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Question Number 83910 by redmiiuser last updated on 07/Mar/20
find all 6 digit numbers which are not  only palindrome but also divisible by 495.
$$\mathrm{find}\:\mathrm{all}\:\mathrm{6}\:\mathrm{digit}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{only}\:\mathrm{palindrome}\:\mathrm{but}\:\mathrm{also}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{495}. \\ $$
Commented by redmiiuser last updated on 08/Mar/20
CAN ANYONE   ANSWER THIS  QUESTION.
$$\mathrm{CAN}\:\mathrm{ANYONE}\: \\ $$$$\mathrm{ANSWER}\:\mathrm{THIS} \\ $$$$\mathrm{QUESTION}. \\ $$
Answered by Kunal12588 last updated on 08/Mar/20
there are 11 such numbers  504405, 513315, 522225, 531135, 540045,  549945, 558855, 567765, 576675, 585585,  594495
$${there}\:{are}\:\mathrm{11}\:{such}\:{numbers} \\ $$$$\mathrm{504405},\:\mathrm{513315},\:\mathrm{522225},\:\mathrm{531135},\:\mathrm{540045}, \\ $$$$\mathrm{549945},\:\mathrm{558855},\:\mathrm{567765},\:\mathrm{576675},\:\mathrm{585585}, \\ $$$$\mathrm{594495} \\ $$
Commented by redmiiuser last updated on 08/Mar/20
GREAT
$${GREAT} \\ $$
Commented by naka3546 last updated on 08/Mar/20
Show  your  workings ,  please .
$${Show}\:\:{your}\:\:{workings}\:,\:\:{please}\:. \\ $$
Commented by Kunal12588 last updated on 08/Mar/20
i don′t think i can find it with any method.  I just used the bruteforce method.
$${i}\:{don}'{t}\:{think}\:{i}\:{can}\:{find}\:{it}\:{with}\:{any}\:{method}. \\ $$$${I}\:{just}\:{used}\:{the}\:{bruteforce}\:{method}. \\ $$
Answered by redmiiuser last updated on 08/Mar/20
there is an easy way  we know   495=9×5×11  as the number must  be divisible by 495 hence  so by 9 and 5 and 11.  to be divisible by 5   the last dight must be  5 and not 0.  again for 9 the sum of  the dights must be divisi6le  by 9.  now for 11 lets take   an example.  594495   here the number is  divisible by 5 and 9.  but the number is divisible  by 11 on applying the  divisiblity trick i get this.  again in 495594  the number is divisible by  11and5and9.  but the change is   the terminal numbers  are decreased by 1   while middle ones   are increased by 1.  hence by applying  this concept we can solve
$${there}\:{is}\:{an}\:{easy}\:{way} \\ $$$${we}\:{know}\: \\ $$$$\mathrm{495}=\mathrm{9}×\mathrm{5}×\mathrm{11} \\ $$$${as}\:{the}\:{number}\:{must} \\ $$$${be}\:{divisible}\:{by}\:\mathrm{495}\:{hence} \\ $$$${so}\:{by}\:\mathrm{9}\:{and}\:\mathrm{5}\:{and}\:\mathrm{11}. \\ $$$${to}\:{be}\:{divisible}\:{by}\:\mathrm{5}\: \\ $$$${the}\:{last}\:{dight}\:{must}\:{be} \\ $$$$\mathrm{5}\:{and}\:{not}\:\mathrm{0}. \\ $$$${again}\:{for}\:\mathrm{9}\:{the}\:{sum}\:{of} \\ $$$${the}\:{dights}\:{must}\:{be}\:{divisi}\mathrm{6}{le} \\ $$$${by}\:\mathrm{9}. \\ $$$${now}\:{for}\:\mathrm{11}\:{lets}\:{take}\: \\ $$$${an}\:{example}. \\ $$$$\mathrm{594495}\: \\ $$$${here}\:{the}\:{number}\:{is} \\ $$$${divisible}\:{by}\:\mathrm{5}\:{and}\:\mathrm{9}. \\ $$$${but}\:{the}\:{number}\:{is}\:{divisible} \\ $$$${by}\:\mathrm{11}\:{on}\:{applying}\:{the} \\ $$$${divisiblity}\:{trick}\:{i}\:{get}\:{this}. \\ $$$${again}\:{in}\:\mathrm{495594} \\ $$$${the}\:{number}\:{is}\:{divisible}\:{by} \\ $$$$\mathrm{11}{and}\mathrm{5}{and}\mathrm{9}. \\ $$$${but}\:{the}\:{change}\:{is}\: \\ $$$${the}\:{terminal}\:{numbers} \\ $$$${are}\:{decreased}\:{by}\:\mathrm{1}\: \\ $$$${while}\:{middle}\:{ones}\: \\ $$$${are}\:{increased}\:{by}\:\mathrm{1}. \\ $$$${hence}\:{by}\:{applying} \\ $$$${this}\:{concept}\:{we}\:{can}\:{solve} \\ $$
Commented by redmiiuser last updated on 08/Mar/20
i am sorry it must be  549945 not 495594
$${i}\:{am}\:{sorry}\:{it}\:{must}\:{be} \\ $$$$\mathrm{549945}\:{not}\:\mathrm{495594} \\ $$

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