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A-certain-personal-identity-number-pin-is-a-sequence-of-four-digits-where-each-digit-is-selected-from-the-digits-0-to-9-How-many-such-pins-have-at-least-two-diffrent-digits-and-are-palindromes-Pa




Question Number 177633 by aurpeyz last updated on 07/Oct/22
A certain personal identity number (pin)  is a sequence of four digits where each digit  is selected from the digits 0 to 9. How many  such pins have at least two diffrent digits  and are palindromes. (Palindrome is a cha−  racter that reads the same forward and   backward e.g 1221)
$$\mathrm{A}\:\mathrm{certain}\:\mathrm{personal}\:\mathrm{identity}\:\mathrm{number}\:\left(\mathrm{pin}\right) \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{of}\:\mathrm{four}\:\mathrm{digits}\:\mathrm{where}\:\mathrm{each}\:\mathrm{digit} \\ $$$$\mathrm{is}\:\mathrm{selected}\:\mathrm{from}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{0}\:\mathrm{to}\:\mathrm{9}.\:\mathrm{How}\:\mathrm{many} \\ $$$$\mathrm{such}\:\mathrm{pins}\:\mathrm{have}\:\mathrm{at}\:\mathrm{least}\:\mathrm{two}\:\mathrm{diffrent}\:\mathrm{digits} \\ $$$$\mathrm{and}\:\mathrm{are}\:\mathrm{palindromes}.\:\left(\mathrm{Palindrome}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cha}−\right. \\ $$$$\mathrm{racter}\:\mathrm{that}\:\mathrm{reads}\:\mathrm{the}\:\mathrm{same}\:\mathrm{forward}\:\mathrm{and}\: \\ $$$$\left.\mathrm{backward}\:\mathrm{e}.\mathrm{g}\:\mathrm{1221}\right) \\ $$
Commented by Strengthenchen last updated on 07/Oct/22
sum of kinds conbination is 10×10×10×10=10000  have two different means:  [prove:if there have 1 elements in 4 squence,only have 1 solution             2                           4          2×2×2×2 = 16  [16]             3                           4          3×3×3×3 = 81  [81]             4                           4         4×4×4×4 =264  [255]             5                           4                                      625                prove  finished.]   simply if choose one number in the first spot,in the  second only have  9 number is different,the third spot have 8,thus,  there have 10×9×8=720.don′t have different number.  10000−720=9230  kinds things has same number  and if want to nerrow mumber is seam,if the first spot   is  a,then aa00−aa99 have 99.as this baa1−baa9?minus  first spot solutions,have 9 kinds,so on get  10×100+9×10+8×1=1098kinds  .
$${sum}\:{of}\:{kinds}\:{conbination}\:{is}\:\mathrm{10}×\mathrm{10}×\mathrm{10}×\mathrm{10}=\mathrm{10000} \\ $$$${have}\:{two}\:{different}\:{means}: \\ $$$$\left[{prove}:{if}\:{there}\:{have}\:\mathrm{1}\:{elements}\:{in}\:\mathrm{4}\:{squence},{only}\:{have}\:\mathrm{1}\:{solution}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{2}\:=\:\mathrm{16}\:\:\left[\mathrm{16}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{3}×\mathrm{3}×\mathrm{3}×\mathrm{3}\:=\:\mathrm{81}\:\:\left[\mathrm{81}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\mathrm{4}×\mathrm{4}×\mathrm{4}×\mathrm{4}\:=\mathrm{264}\:\:\left[\mathrm{255}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{625}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left.