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mathmath333

  • one year ago

Find the number of 6 digit number that can be found using the digits 1,2,3,4,5,6 once such that the six digit number is divisible by its unit digit .(The unit digit is not 1.)

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Find the number of 6 digit number that can be found using the digits }\hspace{.33em}\\~\\ & \normalsize \text{1,2,3,4,5,6 once such that the six digit number is divisible by its }\hspace{.33em}\\~\\ & \normalsize \text{unit digit .(The unit digit is not 1.)}\hspace{.33em}\\~\\ & a.)\ 620 \hspace{.33em}\\~\\ & b.)\ 456 \hspace{.33em}\\~\\ & c.)\ 520 \hspace{.33em}\\~\\ & d.)\ 528 \hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    i am not able to interpret the que

  3. hartnn
    • one year ago
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    A smaller version of the same problem is to consider only 3 digits, 1,2,3 lets first find how many total 3 digit number can be formed using 1,2,3 only once and digit place cannot be 1 so _ _ _ out of these 3 places, the last digit(units place) has 2 options (2,3) middle place also has 2 options (1, digit not used in units place) and hundred's place only has one option. so total options = 2+2 = 4 -------------- now lets break it last digit = 2 _ _ 2 only 2*1 possible numbers (132, 312) and both are divisible by 2 _ _ 3 only 2*1 possible numbers (123, 213) and both are divisible by 3 so for 3 digit case, there are total 4 such numbers and all 4 of them are divisible by their unit's digit

  4. hartnn
    • one year ago
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    *** total options = 2*2*1 = 4

  5. hartnn
    • one year ago
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    for a 6 digit case, |dw:1441517617329:dw| so total number of 6 digit numbers with all digit repeating only once and 1 is not in unit's place is \(\large 5\times 5 \times 4 \times 3 \times 2\times 1 = ..\)

  6. hartnn
    • one year ago
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    now lets break it the units digit is only 6 _ _ _ _ _ 6 so there are 5*4*3*2*1 such numbers. (same is true for units digit = 5 or 4 or 3 or 2 )

  7. hartnn
    • one year ago
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    easiest case is for units digit = 2 and 5 because all the numbers ending in 2 or 5 are divisible by 2 or 5 respectively

  8. hartnn
    • one year ago
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    upon further thoughts since 1+2+3+4+5+6 =21 which is divisible by 3 all the 600 numbers will be divisible by 3! which also means, unit's digit = 3 and 6 are also the easiest :P what remains is just 4

  9. hartnn
    • one year ago
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    for 4 we consider last 2 digits ____x4 x= 1, ____14 will not be divisible by 4 x= 2, ____24 will be divisible by 4 x= 3, ____34 will not be divisible by 4 x= 5, ____54 will not be divisible by 4 x= 6, ____64 will be divisible by 4 hence only 2/5 th of 120 numbers with unit digit as 4, will be divisible by 4

  10. hartnn
    • one year ago
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    unit digit = 2 ............120 unit digit = 3 ............120 unit digit = 4 ............48 unit digit = 5 ............120 unit digit = 6 ............120 120*4+48 = 528 :)

  11. hartnn
    • one year ago
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    go through the steps slowly... and let me know if you have any doubt in nay step..

  12. mathmath333
    • one year ago
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    hey thanks!

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