Question 1204315: A three-digit number ABC is divided by the two-digit number AC. The quotient is 11 with no remainder. What is the largest possible number ABC?
Found 4 solutions by math_tutor2020, MathLover1, ikleyn, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
The three digit number ABC means
A = hundreds digit
B = tens digit
C = units or ones digit
ABC is more formally written as 100A+10B+C
AC becomes 10A+C
Divide those values and we get a quotient of 11 and no remainder.
(100A+10B+C)/(10A+C) = 11
100A+10B+C = 11(10A+C)
100A+10B+C = (10+1)(10A+C)
100A+10B+C = 10(10A+C)+1(10A+C)
100A+10B+C = 100A+10C+10A+C
10B+C = 10C+10A+C
10B = 10C+10A
0 = 10A-10B+10C
10(A-B+C) = 0
A-B+C = 0
A = B-C
To make number ABC as large as possible, we need A as large as possible.
At the same time, we need B to be as large as possible as well.
For unique single digits B and C, B-C maxes out when these digits are as far away from each other as possible, and when B > C.
That happens when B = 9 and C = 0
So A = B-C = 9-0 = 9
But A = 9 and B = 9 overlap.
We assume that A and B are different values. Otherwise the number ABC would be AAC or BBC.
Let's go for B = 9 and C = 1 instead.
A = B-C = 9-1 = 8
This would allow A,B,C to be different integers.
The number ABC = 891 is the largest possible value so that ABC/AC = 11 without a remainder (i.e. remainder is 0).
More specifically,
891/81 = 11
Answer: 891
Answer by MathLover1(20850)  (Show Source):
You can put this solution on YOUR website!
write a three-digit number  as:
and the number  (divisor) is:
We know that  with no remainder,
or
Remember that  ,  ,  , being digits of an integer number, must be whole numbers between  and  , and that isn’t zero.
 must be not  as well, otherwise  would have only two digits and not three.
as you can see,  , , and  are a multiple of  .
It’s easy to verify that the only value of that satisfies this condition is  ; all other values for  between  and  yield a value for  that is not divisible by .
Thus, if  , we can replace  for and write:
dividing both terms by :
The only values that satisfy that condition are:
. Let’s summarize:
then
check:
In fact,  with no remainder.
Answer by ikleyn(52878)  (Show Source):
You can put this solution on YOUR website! .
The @MathLover1 "solution" is one-to-one copy-paste from this web-page,
https://www.quora.com/A-three-digit-number-ABC-is-divided-by-the-two-digit-number-AC-The-quotient-is-13-with-no-remainder-What-is-the-number-ABC-if-c-0
with no reference, naturally.
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
Although a formal algebraic solution is good, and the student should be able to understand it, a very different and much easier path to the answer is possible.
Given that the 3-digit number ABC divided by the 2-digit number AC gives a quotient of 11 with no remainder, we know that AC*11 = ABC. So look at that multiplication in the way we learn multiplication in grade school.
A C
X 1 1
------
A C
A C
-------
A B C
The problem asks for the largest possible value of the 3-digit number ABC. So let's see if the condition can be satisfied if A is 9.
9 C
X 1 1
------
9 C
9 C
-------
9 B C
We can see that with A = 9, C must be 0, giving us
9 0
X 1 1
------
9 0
9 0
-------
9 9 0
So A = B = 9 and C = 0. Generally, in problems like this, it is specified that different letters represent different digits. However, that is not specified in the statement of this problem. So the answer could be
ANSWER: ABC = 990
Assuming any information that is not given in a problem is never good mathematics. However, if we assume that the letters represent different digits, then A = 9 doesn't work. So again looking for the largest possible value of the 3-digit number ABC, we try A = 8:
8 C
X 1 1
------
8 C
8 C
-------
8 B C
Here we can see that B can be at most 9, which means C can be at most 1. And since we want ABC to be the largest possible, we choose C = 1, giving us
8 1
X 1 1
------
8 1
8 1
-------
8 9 1
And in this case we have
ANSWER: ABC = 891
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And then here is another solution VERY different from the others, and MUCH faster.
Using the divisibility rule for 11 in the problem, we know immediately that
A + C = B
Then, knowing that A, B, and C are single digit integers, we immediately know that B is 9 and A + C is 9.
Then, knowing that we want ABC to be as large as possible, we quickly find ABC = 990 if A and B can be the same digit, or ABC = 891 if they can't be the same.
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