What will be the largest number that divides 100 and 25, and leaves 3 as remainder in each case?

We need to find the largest number that divides 100 and 25, and leaves 3 as the remainder. 

To find this we will use the division algorithm which is:

$\mathrm{Dividend}\left(\mathrm{D}\right) = \mathrm{Divisor}\left(\mathrm{d}\right) \times \mathrm{Quotient}\left(\mathrm{q}\right) + \mathrm{Remainder}\left(\mathrm{r}\right)$

Now, let us consider for D=100:

$100 = \mathrm{d}·\mathrm{q} + 3$

$\Rightarrow \mathrm{d}·\mathrm{q}=100-3$

$\Rightarrow \mathrm{d}·\mathrm{q}=97$

Since, 97 is a prime number there are only two possible solutions for d and q, which is:

$\mathrm{d}=97 \mathrm{and} \mathrm{q}=1$ or $\mathrm{d}=1 \mathrm{and} \mathrm{q}=97$      ...(i)

Now, for D=25:

$25 = \mathrm{d}·\mathrm{q} + 3$

$\Rightarrow \mathrm{d}·\mathrm{q} = 25 - 3$

$\Rightarrow \mathrm{d}·\mathrm{q} = 22$

Here, the possible outcome can be:

1) $\mathrm{d}=22 \mathrm{and} \mathrm{q}=1$    ...(ii)

2) $\mathrm{d}=1 \mathrm{and} \mathrm{q}=22$        

3) $\mathrm{d}=2 \mathrm{and} \mathrm{q}=11$

4) $\mathrm{d}=11 \mathrm{and} \mathrm{q}=2$

Therefore, from (i) and (ii) we can say that the only case which satisfies the given condition is $\mathrm{q}=1$.