Question Number 177633 by aurpeyz last updated on 07/Oct/22
$$\mathrm{A}\:\mathrm{certain}\:\mathrm{personal}\:\mathrm{identity}\:\mathrm{number}\:\left(\mathrm{pin}\right) \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{of}\:\mathrm{four}\:\mathrm{digits}\:\mathrm{where}\:\mathrm{each}\:\mathrm{digit} \\ $$$$\mathrm{is}\:\mathrm{selected}\:\mathrm{from}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{0}\:\mathrm{to}\:\mathrm{9}.\:\mathrm{How}\:\mathrm{many} \\ $$$$\mathrm{such}\:\mathrm{pins}\:\mathrm{have}\:\mathrm{at}\:\mathrm{least}\:\mathrm{two}\:\mathrm{diffrent}\:\mathrm{digits} \\ $$$$\mathrm{and}\:\mathrm{are}\:\mathrm{palindromes}.\:\left(\mathrm{Palindrome}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cha}−\right. \\ $$$$\mathrm{racter}\:\mathrm{that}\:\mathrm{reads}\:\mathrm{the}\:\mathrm{same}\:\mathrm{forward}\:\mathrm{and}\: \\ $$$$\left.\mathrm{backward}\:\mathrm{e}.\mathrm{g}\:\mathrm{1221}\right) \\ $$
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 .](Q177647.png)
$${sum}\:{of}\:{kinds}\:{conbination}\:{is}\:\mathrm{10}×\mathrm{10}×\mathrm{10}×\mathrm{10}=\mathrm{10000} \\ $$$${have}\:{two}\:{different}\:{means}: \\ $$$$\left[{prove}:{if}\:{there}\:{have}\:\mathrm{1}\:{elements}\:{in}\:\mathrm{4}\:{squence},{only}\:{have}\:\mathrm{1}\:{solution}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{2}\:=\:\mathrm{16}\:\:\left[\mathrm{16}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{3}×\mathrm{3}×\mathrm{3}×\mathrm{3}\:=\:\mathrm{81}\:\:\left[\mathrm{81}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\mathrm{4}×\mathrm{4}×\mathrm{4}×\mathrm{4}\:=\mathrm{264}\:\:\left[\mathrm{255}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{625}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left.{prove}\:\:{finished}.\right] \\ $$$$\:{simply}\:{if}\:{choose}\:{one}\:{number}\:{in}\:{the}\:{first}\:{spot},{in}\:{the}\:\:{second}\:{only}\:{have} \\ $$$$\mathrm{9}\:{number}\:{is}\:{different},{the}\:{third}\:{spot}\:{have}\:\mathrm{8},{thus}, \\ $$$${there}\:{have}\:\mathrm{10}×\mathrm{9}×\mathrm{8}=\mathrm{720}.{don}'{t}\:{have}\:{different}\:{number}. \\ $$$$\mathrm{10000}−\mathrm{720}=\mathrm{9230}\:\:{kinds}\:{things}\:{has}\:{same}\:{number} \\ $$$${and}\:{if}\:{want}\:{to}\:{nerrow}\:{mumber}\:{is}\:{seam},{if}\:{the}\:{first}\:{spot} \\ $$$$\:{is}\:\:{a},{then}\:{aa}\mathrm{00}−{aa}\mathrm{99}\:{have}\:\mathrm{99}.{as}\:{this}\:{baa}\mathrm{1}−{baa}\mathrm{9}?{minus} \\ $$$${first}\:{spot}\:{solutions},{have}\:\mathrm{9}\:{kinds},{so}\:{on}\:{get}\:\:\mathrm{10}×\mathrm{100}+\mathrm{9}×\mathrm{10}+\mathrm{8}×\mathrm{1}=\mathrm{1098}{kinds} \\ $$$$. \\ $$
Commented by mr W last updated on 07/Oct/22
$${it}'{s}\:{wrong}\:{sir}. \\ $$$${the}\:{pins}\:{should}\:{be}\:{palindromes},\:{i}.{e}. \\ $$$${in}\:{form}\:{of}\:{abba}.\:{so}\:{we}\:{only}\:{need}\:{to} \\ $$$${find}\:{how}\:{many}\:{numbers}\:{ab}\:{we}\:{can}\: \\ $$$${get}.\:{it}'{s}\:{easy}\:{to}\:{see}\:{we}\:{have}\:\mathrm{90}\:{such} \\ $$$${numbers},\:{namely}: \\ $$$${ab}=\mathrm{10},\:\mathrm{11},\:\mathrm{12},\:...,\:\mathrm{98},\:\mathrm{99}.\: \\ $$$${totally}\:\mathrm{90}\:{numbers}.\:{so}\:{the}\:{answer}\:{is} \\ $$$$\mathrm{90}. \\ $$$$\left({it}'{s}\:{assumed}\:{that}\:{the}\:{numbers}\:{may}\:\right. \\ $$$$\left.{not}\:{begin}\:{with}\:\mathrm{0}\right) \\ $$
Commented by Strengthenchen last updated on 08/Oct/22
$${oooops}!\:{Learning}\:{english}\:{well}\:{is}\:{important}! \\ $$
Commented by Tawa11 last updated on 08/Oct/22
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
Answered by mr W last updated on 07/Oct/22
$${XXXX}\:\Rightarrow\:\mathrm{9}\:{numbers}\:\left(\mathrm{0000}\:{excluded}\right) \\ $$$${XYYX}\:\Rightarrow\mathrm{9}×\mathrm{9}=\mathrm{81}\:{numbers}\:\left(\mathrm{0}{YY}\mathrm{0}\:{excluded}\right) \\ $$$$\Rightarrow{totally}\:\mathrm{9}+\mathrm{81}=\mathrm{90}\:{pins} \\ $$
Commented by mr W last updated on 08/Oct/22
$${in}\:{sense}\:{of}\:{the}\:{question}\:{you}\:{are}\:{right} \\ $$$${sir}.\:{thanks}! \\ $$