## 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.)

1. 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}}

2. mathmath333

i am not able to interpret the que

3. 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

4. hartnn

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

5. 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 $$\large 5\times 5 \times 4 \times 3 \times 2\times 1 = ..$$

6. 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 )

7. 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

8. 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

9. 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

10. 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 :)

11. hartnn

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

12. mathmath333

hey thanks!