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- mathmath333

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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- mathmath333

- katieb

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- mathmath333

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

- mathmath333

i am not able to interpret the que

- hartnn

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

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- hartnn

***
total options = 2*2*1 = 4

- hartnn

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
\(\large 5\times 5 \times 4 \times 3 \times 2\times 1 = ..\)

- hartnn

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 )

- hartnn

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

- hartnn

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

- hartnn

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

- hartnn

unit digit = 2 ............120
unit digit = 3 ............120
unit digit = 4 ............48
unit digit = 5 ............120
unit digit = 6 ............120
120*4+48 = 528 :)

- hartnn

go through the steps slowly...
and let me know if you have any doubt in nay step..

- mathmath333

hey thanks!

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