Exactly one of the digits of an integer n is 0. Deleting this zero yields an integer with 9r=n. What is the number of digits in the largest such n?
A. 2
B. 3
C. 4
D. 5
E. 6
A. 2
B. 3
C. 4
D. 5
E. 6
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...
doesn't come that easily
