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Question Number 137745 by bemath last updated on 06/Apr/21
$$ \\ $$Each of the digits 2, 4, 6, and 8 can be used once and once only in writing a four-digit number. What is the sum of all such numbers that are divisible by 11?
Answered by TheSupreme last updated on 06/Apr/21
![ABCD is div for 11 if A+C−B−D=0 or A+C−B−D=11 A+C=B+D i: 8+2=4+6 is the only possible solution A=(2,8),(4,6) B=(4,6),(2,8) C=(2,8),(4,6) D=(4,6),(2,8) n=4×2=8 Σ_(i=1) ^8 1000A_i +100B_i +10C_i +D_i = =1000ΣA_i +100ΣB_i +10ΣC_i +ΣD_i ΣA_i =40 [2+2+4+4+6+6+8+8] =1111×40=44440](Q137749.png)
$${ABCD}\:{is}\:{div}\:{for}\:\mathrm{11}\:{if} \\ $$$${A}+{C}−{B}−{D}=\mathrm{0}\:{or}\:{A}+{C}−{B}−{D}=\mathrm{11} \\ $$$${A}+{C}={B}+{D} \\ $$$${i}:\:\mathrm{8}+\mathrm{2}=\mathrm{4}+\mathrm{6}\:{is}\:{the}\:{only}\:{possible}\:{solution} \\ $$$${A}=\left(\mathrm{2},\mathrm{8}\right),\left(\mathrm{4},\mathrm{6}\right) \\ $$$${B}=\left(\mathrm{4},\mathrm{6}\right),\left(\mathrm{2},\mathrm{8}\right) \\ $$$${C}=\left(\mathrm{2},\mathrm{8}\right),\left(\mathrm{4},\mathrm{6}\right) \\ $$$${D}=\left(\mathrm{4},\mathrm{6}\right),\left(\mathrm{2},\mathrm{8}\right) \\ $$$${n}=\mathrm{4}×\mathrm{2}=\mathrm{8} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{8}} {\sum}}\mathrm{1000}{A}_{{i}} +\mathrm{100}{B}_{{i}} +\mathrm{10}{C}_{{i}} +{D}_{{i}} = \\ $$$$=\mathrm{1000}\Sigma{A}_{{i}} +\mathrm{100}\Sigma{B}_{{i}} +\mathrm{10}\Sigma{C}_{{i}} +\Sigma{D}_{{i}} \\ $$$$\Sigma{A}_{{i}} =\mathrm{40}\:\left[\mathrm{2}+\mathrm{2}+\mathrm{4}+\mathrm{4}+\mathrm{6}+\mathrm{6}+\mathrm{8}+\mathrm{8}\right] \\ $$$$=\mathrm{1111}×\mathrm{40}=\mathrm{44440} \\ $$
Commented by otchereabdullai@gmail.com last updated on 06/Apr/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$