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Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862.

My Approach:

single digit no formed = 2,4,6,8

sum= 2+4+6+8= 20

two digit= 24,26,28,42,46,48,62,64,68,82,84,86

sum= 660

three digit no=246+264+426+462+624+642=2664

268+286+628+682+826+862=3552

248+284+428+482+824+842=3108

468+486+648+684+846+864=3996

sum of all 3 digit nos =13310

Similarly for all 4 digit numbers.

Community
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justin takro
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5 Answers5

21

There are:

  • four $1$-digit numbers, whose average is $5$: sum = $20$
  • twelve $2$-digit numbers, whose average is $55$: sum = $660$
  • twenty-four $3$-digit numbers, whose average is $555$: sum = $13320$
  • twenty-four $4$-digit numbers, whose average is $5555$: sum = $133320$

To see why these averages are correct, you can pair off each number uniquely with its 'complement' obtained by exchanging $2$ with $8$ and $4$ with $6$. The average of each of these pairs is $5\dots5$.

TonyK
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    This solution is missing why there are 4/12/24/24 numbers with 1/2/3/4 digits. Some mentioning to $\frac{4!}{(4-i)!}$ and why it is so. – Timbo Aug 25 '15 at 15:27
  • @Timbo: Feel free to post your own answer. This is how I chose to post mine $-$ I think interested readers can work such details out for themselves. – TonyK Aug 25 '15 at 23:10
  • @Timbo u added this part which is missing but how u get this?please explain in detail. – justin takro Aug 26 '15 at 05:30
  • @TonyK i was actually left speechless after seeing this mathematical brilliance but except for the clear patterns of averages (5,55,555,5555) i find the sums (20,660,13320,133320) following a neat pattern also...is it so?? Do you know what it is..?? – Freelancer Nov 04 '15 at 15:30
10

To sum all the numbers with four digits, note that for any given digit in any given position, it appears in the sum six times. That is, for example, there are six ways to complete a number if we know that its third digit is $4$. Then, the sum is $$6(2222+4444+6666+8888)=6(2+4+6+8)1111$$

ajotatxe
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  • How 6?Could not note that point.could u explain that point in detail a bit? – justin takro Aug 25 '15 at 10:51
  • @justintakro Whatever the first digit is, there are 3 remaining choices for the second digit, 2 for the third, and 1 choice left for the last. – Mike Aug 25 '15 at 11:06
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    @Mike why whatever first digit?It can be arranged in 4 ways,then 3 ways for second digit then 2 ways for 3rd and then 1 way for last. – justin takro Aug 26 '15 at 05:25
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    @justintakro For example, try to figure out how many 4 digit numbers have an 8 as the first digit. There are 6 permutations for the remaining digits, so an 8 in the first digit needs to be added in 6 times. That's where the 6 comes in. The same logic applies for calculating how many 4 digit numbers have an 8 as the second, third, or fourth digit, which is where the 8888 comes from. Ditto for choosing something other than 8. – Mike Aug 26 '15 at 06:18
1

I guess this might help.

sum of number by taking $n$ digits all at a time $= (n-1!)(x_1+ x_2 + x_3 + \cdots + x_n)$

so if we want to find the sum of all the number formed by $2,4,6$ and $8$ without repetition, "without repetition". [ ( 2 + 4 + 6 + 8 )( 3! ) = 120. ] Now multiply this with the place values = $120 ( 10^0 + 10^1 + 10^2 + 10^3 )$

if we want to find the sum of all the number formed by $2,4,6$ and $8$ without repetition, "with repetition", then

$( 2 + 4 + 6 + 8 )( 4*4*4 )$ (keeping $2$ at units place _ _ _ $2$), and the next step is same.

Singhal
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dovahkiin
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0

Simple formula :- (n-1)! * Sum of digits * 1111 n times. That is, (4-1)! * (2+4+6+8)* 1111 = 6*20*1111 = 1,33,320

user567831
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Consider an example by taking digits as $3,4,6,8$. total combinations possible$=4\times 3\times 2=24$ numbers possible no of digits$=4$ hence $\dfrac{24}{4}$ each digit comes $6$ times in ones, tens and hundred's place at ones place=$6\times (3+4+6+8)=126\equiv6$. at ten's place$=6\times (3+4+6+8)+12(\text{carry})=138\equiv 8$. At hundred's place$=6\times (3+4+6+8)=139\equiv 9$ so, $3$ will come at thousands place and $1$ at ten thousands place, Hence, the number is $139864$.

BLAZE
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user323895
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