{prove}\:\:{finished}.\right] \\ $$$$\:{simply}\:{if}\:{choose}\:{one}\:{number}\:{in}\:{the}\:{first}\:{spot},{in}\:{the}\:\:{second}\:{only}\:{have} \\ $$$$\mathrm{9}\:{number}\:{is}\:{different},{the}\:{third}\:{spot}\:{have}\:\mathrm{8},{thus}, \\ $$$${there}\:{have}\:\mathrm{10}×\mathrm{9}×\mathrm{8}=\mathrm{720}.{don}'{t}\:{have}\:{different}\:{number}. \\ $$$$\mathrm{10000}−\mathrm{720}=\mathrm{9230}\:\:{kinds}\:{things}\:{has}\:{same}\:{number} \\ $$$${and}\:{if}\:{want}\:{to}\:{nerrow}\:{mumber}\:{is}\:{seam},{if}\:{the}\:{first}\:{spot} \\ $$$$\:{is}\:\:{a},{then}\:{aa}\mathrm{00}−{aa}\mathrm{99}\:{have}\:\mathrm{99}.{as}\:{this}\:{baa}\mathrm{1}−{baa}\mathrm{9}?{minus} \\ $$$${first}\:{spot}\:{solutions},{have}\:\mathrm{9}\:{kinds},{so}\:{on}\:{get}\:\:\mathrm{10}×\mathrm{100}+\mathrm{9}×\mathrm{10}+\mathrm{8}×\mathrm{1}=\mathrm{1098}{kinds} \\ $$$$. \\ $$
Commented by mr W last updated on 07/Oct/22
it′s wrong sir.  the pins should be palindromes, i.e.  in form of abba. so we only need to  find how many numbers ab we can   get. it′s easy to see we have 90 such  numbers, namely:  ab=10, 11, 12, ..., 98, 99.   totally 90 numbers. so the answer is  90.  (it′s assumed that the numbers may   not begin with 0)
$${it}'{s}\:{wrong}\:{sir}. \\ $$$${the}\:{pins}\:{should}\:{be}\:{palindromes},\:{i}.{e}. \\ $$$${in}\:{form}\:{of}\:{abba}.\:{so}\:{we}\:{only}\:{need}\:{to} \\ $$$${find}\:{how}\:{many}\:{numbers}\:{ab}\:{we}\:{can}\: \\ $$$${get}.\:{it}'{s}\:{easy}\:{to}\:{see}\:{we}\:{have}\:\mathrm{90}\:{such} \\ $$$${numbers},\:{namely}: \\ $$$${ab}=\mathrm{10},\:\mathrm{11},\:\mathrm{12},\:…,\:\mathrm{98},\:\mathrm{99}.\: \\ $$$${totally}\:\mathrm{90}\:{numbers}.\:{so}\:{the}\:{answer}\:{is} \\ $$$$\mathrm{90}. \\ $$$$\left({it}'{s}\:{assumed}\:{that}\:{the}\:{numbers}\:{may}\:\right. \\ $$$$\left.{not}\:{begin}\:{with}\:\mathrm{0}\right) \\ $$
Commented by Strengthenchen last updated on 08/Oct/22
oooops! Learning english well is important!
$${oooops}!\:{Learning}\:{english}\:{well}\:{is}\:{important}! \\ $$
Commented by Tawa11 last updated on 08/Oct/22
Great sirs
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
Answered by mr W last updated on 07/Oct/22
XXXX ⇒ 9 numbers (0000 excluded)  XYYX ⇒9×9=81 numbers (0YY0 excluded)  ⇒totally 9+81=90 pins
$${XXXX}\:\Rightarrow\:\mathrm{9}\:{numbers}\:\left(\mathrm{0000}\:{excluded}\right) \\ $$$${XYYX}\:\Rightarrow\mathrm{9}×\mathrm{9}=\mathrm{81}\:{numbers}\:\left(\mathrm{0}{YY}\mathrm{0}\:{excluded}\right) \\ $$$$\Rightarrow{totally}\:\mathrm{9}+\mathrm{81}=\mathrm{90}\:{pins} \\ $$
Commented by mr W last updated on 08/Oct/22
in sense of the question you are right  sir. thanks!
$${in}\:{sense}\:{of}\:{the}\:{question}\:{you}\:{are}\:{right} \\ $$$${sir}.\:{thanks}! \\ $$

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