Sufficient 9391

In Kocourkov, they use coins with only two values expressed in Kocourkov crowns by positive integers. With a sufficient number of such coins, it is possible to pay any integer amount greater than 53 cats’
crowns accurately and without return. However, we cannot pay 53 cat crowns without a refund.

Find out what values ​​could have been on the cat’s coins. Identify at least two solutions.

Correct answer:

a₁ =  3
b₁ =  28
a₂ =  4
b₂ =  19
a₃ =  7
b₃ =  10
a₄ =  2
b₄ =  55

Step-by-step explanation:

$$\begin{gathered} 53\ne a_x+b_y \\ 53<a_x+b_y \\ {a}_1=3 \end{gathered}$$

$$\begin{gathered} b_1=28 \\ {t}_1=18\cdot a_1+0\cdot b_1=18\cdot 3+0\cdot 28=54 \\ {t}_2=9\cdot a_1+1\cdot b_1=9\cdot 3+1\cdot 28=55 \\ {t}_3=0\cdot a_1+2\cdot b_1=0\cdot 3+2\cdot 28=56 \\ \dots \\ {b}_1=28 \end{gathered}$$

$a_2=4$

$$\begin{gathered} b_2=19 \\ {t}_4=4\cdot a_2+2\cdot b_2=4\cdot 4+2\cdot 19=54 \\ {t}_5=9\cdot a_2+1\cdot b_2=9\cdot 4+1\cdot 19=55 \\ {t}_6=14\cdot a_2+0\cdot b_2=14\cdot 4+0\cdot 19=56 \\ {t}_7=0\cdot a_2+3\cdot b_2=0\cdot 4+3\cdot 19=57 \\ \dots \end{gathered}$$

$a_3=7$

$b_3=10$

$a_4=2$

$$\begin{gathered} b_4=55 \\ {t}_8=27\cdot a_4+0\cdot b_4=27\cdot 2+0\cdot 55=54 \\ {t}_9=0\cdot a_4+1\cdot b_4=0\cdot 2+1\cdot 55=55 \\ {t}_{10}=28\cdot a_4+0\cdot b_4=28\cdot 2+0\cdot 55=56 \\ \dots \end{gathered}$$

Try another example