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(A) number is divisible by $88$ if it is divisible by both $8$ and $11$,since $88 = 8 	imes 11$ and $\gcd(8, 11) = 1$.First,apply the divisibility te

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$8, 2$

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(A) number is divisible by $88$ if it is divisible by both $8$ and $11$,since $88 = 8 	imes 11$ and $\gcd(8, 11) = 1$.First,apply the divisibility test for $8$. $A$ number is divisible by $8$ if its last three digits are divisible by $8$. The last three digits are $23y$. For $23y$ to be divisible by $8$,we check $230 \div 8 = 28$ with remainder $6$. Adding $2$ to $230$ gives $232$,which is $8 	imes 29$. Thus,$y = 2$ is the smallest value.Now the number is $5x4232$. Apply the divisibility test for $11$. $A$ number is divisible by $11$ if the difference between the sum of digits at odd places and the sum of digits at even places is either $0$ or a multiple of $11$.Sum of digits at odd places (from right): $2 + 2 + x = x + 4$.Sum of digits at even places (from right): $3 + 4 + 5 = 12$.Difference: $(x + 4) - 12 = x - 8$.For the number to be divisible by $11$,$x - 8 = 0$,which gives $x = 8$.Therefore,the least values are $x = 8$ and $y = 2$.

Find the least values of $x$ and $y$ so that the number $5x423y$ is divisible by $88$.

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