Bases

Let the three-digit base 5 number be represented as $\overline{abc}_5$. Then, the value of this number in base 10 is $a\cdot 5^2+b\cdot 5^1+c\cdot 5^0$. We want this value to be expressed in base 7 with exactly two digits.

Let the two-digit base 7 number be represented as $\overline{de}_7$. Then, the value of this number in base 10 is $d\cdot 7^1+e\cdot 7^0$. We want this value to be equal to $a\cdot 5^2+b\cdot 5^1+c\cdot 5^0$.

Converting from base 5 to base 10, we have:

$$\begin{align*} \overline{abc}_5 &= a\cdot 5^2+b\cdot 5^1+c\cdot 5^0 \ &= 25a+5b+c. \end{align*}$$

Converting from base 7 to base 10, we have:

$$\begin{align*} \overline{de}_7 &= d\cdot 7^1+e\cdot 7^0 \ &= 7d+e. \end{align*}$$

Since we want these two values to be equal, we have:

$$\begin{align*} 25a+5b+c &= 7d+e \ 25a &= 7d+e-5b-c. \end{align*}$$

Since $0\le a,b,c\le 4$ and $0\le d,e\le 6$, we have:

$$\begin{align*} 0 &\le 7d+e-5b-c \le 6\cdot 7+6 = 48 \ 0 &\le 25a \le 4\cdot 25 = 100. \end{align*}$$

This means that $7d+e-5b-c$ can take on one of 49 possible values and $a$ can take on one of 4 possible values. Therefore, the total number of integers that can be expressed in base 5 using exactly three digits and expressed in base 7 using exactly two digits is $49\cdot 4 = \boxed{196}$.

i dont think it would be 196

How many integers can be expressed in base 5 using exactly three digits

the digits are 0,1,2,3,4   repeats allowed but cannot start with 0

smallest = 100 base 5 = 5^2 base 10 = 25

biggest = (1000-1)base 5 = 5^3-1 base10 = 124 base 10

25<= [number base 10] <= 124

and expressed in base 7 using exactly two digits?

smallest = 10 base 7 = 7 base 10

biggest = 100-1  base 7= 7^2-1 base 10 = 48

7<= [number base 10] <= 48

The intersection is

25<= [number base 10] <= 48

48-25+1 = 44