Latest Elitmus Aptitude Question SOLUTION: How many six digit numbers can be formed using the digits 1to 6,without repetition such that the no is divisible by the digit at its units place? a.402 B.528 C.648 D.720

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(#M40151799) ELITMUS Numerical Ability question Number system Keep an EYE

How many six digit numbers can be formed using the digits 1to 6,without repetition such that the no is divisible by the digit at its units place?
a.402 B.528 C.648 D.720

Asked In Elitmus (9 years ago)

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siddharth prakash singh  (9 years ago)

ans is c.648
sol: when unit digit is 1.
_ _ _ _ _ 1. remaining 5 digit can be filled by 5!=120 ways.
when unit digit is 2.
_ _ _ _ _ 2. remaining 5 digit can be filled by 5!=120 ways.
when unit digit is 3.
_ _ _ _ _ 3.(as sum of all digit is divisible by 3) remaining 5 digit can be filled by 5!=120 ways.
when unit digit is 4.
_ _ _ _ _ 4. tens place can be filled by only 2 ways i.e wid 2 and 6 only. remaining 4 digit can be filled by 4!=24 ways. so total n of ways for divisible by 4=2*24=48
when unit digit is 5.
_ _ _ _ _ 5. remaining 5 digit can be filled by 5!=120 ways.
when unit digit is 6.
_ _ _ _ _ 6. remaining 5 digit can be filled by 5!=120 ways.
so total no of ways=120*5 +48= 648 ways

Is this solution Helpfull? Yes (41) | No (7)  

siddharth prakash singh  (9 years ago)

@SOUPARNO MUKHERJEE. if 6 will be at unit place then it will be divided by 6, as for divisibility of a no by 6, no should be divisible 2 & 3 both.
since 6 is a even no, and present at unit place then the whole no will be divisible by 6, and as sum of all digit frm 1 to 6 is 21 which is divisible by 3 hence the no will be divisible by 3.

Is this solution Helpfull? Yes (3) | No   

Rohan Kumar  (8 years ago)

One way to get the answer:
XXXXX1 is always divisible by 1, so we have 5! numbers.
XXXXX2 is always divisible by 2, so we have 5! numbers.
XXXXX3 is always divisible by 3 (sum of digits is always 21), so we have 5! numbers.
XXXXY4 is divisible by 4 only if Y is 2 or 6, so we have 2 * 4! numbers.
XXXXX5 is always divisible by 5, so we have 5! numbers.
XXXXX6 is always divisible by 6 (even number divisible by 3), so we have 5! numbers.

So total number of numbers with required property = 5 * 5! + 2* 4! = 600 + 48 = 648 numbers.

Another way to get the answer:
Almost all number have the required property except for:
XXXXY4, where Y is 1 or 3 or 5.
So 3 * 4! number do not have the required property.
Number of 6 digits numbers = 6!

So total number of numbers with required property = 6! - 3 * 4! = 720 - 72 = 648 numbers.

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Souparno Mukherjee  (9 years ago)

@ SIDDHARTH what logic did you apply for digit 6 at unit's place?

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Souparno Mukherjee  (9 years ago)

Thankyou @ Siddharth prakash singh

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Kaustav Chaki  (8 years ago)

Ans is c.648
Sol:
taking units digit as 1: __ __ __ __ __ 1 all the numbers are divisible by 1 hence 5! ways=120
taking units digit as 2: __ __ __ __ __ 2 all the numbers formed will be divisible by 2 hence 5! ways=120
taking units digit as 3: __ __ __ __ __ 3 sum of the digits 1+2+3+4+5+6=21 hence all numbers formed will be divisible by 3 hence 5! ways=120
taking units digits as 4: __ __ __ __ __ 4 last two digits must be divisible by 4 hence 24 and 64 being the only possibilities hence 4!*2 =48
taking units digit as 5: __ __ __ __ __ 5 all numbers formed will be divisible by 5 hence 5! ways =120
taking units digit as 6 : __ __ __ __ __ 6 numbers must be evenly divisible by 3. Since all numbers are divisible by 3 and since 6 is even all numbers formed will be divisible by 2 hence all numbers formed will be divisible by 6 hence 5! ways=120
therefore total no of ways=120+120+120+48+120+120=648

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Satyam  (9 years ago)

Correct answer will be 528 BCUZ UPTO 5 528 Is total number,for 6 No should be divisible by both 2 and 3,n this is not posdible in this case.

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Puja Kumari  (8 years ago)

C. 648 is the correct answer

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