Digits Question

Manisha
#1. May 15th, 2008, at 12:34 PM.

Though Options are not given….but according to me….

Possible 3 Digit Nos. are: 345, 354, 435, 453, 543, 534 and hence the sum of these nos. is = 2664

SD
#2. May 15th, 2008, at 12:48 PM.

2664

Amar
#3. May 15th, 2008, at 3:57 PM.

1.B
2.C
3.E
4.C
5.E
6.C

Rajib Banerjee
#4. May 15th, 2008, at 8:56 PM.

345 + 354 + 435 + 453 + 543 + 534 = 2664

Will
#6. May 16th, 2008, at 12:09 AM.

6 different combination.
2664

Atul
#7. May 16th, 2008, at 8:11 PM.

To form the 3 digit number using 3,4,5, total no. of ways to form a number with the given number without repetation :

3*2*1 = 6

Let us suppose three digit nol is ABC (A at hundreth place, B tense place, C at Unit place)

when we form different combinations of the 3 digit numbers. No. of time every no. can appear on a particualar place (Unit, Ten etc) is

6 (Total possible combinations) /3 (Total digits available) = 2

So every no. will come 2 times in all the places (’3′ will come 2 times at A, 2 times at B and 2 times at C in total combinations. similarly 4 and 5). So sum total for ”Unit’s place” Ten’s place and Hunderd’s place individually are:

Unit’s place = 3*2 + 4*2 + 5*2 = 24
Ten’s place = 4*2 + 5*2 + 3*2 = 24
Hundred’s place = 5*2 + 3*2 + 4*2 = 24

forming the final total:

24 / 24 / 24
= 24 / (24 + 2) / 4 (Carry fwd 2 from unit’s place to ten’s place)
= 24 / 26 / 4
= (24 + 2) / 6 / 4 (Carry fwd 2 from ten’s place to Hunderd’s place)
= 26/6/4

Final answer = 2664

The method will be very effective for higher digit nos (example sum of 5 digit no. using 1,2,3,4 and 5).

jaspreet cheema
#8. May 17th, 2008, at 9:44 AM.

Total should be = 345+354+435+453+543+534=2664

Ritula
#9. May 19th, 2008, at 9:41 AM.

Since each digit can appear only once.make the combinations by fixing one no and interchanging the other two .the nos are :
345,354
435,453
534,543
Now add the above nos .The result is 2664

Charanya
#10. May 19th, 2008, at 8:04 PM.

there is a simple formula for it -

(111)(n-1)(sum of the digits)
= (111)(2)(12)
= 2664

manisha
#11. May 20th, 2008, at 12:07 AM.

@ Charanya

What is the basic concept behind this formula? whr all it can be actually used?

Vertica
#12. May 20th, 2008, at 3:22 AM.

Hey Atul

Can this method be used for any kind of problem similar to give here?
Nice concept..do you have some more tricks/ logical solutions…i would appreciate if you would share it!

Charanya
#13. May 21st, 2008, at 1:40 PM.

@manisha…
its a simple formula which someone taught me …

(111..n times)(n - 1) (Sum of digits)

The formula varies when one of the digits is zero.

I’ll post it ASAP.

Atul
#14. June 2nd, 2008, at 3:23 PM.

@vertica,
Apologies for the delay response .. Yes this method is applicable for all the scenarios.

@Charanya / manisha,
Nice formula .. there is a slight change is the formula to make it generic:

suppose n is the no. of digits given. Our scenario it is 3 (digits given 3,4,5)

Formula will be:

(n times 1) * {fact (n-1)} * (sum of digits)

explaination:
First term : ‘n times 1′

for n = 3 this means ‘3 times 1′ i.e. 111
for n = 4 this will be ‘4 times 1′ i.e. 1111

Second term : fact(n-1) ……..(factorial of (n-1)

for n = 3 this means fact(3-1) = fact (2) = 2
for n = 4 it will be !(4-1) = !3 = 6

Third term : Sum of the digits

in our case = 3+4+5 = 12

so the solution of our problem with this formula will be as Charanya explained:

= (3 times 1) (fact(3-1) (3+4+5)
= (111)(!2)(12) = 111*2*12
= 2664

This formula can be used in the scenarios were ‘0′ is also present in the given digits. Hope it will help.

manisha
#15. June 3rd, 2008, at 11:48 AM.

@ Atul

Great Explanation !!! Thanks!!

kapil dhawan
#16. June 4th, 2008, at 9:25 AM.

is this method simple or formula
the possible no. of 3,4,5 are
345
354
435
453
534
543
____
2664
____

Navya
#17. June 13th, 2008, at 2:40 PM.

@ Kapil

Your md is simpler fr this sum.. But this formula enhances our capability ot solvesuch sums even with more no of digits… and generic formulae always help .. thanks atul for the formula and explanation.. very helpful