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Question Number 49930 by OTCHRRE ABDULLAI last updated on 12/Dec/18

Please help me solve this question  How many 4 digits number with all  digits   different are tbere between 1000 and 9999  such tbat the difference between the first  and the last digit is plus or ninus 3.

$${Please}\:{help}\:{me}\:{solve}\:{this}\:{question} \\ $$$${How}\:{many}\:\mathrm{4}\:{digits}\:{number}\:{with}\:{all}\:\:{digits}\: \\ $$$${different}\:{are}\:{tbere}\:{between}\:\mathrm{1000}\:{and}\:\mathrm{9999} \\ $$$${such}\:{tbat}\:{the}\:{difference}\:{between}\:{the}\:{first} \\ $$$${and}\:{the}\:{last}\:{digit}\:{is}\:{plus}\:{or}\:{ninus}\:\mathrm{3}. \\ $$$$ \\ $$

Answered by mr W last updated on 12/Dec/18

number pairs with difference 3:  0/3, 1/4, 2/5,...,6/9⇒7 pairs    (2×7−1)×8×7=13×8×7=728    ⇒728 such numbers.    −1 because numbers starting with 0  must be removed.

$${number}\:{pairs}\:{with}\:{difference}\:\mathrm{3}: \\ $$$$\mathrm{0}/\mathrm{3},\:\mathrm{1}/\mathrm{4},\:\mathrm{2}/\mathrm{5},...,\mathrm{6}/\mathrm{9}\Rightarrow\mathrm{7}\:{pairs} \\ $$$$ \\ $$$$\left(\mathrm{2}×\mathrm{7}−\mathrm{1}\right)×\mathrm{8}×\mathrm{7}=\mathrm{13}×\mathrm{8}×\mathrm{7}=\mathrm{728} \\ $$$$ \\ $$$$\Rightarrow\mathrm{728}\:{such}\:{numbers}. \\ $$$$ \\ $$$$−\mathrm{1}\:{because}\:{numbers}\:{starting}\:{with}\:\mathrm{0} \\ $$$${must}\:{be}\:{removed}. \\ $$

Commented by OTCHRRE ABDULLAI last updated on 12/Dec/18

God richly bless you Mr.w for your   help  am much greatful

$${God}\:{richly}\:{bless}\:{you}\:{Mr}.{w}\:{for}\:{your}\: \\ $$$${help}\:\:{am}\:{much}\:{greatful} \\ $$

Commented by mr W last updated on 12/Dec/18

thank you sir!

$${thank}\:{you}\:{sir}! \\ $$

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