Latest Elitmus Aptitude Question SOLUTION: Q. How many 3 digit no sum will have sum 18 Option a) 51 b) 54 c) 61 d) 64

b. 54
To get sum of 3 digits equal to 18, Sum of 1st and/or 2nd digit must be greater than or equal to 9.On observing such 3 digit numbers in the range of hundred we get,
(100-200)=189,198
(201-300)=279,288,297
(301-400)=369,378,387,396
(401-500)=459,468,477,486,495
.
.
.
(900-999)=909,918,927,936,945,954,963,972,981,990

So Total numbers=2+3+4+5+.....+10=54

Please provide any short answers because we don't have this much time in exams..

sum=18. a=9,9,0(sp.case as 099 !=3digit no.) b=882 c=774 d=666
9,8,1 873 765
9,7,2...9,6,3....9,5,4 864...8,5,5
7 nos hav al digits distinct(e.g=981)4 hav 2 digits sam & 1 hav 3 same(i.e 666)
so no. of digits=7*3! +4*3!/2 + 3!/3!=55 -1=54(sp.)...ans

b). 54

(Dont get scared it is easy)

first we should list down all the combinations of nums, sum which should be 18.

start from last 3 digit num to low (i.e from 999 to 800, 700..500).

9,(0,9)

8,(9,1),(8,2),(7,3),(6,4),(5,5) :-> 8 + (10) = 18

7,(9,2),(3,8)D, (7,4),(6,5) : Here (3,8)D is duplicate because it forms (7,3,8),
but we have (8,7,3) in series which start with 8

6,(9,3),(8,4)D,(7,5)D,(6,6)

5,(9,4),(8,5)D,(7,6)D

=> all you have to do here is that, get all passible combinations of 3 digits from 1..9 which
forms a sum of 18.(just like 2 dice problems in probability, here we can assume that a dice which have 1..9 faces, then find all the cases of getting sum 18 when it throwed 3 times)

Upto 500 is enough because all the combinations which are in less than 500 are already covered.

remove duplicates:

Use the arrangement method.

9,(0,9) = 990, 909 => 2 or ( 2.2!/2! = 2 => bocz we can't set 0 in
first position)

8,(9,1),(8,2),(7,3),(6,4),(5,5) = 3! + 3!/2! + 3! + 3! + 3!/2! => 24 (becz 2 8s, two 5s)

7,(9,2),(7,4),(6,5) = 3! + 3!/2! + 3! => 15

6,(9,3),(6,6) = 3! + 3!/3! => 7 (we have three 6s).

5,(9,4) = 3! => 6


finally Sum all of them => 2 + 24 + 15 + 7 + 6 = 54

it looks little bit heavy, but one you practice you'll get the logic, and you can solve these
kind of questions easily next time.

hmm. get it.. in case of 900(*) .. copy..sum=18.a=9,9,0 b=882 c=774 d=666
9,8,1 873 765
9,7,2...9,6,3....9,5,4 864...8,5,5
7 nos hav al digits distinct(e.g=981)4 hav 2 digits sam & 1 hav 3 same(i.e 666)
so no. of digits=7*3! +4*3!/2 + 3!/3!=55-1(*) as (099 is not a 3 digit no) ...ans

..ye le paste

i think 18 is the answer

the number of ways should have always divisible by 3,2 and 6 as there are 3*2*1 ways of arranging a 3 digit number.in this particular question only one option satisfied the condition

answer-61

quick solution:
spot the pattern here.
fixing 3rd digit to 9:- 9_ _ : we get 10 numbers from 990 to 909
fixing 3rd digit to 8:- 8_ _ : we get 9 numbers from 891 to 819
fixing 3rd digit to 7:- 7_ _ : we get 8 numbers from 792 to 729
.
.
.
fixing 3rd digit to 1:- 1_ _ : we get 2 numbers from 198 to 189
this is triangular series: so n(n+1)/2 :- 10+9+8+7...1
0_ _ not valid so subtract 1.
10(11)/2 -1 = 54 answer

61 is answer

sum=18.a=9,9,0 b=882 c=774 d=666
9,8,1 873 765
9,7,2...9,6,3....9,5,4 864...8,5,5
7 nos hav al digits distinct(e.g=981)4 hav 2 digits sam & 1 hav 3 same(i.e 666)
so no. of digits=7*3! +4*3!/2 + 3!/3!=55 ...ans

awsum sala!!!55 to option hee nahi hai

54 is d answer.

ans-b.frame 3 digit no.s wth keepng the 1st digit fixed whose sum is equal to 18.
for ex-with 1 only 2 such no.s are possible.such as 189 & 198.wth 2 only 3 are possible.wth this we get the series.2+3+4+5+6+7+8+9+10=54