I started off by trying a more general/abstract approach but it gets cumbersome (too many different cases) so I opted for a slightly less elegant route. Two restrictions are important in answering the question:
1. The units digit cannot be zero:
abc0 = 10*abc, which can't equal abc*9
2. The zero digit must be the second digit from the left, i.e., a0b, a09bc, a0bcdefg, etc.
proof: let a...bcd0e...f be a number with where a...bc and d...e represent, respectively, sequences of n and m arbitrary non-zero digits with n > 2, m > 1. Since a...bc0d...e = 9*a...bcd...e, we have that
mod10(9*c + lastCarry) = 0
where mod10(n) represents the remainder of dividing n by 10 and lastCarry the carry over from the previous step of the multiplication.
The carry over for the following step will be c. Note that lastCarry also needs to be c. A couple of examples to help clarify:
mod10(9*4 + 4) = 0
mod10(9*7 + 7) = 0
Therefore,
mod10(9*b + c) = c
The only single-digit number for which the equation above holds is 0 and therefore b = 0
Using the two statements above, we can derive the following:
a) The units digit must be 5, since 0 and 5 are the sole single-digit numbers for which mod10(9*u) = u and we have ruled 0 out.
a0b...pth5
b) Now the hundreds digit h. Let's assume h!=0, since we're interested in determining how many non-zero digits there are after the second digit from the left.
mod10(9*h + 4) = h (note the carry over is 4, from the units step of the multiplication, 9*5 = 45)
For the equation to hold, h must be either 2 or 7. This step of the multiplication will yield a carry over of 2 (h=2) or 6 (h=7)
b1) h = 2
The multiplication step for the thousands digit:
a0b...pt25
mod10(9*t + 2) = t
For the equation to hold, t must be either 1 or 6. This step of the multiplication will yield a carry over of 1 (t=1) or 5 (t=6)
Before proceeding further, note that for the equation mod10(9*s + r) = s(s, r integers, s!=r), r must be even or s must be zero.
Since 1 and 5 are both odd, p must be zero and we can stop. mod10(9*p + 1) = p
mod10(9*p + 5) = p
The numbers:
10125 50625
b2) h = 7
The multiplication step for the thousands digit:
a0b...pt75
mod10(9*t + 6) = t
For the equation to hold, t must be either 3 or 8. This step of the multiplication will yield a carry over of 3 (t=3) or 7 (t=8)
Following the same reasoning as in b1), since 3 and 7 are both odd, p must be zero and we can stop.
mod10(9*p + 3) = p
mod10(9*p + 7) = p
The numbers:
30375 70875
Answer is therefore D
PS: Although irrelevant in answering the question, the other numbers meeting the requirements:
405 2025 6075
told you... not very elegant. hope the explanation is clear enough, which I very much doubt...
A Zero is Deleted
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Source: Beat The GMAT — Problem Solving |
Good work. This is a difficult question which may not appear on the GMAT but a subtle variation of it might. I post the solution as written without any alteration ( I found a small problem with a portion of it in going over it but I leave that for all to figure out) and it must be studied for how to attack such problem. Right now my head is pounding with problems and solutins to read your solution. Will go over it later. But you did one hell of a job. I owe you a beer! In my attempt at solution I did not get the answer right answer. Below is the OA.avenus wrote:I started off by trying a more general/abstract approach but it gets cumbersome (too many different cases) so I opted for a slightly less elegant route. Two restrictions are important in answering the question:
1. The units digit cannot be zero:
abc0 = 10*abc, which can't equal abc*9
2. The zero digit must be the second digit from the left, i.e., a0b, a09bc, a0bcdefg, etc.
proof: let a...bcd0e...f be a number with where a...bc and d...e represent, respectively, sequences of n and m arbitrary non-zero digits with n>=2, m>=1. Since a...bc0d...e = 9*a...bcd...e, we have that
mod10(9*c + lastCarry) = 0
where mod10(n) represents the remainder of dividing n by 10 and lastCarry the carry over from the previous step of the multiplication.
The carry over for the following step will be c. Note that lastCarry also needs to be c. A couple of examples to help clarify:
mod10(9*4 + 4) = 0
mod10(9*7 + 7) = 0
Therefore,
mod10(9*b + c) = c
The only single-digit number for which the equation above holds is 0 and therefore b = 0
Using the two statements above, we can derive the following:
a) The units digit must be 5, since 0 and 5 are the sole single-digit numbers for which mod10(9*u) = u and we have ruled 0 out.
a0b...pth5
b) Now the hundreds digit h. Let's assume h!=0, since we're interested in determining how many non-zero digits there are after the second digit from the left.
mod10(9*h + 4) = h (note the carry over is 4, from the units step of the multiplication, 9*5 = 45)
For the equation to hold, h must be either 2 or 7. This step of the multiplication will yield a carry over of 2 (h=2) or 6 (h=7)
b1) h = 2
The multiplication step for the thousands digit:
a0b...pt25
mod10(9*t + 2) = t
For the equation to hold, t must be either 1 or 6. This step of the multiplication will yield a carry over of 1 (t=1) or 5 (t=6)
Before proceeding further, note that for the equation mod10(9*s + r) = s(s, r integers, s!=r), r must be even or s must be zero.
Since 1 and 5 are both odd, p must be zero and we can stop. mod10(9*p + 1) = p
mod10(9*p + 5) = p
The numbers:
10125 50625
b2) h = 7
The multiplication step for the thousands digit:
a0b...pt75
mod10(9*t + 6) = t
For the equation to hold, t must be either 3 or 8. This step of the multiplication will yield a carry over of 3 (t=3) or 7 (t=8)
Following the same reasoning as in b1), since 3 and 7 are both odd, p must be zero and we can stop.
mod10(9*p + 3) = p
mod10(9*p + 7) = p
The numbers:
30375 70875
Answer is therefore D
PS: Although irrelevant in answering the question, the other numbers meeting the requirements:
405 2025 6075
told you... not very elegant. hope the explanation is clear enough, which I very much doubt...
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Let the 0 be in the mth position (start counting at the 1's place).
Then n can be written as a x 10^m + b, with b < 10^m-1, and b has no
0's. Removing the 0 yields r = 10^m-1 + b. Since n = 9r, we obtain
a10^m-1= 8b. If m >= 5, then b = (1250)a x 10^m-5 which is not possible because b cannot have 0's. Therefore, m <= 4. Since b < 10^m-1, we have a10^m-1 = 8b < 8 x 10^m-1, so a < 8. Therefore, a is a single digit. Since m <= 4, there can be at most 5 digits in n. An example of such an n is 70875, and the above reasoning shows that it is the largest. The answer is d.
of course... :roll: that's what I was aiming for, but got off on the wrong foot and gave it up to pursue a more mundane, clumsier approach. After glancing at the first line of the OE, I was able to wrap it up on my own, though...sometimes the 
doesn't come that easily
.
doesn't come that easily.
