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Question Number 166102 by Tawa11 last updated on 13/Feb/22

Answered by mr W last updated on 13/Feb/22

we have totally 12 rolls for the numbers  1,2,3,4,5,6, the sum of them is 47.   only one number occurs 3 times. it′s  easy to get that an other number must  occur only one time and all other  numbers must occur 2 times each.  say the number X occurs 3 times,  and the number Y occurs one time,  then we have  2(1+2+3+4+5+6)+X−Y=47  X−Y=5  the only possibility is X=6, Y=1.  that means 6 occurs 3 times, 1 occurs  one time, all others occur 2 times each.

$${we}\:{have}\:{totally}\:\mathrm{12}\:{rolls}\:{for}\:{the}\:{numbers} \\ $$$$\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\:{the}\:{sum}\:{of}\:{them}\:{is}\:\mathrm{47}.\: \\ $$$${only}\:{one}\:{number}\:{occurs}\:\mathrm{3}\:{times}.\:{it}'{s} \\ $$$${easy}\:{to}\:{get}\:{that}\:{an}\:{other}\:{number}\:{must} \\ $$$${occur}\:{only}\:{one}\:{time}\:{and}\:{all}\:{other} \\ $$$${numbers}\:{must}\:{occur}\:\mathrm{2}\:{times}\:{each}. \\ $$$${say}\:{the}\:{number}\:{X}\:{occurs}\:\mathrm{3}\:{times}, \\ $$$${and}\:{the}\:{number}\:{Y}\:{occurs}\:{one}\:{time}, \\ $$$${then}\:{we}\:{have} \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}\right)+{X}−{Y}=\mathrm{47} \\ $$$${X}−{Y}=\mathrm{5} \\ $$$${the}\:{only}\:{possibility}\:{is}\:{X}=\mathrm{6},\:{Y}=\mathrm{1}. \\ $$$${that}\:{means}\:\mathrm{6}\:{occurs}\:\mathrm{3}\:{times},\:\mathrm{1}\:{occurs} \\ $$$${one}\:{time},\:{all}\:{others}\:{occur}\:\mathrm{2}\:{times}\:{each}. \\ $$

Commented by Tawa11 last updated on 13/Feb/22

God bless you sir. I appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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