Number Property, High Difficulty Level Problem

[bretania]

Guys, let's all try and solve this problem. For those who know already this kinds of problems, please share.

How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

(A) 31 (B) 51(C) 56(D) 62(E) 934.

[seahawk]

Whenever you divide these numbers by 9 the remainder should be 5. (Correct me if I am wrong).
so its 5,14,23,......9995
Do the maths

[Crack_Mat]

Originally Posted by bretania

Guys, let's all try and solve this problem. For those who know already this kinds of problems, please share.

How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

(A) 31 (B) 51(C) 56(D) 62(E) 934.

I think answer should be 1110...no choice given
anybody with same answer?

[element123]

Originally Posted by seahawk

Whenever you divide these numbers by 9 the remainder should be 5. (Correct me if I am wrong).
so its 5,14,23,......9995
Do the maths

Great logic..
but i think the Q asks the numbers whose digits total 5 without the need of a carry over.

lets take 4 digit numbers:
a number can have:
1) 4,1,0,0: number of ways of arranging: 8
2) 3,2,0,0: number of ways: 8
3) 2,2,1,0: 9
4) 1,1,3,0: 9
5) 5000: 1
Total number of 4 digit numbers: 35

similarly,
total 3 digit number:15
total 2 digit number: 5
1 digit number: 1

therefore total = 56

[bretania]

Element 123, sorry to ask this.. but could you kindly explain how you came up with 8 arrangements for
1) 4,1,0,0 or for 2) 3,2,0,0?

I think I am really messing up my combination basics here. Sorry and really appreciate if you could throw light on this.

And you're right, the OA here is C (56)

[desret man]

Originally Posted by bretania

Element 123, sorry to ask this.. but could you kindly explain how you came up with 8 arrangements for
1) 4,1,0,0 or for 2) 3,2,0,0?

I think I am really messing up my combination basics here. Sorry and really appreciate if you could throw light on this.

And you're right, the OA here is C (56)

for the arrangements 4100 is 8 because
0041 , 0014 , 0410 , 0140 , 0104 , 0401 . 4100 and 1400
I hope it becomes clear

[element123]

Originally Posted by bretania

Element 123, sorry to ask this.. but could you kindly explain how you came up with 8 arrangements for
1) 4,1,0,0 or for 2) 3,2,0,0?

I think I am really messing up my combination basics here. Sorry and really appreciate if you could throw light on this.

And you're right, the OA here is C (56)

I made a mistake for calculating the number of digits for 4,1,0,0 and 3,2,0,0. so here goes the expln:
For 4,1,0,0:
number of combinations: 4!/2! = 12
of these some start with 0 or 00...we need to eliminate those as they are not 4 digit numbers:
digits starting with 00: 2 (0041, 0014)
digits starting with 0: 4 (0410,0401,0140,0104)

total = 12-2-4 = 6

similarly for 3,2,0,0 = 6


there is one more combination for 4 digits: 1,1,1,2:
for this the total number is: 4!/3! = 4

therefore the total is 56.

[Crack_Mat]

Originally Posted by element123

Great logic..
but i think the Q asks the numbers whose digits total 5 without the need of a carry over.

oh that is right!!

[bretania]

Originally Posted by element123

I made a mistake for calculating the number of digits for 4,1,0,0 and 3,2,0,0. so here goes the expln:
For 4,1,0,0:
number of combinations: 4!/2! = 12
of these some start with 0 or 00...we need to eliminate those as they are not 4 digit numbers:
digits starting with 00: 2 (0041, 0014)
digits starting with 0: 4 (0410,0401,0140,0104)

total = 12-2-4 = 6

similarly for 3,2,0,0 = 6


there is one more combination for 4 digits: 1,1,1,2:
for this the total number is: 4!/3! = 4

therefore the total is 56.

Element, thanks again!

[CyberSpy]

Element, instead of considering 2 digit nos and 3 digit nos seperately this problem can simply be solved by considering XXXX and filling these four places with the required digits. I am able to wok this out as follows,

1,4

4x3 = 12 possible values.

Similarly for 3,2 12 values
Those nos with zeros preceding them are the reqd 2 digit and 3 digit nos. you dont have to calculate them seperately. All you have to do id to do away with the subtraction part.



For 1,2,2

1220
1202
1022
2120
0122
0212
2102
2012
2210
2021
2201
0221

i.e 4x3= 12(4 positions for 1 and 3 postions for the rest of the numbers for each positionsof 1).Can someone give me a better explanation of this? Im not satisfied with mine.
Similar 12 combinations for 1,1,3

4 combinations each for 1,1,1,2 and 0,0,0,5

Adding these we get 56.(12+12+12+12+4+4).