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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

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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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