multiple between the bounds.

Therefore, there are $30 - 6 + 1 = \boxed{25}$ 3-digit numbers in base 6 that are divisible by 7.

greatest 3 digit number in base 6= 555 = 5( $6^{2}$ ) + 5(6)+5 = 215 in decimals.
smallest 3 digit number in base 6= 100 = 36 in decimals.
number of numbers between 36 and 215 which are divisible by 7 = 25.

In 10 base and 6 base, both count in the difference of 1. Now, in 10 base there are $10$ one-digit numbers, $9*10 = 90$ two-digit numbers and $9*100 = 900$ three-digit numbers. For 6 base, $6$ one-digit numbers, $5*6 = 30$ two-digit numbers and $5*36 =180$ three-digit numbers. Now in 6 base 1000 denotes $6*6*6 = 216$ of 10 base counting. [ At 10 base 1000 denotes $10*10*10 =1000$ , this happens for all series of numbers with any base as the count differs by 1 ]. And 100 denotes $6*6 = 36$ in 10 base. Now in 10 base there are ( $5*7= 35 <36$ ) 5 integers divisible by 7 within 36 and ( $30*7 = 210$ ) 30 integers divisible by 7 within 216. Thus there are $30 - 5 = 25$ integers between 36-216 divisible by 7. So in 6 base numbers there are 25 three digit integers which are divisible by 7 of ten base numbers.

Three-digit numbers in base 6 range from 36 to 216. The smallest multiple of 7 in this range is $7 \times 6 = 42$ and the largest is $7 \times 30 = 210$ . Hence, there are $\boxed{25}$ three-digit numbers in base 6 which are divisible by 7.

largest 3 digit no. with base 6 is 555 which equals (36+6+1)*5=215 smallest 3 digit no. in base 6 is 100 which equals 36 then no. of three digit no. divisible by 7 are [215/7]-[36/7]=30-5=25

Base 6 : Given that numbers count starts from 100 to 105,..... 110 to 115.........and so on till 550. Now, from 150 numbers, every group of 6 has 1 numbers which is divisible by 7 .

So total 3-digit numbers are = [25]

1. Smallest 3 digit base6 number = 100 = 36 in base 10
2. Largest 3 digit base6 number = 555 = 215 in base 10
3. Difference between the two in base 10 = 179
4. Divide 179 by 7 gives you 25 as quotient.

Let us first convert $100_6$ , the smallest $3$ -digit number in base $6$ , and $555_6$ , the largest base- $6$ number, to base $10$ for greater clarity. $100_6$ is equal to $36_10$ , and $555_6$ is equal to $215_10.$ From here, we can divide both integers by $7$ to find the number numbers that are divisible by $7.$ The least multiple of $7$ greater than $36$ is $6\times7=42,$ and the largest less than $215$ is $30\times7=210.$ All we must do now is count the number of integers in this list. The list simplifies to $6(7), 7(7), 8(7), 9(7),..., 28(7), 29(7), 30(7).$ Thus, we have $30-6+1=\boxed{25}.$

Digits in base 6: 0, 1, 2, 3, 4, 5

Smallest 3-digit number: $100 (base 6) = 36 (base 10)$

Greatest 3-digit number: $555 (base 6) = 215 (base 10)$

The smallest number divisible by 7 which is greater than 36 is 42, that equals $6*7$ .

The greatest number divisible by 7 which is less than 215 is 210, that equals $30*7$ .

Between 6 and 30 there are $30 - 6 + 1$ numbers, or 25 numbers from which we can choose. Thus, there are 25 numbers in base 6 that have 3 digits and divide 7.

100 and 1000 in base 6 are $6^2=36$ and $6^3=216$ respectively. We need to find how many numbers in this range are divisible by 7. Now smallest and biggest multiples of 7 in this range are 42 and 210 respectively. Now $\frac{210-42}{7}=24$ . So the answer is 24 + 1= 25 as you have to count the 42 at the beginning as well.

3 digit numbers in base 6 start from 100( decimal 36) to 555( decimal 215). So there are 25 nos in between which are divisible by 7. i.e from 7 6 to 7 30.

Base 6 means she choses numbers from 0 to 5 Numbers start from 100 to 105, 110 to 115 ..... upto 550 to555 Out of 150 numbers, every group of 6 as mentioned has 1 numbers exactly divisible by 7 (property of modulus) So total numbers = 25

All 3-digit numbers in base 6 range from $100_{6}$ to $555_{6}$ . We'll work in base 10 though : $100_{6}=36$ and $555_{6}=5*36+5*6+5=215$ . From 1 to 35 there are 5 multiples of 7 , from 1 to 215 there are $\lfloor{\frac{215}{7}}\rfloor = 30$ multiples . Therefore from 36 to 215 there are $30-5 = \boxed{25}$ multiples of 7 .

This question is asking for the number of $3$ - digit numbers in base $6$ that is divisible by $7$ . This is equal to: $36(k) + 6(n) + m \equiv 0 (\bmod 7)$ , where $k,n,m$ being positive integers satisfying: $5 \geq k \geq 1$ and $5 \geq n,m \geq 0$ .

This expression is simply equal to $1(k) -1(n) +m \equiv 0 (\bmod 7)$ . Note that the values of $k$ and $n$ will determine the value of $m$ . Hence, total number of solutions to the above is simply $5 \text{ (ways to choose } k) \times 6 \text{ (ways to choose } n) = 30$ .

However, the cases whereby $k - n = 1$ do not work because $m \not = 6$ . There are $5$ solutions that are in this group of cases ( $(5,4), (4,3), (3,2), (2,1), (1,0)$ . Thus, our final answer is $30 - 5 = \boxed{25}$ .

A 3-digit number $\large abc_6$ , such that $\large 1 \le a \le 5, 0 \le b \le 5,0 \le c \le 5$ , can be converted to Base-10 as:

$\large abc_6=(6^2)a+(6)b+c=(36a+6b+c)_{10}$ .

Now from inequalities we mentioned earlier we have that $\large 36 \le 36a+6b+c$ \le 215). Hence $\large 36a+6b+c$ may take on multiples of ranging from a minimum of $\large 7 \times 6$ to a maximum multiple of $\large 7 times 30$ . Counting all the integers in the sequence $\large 6,7,8,9,...,30$ is equivalent to counting the number of terms in the sequence $\large 1,2,3,4,...,25$ . Therefore there exist $\large 25$ 3-digit numbers in Base-6 divisible by 7.