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diff --git a/racket/leetcode/lc-9-palindromic-number.rkt b/racket/leetcode/lc-9-palindromic-number.rkt
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+++ b/racket/leetcode/lc-9-palindromic-number.rkt
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+#lang racket
+(define/contract (is-palindrome x)
+  (-> exact-integer? boolean?)
+  ; get the easy cases out of the way first
+  ; negative numbers are not palindromes, single-digit numbers are
+  (cond [(x . < . 0) #false]
+        [(x . < . 10) #true]
+        [else
+         ; order-of-magnitude returns the scientific notation exponent
+         ; so add 1 to get the number of digits
+         (define digits
+           (add1 (order-of-magnitude x)))
+         ; figure out how many digits we need to trim to find the mirrored halves
+         ; if there are an even number of digits 2n, we will remove n of them from the right
+         ; if there are an odd number of digits 2n+1, we will remove n+1 of them from the right
+         (define half-digits
+           (cond [(even? digits) (/ digits 2)]
+                 [else (/ (add1 digits) 2)]))
+         ; divide x by a power of 10 to get the digits to match
+         (define front-half
+           (quotient x (expt 10 half-digits)))
+         ; reverse the back half with repeated divisions by 10
+         (define back-half-reversed
+           (for/fold ([reversed 0]
+                      [remaining (remainder x (expt 10 half-digits))]
+                      ; build up the sum of the digits in reversed and return it at the end
+                      #:result reversed)
+                     ; if we have an odd number of digits, we don't need to match the middle one
+                     ([n (in-range (if (even? digits)
+                                       half-digits
+                                       (sub1 half-digits)))])
+             ; shift all the accumulated digits in the mirror to the left one place
+             ; and add the next one to the right,
+             ; then chop the right-most digit off the original
+             (values (+ (* 10 reversed) (remainder remaining 10))
+                     (quotient remaining 10))))
+         ; finally, check to see if the mirrored right is equal to the original left
+         (= front-half back-half-reversed)]))