18-04-09 Sum of all numbers

This is a bit dirty but "standard" question if I could call it so

You must have seen it many times

find the sum of all 6 digit numbers that can be formed using

1,1, 3, 4, 6, 9

15 Answers

9
Celestine preetham ·

do you want jee aspirants or jee finished ones to try ????

11
virang1 Jhaveri ·

There will be 5! times for each number in units place .
Sum of the unit = 5!(1+1+3+4+6+9) = 5!(24)

There will be 5! times each number in the tens place
Sum = 50*4!(24)

There will be 5! times each number in the hundreds place
Sum = 400*5*3!(24)

There will be 5! times each number in the thousands place
Sum = 4000*5*3*2!(24)

There will be 5! times each number in the ten thousands place
Sum = 5*4*3*2*1!(24)

There wil be 5! times each number in 1 lakhs place
Sum = 5!(24)*100000
Total
5!(24)*100000+5!(24)*10000 + 5!*1000(24) + 100*5!(24) + 10*5!(24) + 5!(24)
5!(24)[111111]

The sum of all 6 digit numbers that can be formed using

1,1, 3, 4, 6, 9 is = 5!(24)[111111]

11
virang1 Jhaveri ·

Bhaiyya check it out

1
ANKIT MAHATO ·

itna confident hai toh sahi hi hoga !

62
Lokesh Verma ·

well virang.. I think there is a small mistake there..

will the unit digit be repeated that many no of times for every digit>?

24
eureka123 ·

is repetition allowed????????????????????????????????????

62
Lokesh Verma ·

repitition is allowed exactly as many times as the numbers appear

11
virang1 Jhaveri ·

Sir take any one number in the units place
Lets say we take 9
Then the remaining 5 numbers can be rearranged amongst itself in 5!
Therefore all the numbers will be repeated 5! times

Is my approach rite?

11
virang1 Jhaveri ·

Since there are 2 '1's therefore the final answer should be divided by 2!
5!(24)[111111] /2
5!*12*111111

62
Lokesh Verma ·

virang but here there are 2 one's

so that will cause a trouble..

11
virang1 Jhaveri ·

Sir check post #10 you will have to divide the answer by 2

62
Lokesh Verma ·

virang .. i think the treatment will be different for digits 1 and the other digits..

Think once..

11
virang1 Jhaveri ·

ok 1 has to be considered only once

ok
Therefore the total will be
5!(23)*111111/2

9
Celestine preetham ·

ans is 222222 X 6! not as dirty as i thought it to be

fix a place
if occupied by 1 tot nos possible = 5! contributes total 5!
if occ by others (3,4,6,9) nos possible =5!/2 contributes 11X5!

tot per digit = 12X5!

so grand total is 111111 X12X5!

62
Lokesh Verma ·

Good work celestine..

It would have become dirtier if there was a zero.. that is why i did not include a zero in the question..

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