Determine the values of the digits $a, b, c, d, e$ in the following multiplications:
a) $a b c d e 7 \cdot 5=7 a b c d e$
b) $1 a b c d e \cdot 3=a b c d e 1$
# I. solution.

a) Since $5 \cdot 7=35$, then $e=5$ and the remainder is 3,

$5 e+3=5 \cdot 5+3=28$, so $d=8$ and the remainder is 2.

Continuing this process and substituting the obtained values, we get

$c=2, b=4, a=1$.

Indeed, $142857 \cdot 5=714285$.
b) The number ending in 1 and divisible by 3, but less than 27 (since $e$ is at most 9), is clearly 21, so $e=7$ and the remainder is 2.

Thus, $(3 d+2)$ must end in 7, which means $3 d$ must end in 5, leading to $d=5$, and substituting the obtained values and continuing the multiplication, we get

$c=8, b=2, a=4$.

Indeed, $142857 \cdot 3=428571$.

Roboz Ágnes (Bp. VI., Varga Katalin lg. II. o. t.)

II. solution.

a) Let $x$ be the 5-digit number abcde, so $x=10000 a+1000 b+100 c+10 d+e$. Substituting $x$ into the equation, we get:

$$
(10 x+7) 5=700000+x
$$

which simplifies to

$$
49 x=699965
$$

from which

$$
x=14285
$$

and thus $a=1, b=4, c=2, d=8, e=5$.

b) Similarly, let $x$ be the 5-digit number abcde,

$$
(100000+x) \cdot 3=10 x+1
$$

which simplifies to

$$
7 x=299999
$$

from which

$$
x=42857
$$

and thus $a=4, b=2, c=8, d=5, e=7$.