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kevincan
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Jane took a list of all the three digit positive integers Updated on: 04 Aug 2006, 12:56
Originally posted by kevincan on 01 Aug 2006, 07:26.
Last edited by kevincan on 04 Aug 2006, 12:56, edited 2 times in total.
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Jane took a list of all the three digit positive integers and crossed out all numbers that contained her unlucky number q, a positive integer less than 10. What is the value of q?

(1) When Jane added up the 3-digit integers she did not cross out, the sum was divisible by 5.
(2) When Jane added up the 3-digit integers she did not cross out, the sum was divisible by 2.



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Jane took a list of all the three digit positive integers 01 Aug 2006, 08:48
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I think this is E.

Sum of unit digits of all the three digit numbers = 90 * (Sum of digits 0 to 9)

If one digit is crossed then the above sum will be = 90 * (sum of digits that are not crossed out).
So the sum will always be divisible by 5 (Statement 1) and 2 (Statement 2).
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Jane took a list of all the three digit positive integers 01 Aug 2006, 15:02
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kevincan
ps_dahiya
I think this is E.

Sum of unit digits of all the three digit numbers = 90 * (Sum of digits 0 to 9)

If one digit is crossed then the above sum will be = 90 * (sum of digits that are not crossed out).So the sum will always be divisible by 5 (Statement 1) and 2 (Statement 2).

Don't understand this.
Show more

My bad.
Let me try again:
Let x is unlucky number.

Total numbers crossed out will be:
1. When x is at 100s position:
Total such numbers = 100
Sum of those numbers = 100*100*x + 99*50: DROP this because both terms are divisible by 5 as well as 2.

2. When x is at 10s position and NOT at 100s position:
Total such numbers = 8*10 = 80
Sum of those numbers = (45-x) * 10 * 100 + 10 * x * 80 + 8 * 45: DROP this because all terms are divisible by 5 as well as 2.

3. When x is at unit position and NOT at 10s or 100s position:
Total such numbers = 8*9 = 72
Sum of those numbers = (45-x) * 9 * 100 + 10 * x * 72 + x*72: DROP terms in bold because these are divisible by 5 as well as 2.

So we will take only 72x.

St1: If 72x is divisible by 5 then it means x = 5 or x = 0: INSUFF

St2: If 72x is divisible by 2 then x could be anything from 0 to 9.: INSUFF

Combined:
72x is divisible by 10 means x may be 5 or 0.: INSUFF

Is there a shorter way to solve? Whats the OA?
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Jane took a list of all the three digit positive integers 01 Aug 2006, 15:33
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IMO A

Assuming Jane added up the rest of the numbers after she crossed out.
0<q<10

S1. (45-q)*100*99 + (45-q)*10*9 + (45-q)*9 -4950 = 5m
Thus (45-q)*9 = 5m
45-q is multiple of 5
thus q = 5 : sufficient.

S2. (45-q)*100*99 + (45-q)*10*9 + (45-q)*9 -4950= 2n
45-q is multiple of 2
q=1,3,5,7,9 : insufficient..

Hey Kevin.. I kindda lost.. What did I do wrong??
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Jane took a list of all the three digit positive integers 01 Aug 2006, 16:25
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This question sounds challenging
A 100%

Back to simple
If you rule out "1"
the rest of number at 1St digit will be 2,3,4,5,6,7,8,9
for every 100s and 10s position
Sum of this = 2+3+4+5+6+7+8+9 = 44
this is not a real sum of all number but it is a representative
of what is SUM of all number at first digit
If you rule out "2"
SUM = 1+3+4+5+6+7+8+9 = 43
Rule out "3"
SUM = 1+2++4+5+6+7+8+9 = 42

Looks like we see a pattern
4 - 41 , 5- 40 , 6-39 , 7 -38 , 8-37 , 9-36

1-44 2-43 3-42 4-41 5-40
6-39 7-38 8-37 9-36

for the first question that divisible by 5
only "40" which is representative of "5 being ruled out "
can be divisible by 5
so we can conclude that 1 is Sufficient

the second question that divisible by 2
many representative can be divisible by 2
so this is not sufficient

this kind of balance is a pattern of 0-9 to 9-0
if no number is ruled out , sum of this can be divisible by 5 (1+2+...+9)=45 which is divisible by 5
but when any number is taken out , balance is lost
and new result occur

If this question ask what Q if the sum is divisible by 7
you can answer that Q being "3" which is ruled out
If sum is divisible by 13 Q=6
sum is DIV by 41 Q=4

I'm sorry if this is too messy to undenstand
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Jane took a list of all the three digit positive integers 01 Aug 2006, 17:01
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Clue: Focus on units digit of sum
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Jane took a list of all the three digit positive integers 01 Aug 2006, 17:45
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Found my mistake..let me try it again..
Assuming Jane added up the rest of the numbers after she crossed out.
0<q<10
Ways of arrange 3digit integer
8*9*9 thus 100s:9*9, 10s:8*9, units: 8*9

S1. (45-q)*100*(9*9) + (45-q)*10*(9*8) + (45-q)*1*(9*8) = 5m

(45-q)*1*(9*8) is multiple of 5
thus q = 5 sufficient.

S1. (45-q)*100*(9*9) + (45-q)*10*(9*8) + (45-q)*1*(9*8) = 2n
since each term is divisible by 2
q=any number : insufficient..

Thus, A.
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Jane took a list of all the three digit positive integers 04 Aug 2006, 06:36
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kevincan
Jane took a list of all the three digit positive integers and crossed out all numbers that contained her unlucky number q, a positive integer less than 10. What is the value of q?

(1) When Jane added up the numbers she did not cross out, the sum was divisible by 5.
(2) When Jane added up the numbers she did not cross out, the sum was divisible by 2.
Show more


(1) and (2) states that when Jane added up the numbers she did not cross out...Does it mean that the remaining numbers are just the digits in the 3 digit number that is not crossed out or the numbers from 1 to 9 excluding the number she crossed out.

That sort of diverted my attention. If it is referring to the remaining numbers as numbers from 1 to 9 excluding the number she crossed out, then 45-q is what we need to consider. If not the answer will be E. I guess I am going wrong in my approach somewhere.

Can anyone please explain.

Thanks.
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kevincan
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Jane took a list of all the three digit positive integers 04 Aug 2006, 12:54
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Numbers in this case refer to 3-digit integers. Thanks for the feeback- I've made the question clearer



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