Jane took a list of all the three digit positive integers

Question

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.

ps_dahiya · Answer

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).

ps_dahiya · Answer

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?

freetheking · Answer

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??

CHEN · Answer

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

kevincan · Answer

Clue: Focus on units digit of sum

freetheking · Answer

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.

yessuresh · Answer

(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.

kevincan · Answer

Numbers in this case refer to 3-digit integers. Thanks for the feeback- I've made the question clearer

Jane took a list of all the three digit positive integers

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