Latest TCS Aptitude Question SOLUTION: How many five digit multiples of 3 are formed using 0, 1, 2, 3, 4, 6 and 7 without repetition? 1) 312 2) 3125 3) 120 4) 216

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How many five digit multiples of 3 are formed using 0, 1, 2, 3, 4, 6 and 7 without repetition?


1) 312
2) 3125
3) 120
4) 216

Asked In TCS sravani (10 years ago)

Unsolved Read Solution (18)

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Read Solution (18) :

shanmuga sundaram  (10 years ago)

0,1,2,3,4,6,7 - seven digits are given. we need only 5 digits.
if a number to be divisible by 3.the sum of their digits must be divisible by 3.
0+1+2+3+4+6+7 = 23
we have to leave two digits, we have to select 5 out 0f 7 ,7c5 = 35 ways.
minimum pair {0,1},sum = 1
maximum pair {6,7},sum=13
sum varies from 1 to 13.
i.e)23-1 to 23-13 =22 to 10
in that sum of 21,18,15,12 are divisible
pairs that produces sum of 21 is {0,2}
{1,3,4,6,7} 5!=120
pairs that produces sum of 18 are {1,4} & {2,3}
{0,2,3,6,7} 5! - 4!(since the first digit should not be zero)=96
{0,1,4,6,7} similarly 96
pairs that produces sum of 15 are {1,7},{2,6}
{0,2,3,4,6} 96
{0,1,3,4,7} 96
pairs that produces sum of 12 are {4,7}
{0,1,2,3,6} 96

Total no of numbers formed=120+(96*5)=600

Is this solution Helpfull? Yes (23) | No (20)  

SIVA KUMAR  (10 years ago)

For 3 multiple the sum should be divisible by 3
So {0,1,2,3,4,6,7} produces a set {1,3,4,6,7} which sum is 21 divisible by 3
Set {1,3,4,6,7} has 5 combination then 5!= 120 answer is option c)120

Is this solution Helpfull? Yes (17) | No (9)  

poojitha  (10 years ago)

every one's approach is different and you all got different set of answers but, i didn't get the solution from these answers.

Is this solution Helpfull? Yes (12) | No (2)  

lipsa mishra  (10 years ago)

4×4×3×2×2=172

Is this solution Helpfull? Yes (5) | No (9)  

arvind kumar  (10 years ago)

if zero is left out 5!=120 and 3 is left out
then it will be 4*4!=96
total 216
ans (4)

Is this solution Helpfull? Yes (3) | No (10)  

VIGNESH S  (10 years ago)

it is 5 digit number so 1st digit is always start with 1,2,3,4,5... then if a number is divisible by 3 sum of their digits is always divisible by 3...
the following chances are:
1){0,1,2,3.6}=4*4*3*2*1=96
2){1,3,4,6,7}=5*4*3*2*1=120
3){0,3,5,6,7}=4*4*3*2*1=96
4){0,2,3,6,7}=4*4*3*2*1=96
5){0,1,3,4,7}=4*4*3*2*1=96
6)(0,2,4,5,7}=4*4*3*2*1=96
7){1,2,5,6,7}=5*4*3*2*1=120

the total chances are=720
therefore 720 five digits are formed using 0, 1, 2, 3, 4, 6 and 7 without repetition divisible by 3...

Is this solution Helpfull? Yes (3) | No (6)  

rahul  (9 years ago)

if sum of 5 digits number is divisible by 3 then such number is also multiple of 3 .except o we can put any no on 10000's position .
let us consider 10000's place will be fill by one and other 4 position will fill by other remaining digits,remaining digits also rearranged between each other by 5! ways.
similarly we can put the all remaining 5 digits in 10000's place .
so total no of such combination is 6*5!=600

Is this solution Helpfull? Yes (2) | No   

satish  (9 years ago)

We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.

We have six digits: 0,1,2,3,4,5. Their sum=15.

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.

Second step:

We have two set of numbers:
{1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets:

{1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.

{0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=96

120+96=216

Is this solution Helpfull? Yes (2) | No (1)  

lalli kathirvel  (9 years ago)

0,1,2,3,4,6,7 -seven digits
we have fill in five palaces
first place- 5
second place-5
third place-4
fourth place-3
fifth place-2
Therefore 5*5*4*3*2=600

Is this solution Helpfull? Yes (2) | No (1)  

yukti  (10 years ago)

if 0 is used
4*4*3*2*2=192
if 1 is used 5!=120
therefore total ways 312

Is this solution Helpfull? Yes (1) | No (3)  

manu k r  (10 years ago)

312MULTIPLES

Is this solution Helpfull? Yes (1) | No (1)  

aastha huria  (9 years ago)

0+1+2+3+4+6+7=23
the set of5 digits that could be possible is: (1,3,4,6,7) and (2,3,4,6,0)
(1,3,4,6,7) can be arranged in 5! ways= 120
(2,3,4,6,0) can be arranged in 4x4x3x2x1 ways= 96
so no. of digits are= 120+96=216
option d is correct.

Is this solution Helpfull? Yes (1) | No   

SURAJ BALANI  (7 years ago)

Correct me if i m wrong....
Given set,{0,1,2,3,4,6,7}
first set will be {0,1,2,3,6} - ways 96 (4*4*3*2*1)
second set will be {1,3,4,6,7} - ways 120 (5!)
third set will be {1,3,4,7,0}- ways 96 similar as above.
Total ways= 96+96+120
=>312

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Devesh  (6 years ago)

for sum = 21 (i.e divisible by 3 no. are : 1,3,4,6,7,)
for sum = 18 (no are (0,2,3,6,7) or(1,4,6,7,0))

(5*4*3*2) + 2(4*4*3*2) = 312

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ajit jha  (10 years ago)

i agree with you poojitha all the answers are wrong

Is this solution Helpfull? Yes | No (2)  

karthik patel  (8 years ago)

irst step:
We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.

We have six digits: 0,1,2,3,4,5. Their sum=15.

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.

Second step:

We have two set of numbers:
{1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets:

{1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.

{0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=96

120+96=216

Answer: E.

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gaurav  (7 years ago)

the answer has to be by method as used by shanmuga

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Ankit singh  (6 years ago)

Required numbers are = 5! + 5! - 4! = 216

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