division problem

nanase

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I am always clueless in doing this kind of problem
how are you supposed to intelligently guess the numbers?
 
Next, you might list possible pairs (A, C) such that their product ends with C. Examples are 6*4=24 and 3*5 = 15.
 
A*4C = BCC which is a three digit number. Since 4*25 =100 we must have AC>25. So the possibilitles for A and C are 6 & 7 or 6&8 or 6&9 or 7&8 or 7&9 or 8&9 (The 1st numbers I listed are not necessarily for A). Now 4AC is a multiple of 4 and ends in C. How does this help?
 
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I am always clueless in doing this kind of problem
how are you supposed to intelligently guess the numbers?
It helps to simplify the notation a bit, and not much trial and error is needed. (Perhaps none is, but I used some.)

Each of a, b, and c is a positive integer less than 10, and none is equal to the either of the others.

[MATH]x = 100b + 10c + c = 100b + 11c \text { and } y = 40 + c \implies 122 \le x \in \mathbb Z,\ y \in \mathbb Z, \text { and } 41 \le y \le 49.[/MATH]
It is given that [MATH]y = \dfrac{x}{a} \implies ay = x \implies x \le 9 * 49 = 441 \implies 122 \le x \le 433.[/MATH]
[MATH]\text {If } a = 5, \text { integer } y = \dfrac{x}{5} = \dfrac{100b + 11c}{5} = 20b + 2c + \dfrac{c}{5}.[/MATH]
But no permissible value of c will cancel 5 so we have a contradiction. Thus, a is not 5.

Playing around with

[MATH]41a \le ay \le 49a \text { and } ay = x = 100b + 11c \ge 122 \implies[/MATH] we find the following:

[MATH]a \ge 3,[/MATH]
[MATH]b = 1 \iff a = 3 \text { or } 4,[/MATH]
[MATH]a = 6 \implies b = 2,[/MATH]
[MATH]a = 7 \implies b = 2 \text { or } 3,[/MATH]
[MATH]a = 8 \implies b = 3,\text { and}[/MATH]
[MATH]a = 9 \implies b = 3 \text { or } 4,[/MATH]
[MATH]\text {If } a = 4, \text { integer } y = \dfrac{x}{4} = \dfrac{100b + 11c}{4} = 25b + 2c + \dfrac{3c}{4}.[/MATH]
The only permissible value of c that will cancel 4 is 8, which entails that

[MATH]y = 40 + 8 = 48 \text { and } x = 100 * 1 + 11 * 8 = 188, \text { but } 4 * 48 = 192 \ne 18[/MATH][MATH]8[/MATH].

We have a contradiction. Thus, a is not 4. In short, a = 3, 6, 7, 8, or 9.

[MATH]a(40 + c) = ay = x = 100b + 10c + c \implies ac = 10d + c.[/MATH]
Now we run through the multiplication tables and find

[MATH]a = 3 \implies c = 5 \implies x = 155 \text { and } y = 45, \text { but } 3 * 45 = 135 \ne 155,[/MATH]
So a is not 3.

[MATH]a = 7 \implies c = 5 \implies x = 355 \text { and } y = 45, \text { but } 7 * 45 = 315 \ne 355,[/MATH]
So a is not 7. And a = 8 does not work at all.

[MATH]a = 9 \implies c = 5 \implies x = 455 \text { and } y = 45, \text { but } 9 * 45 = 405 \ne 455,[/MATH]
Thus a = 6. So b = 2, and c must equal 2, 4, or 8. But c cannot equal b so c must be 4 or 8.

[MATH]44 * 6 = 264 \ne 244.[/MATH]
So c must equal 8. Let's check.

[MATH]48 * 6 = 288.[/MATH]
 
It helps to simplify the notation a bit, and not much trial and error is needed. (Perhaps none is, but I used some.)

Each of a, b, and c is a positive integer less than 10, and none is equal to the either of the others.

[MATH]x = 100b + 10c + c = 100b + 11c \text { and } y = 40 + c \implies 122 \le x \in \mathbb Z,\ y \in \mathbb Z, \text { and } 41 \le y \le 49.[/MATH]
It is given that [MATH]y = \dfrac{x}{a} \implies ay = x \implies x \le 9 * 49 = 441 \implies 122 \le x \le 433.[/MATH]
[MATH]\text {If } a = 5, \text { integer } y = \dfrac{x}{5} = \dfrac{100b + 11c}{5} = 20b + 2c + \dfrac{c}{5}.[/MATH]
But no permissible value of c will cancel 5 so we have a contradiction. Thus, a is not 5.

Playing around with

[MATH]41a \le ay \le 49a \text { and } ay = x = 100b + 11c \ge 122 \implies[/MATH] we find the following:

[MATH]a \ge 3,[/MATH]
[MATH]b = 1 \iff a = 3 \text { or } 4,[/MATH]
[MATH]a = 6 \implies b = 2,[/MATH]
[MATH]a = 7 \implies b = 2 \text { or } 3,[/MATH]
[MATH]a = 8 \implies b = 3,\text { and}[/MATH]
[MATH]a = 9 \implies b = 3 \text { or } 4,[/MATH]
[MATH]\text {If } a = 4, \text { integer } y = \dfrac{x}{4} = \dfrac{100b + 11c}{4} = 25b + 2c + \dfrac{3c}{4}.[/MATH]
The only permissible value of c that will cancel 4 is 8, which entails that

[MATH]y = 40 + 8 = 48 \text { and } x = 100 * 1 + 11 * 8 = 188, \text { but } 4 * 48 = 192 \ne 18[/MATH][MATH]8[/MATH].

We have a contradiction. Thus, a is not 4. In short, a = 3, 6, 7, 8, or 9.

[MATH]a(40 + c) = ay = x = 100b + 10c + c \implies ac = 10d + c.[/MATH]
Now we run through the multiplication tables and find

[MATH]a = 3 \implies c = 5 \implies x = 155 \text { and } y = 45, \text { but } 3 * 45 = 135 \ne 155,[/MATH]
So a is not 3.

[MATH]a = 7 \implies c = 5 \implies x = 355 \text { and } y = 45, \text { but } 7 * 45 = 315 \ne 355,[/MATH]
So a is not 7. And a = 8 does not work at all.

[MATH]a = 9 \implies c = 5 \implies x = 455 \text { and } y = 45, \text { but } 9 * 45 = 405 \ne 455,[/MATH]
Thus a = 6. So b = 2, and c must equal 2, 4, or 8. But c cannot equal b so c must be 4 or 8.

[MATH]44 * 6 = 264 \ne 244.[/MATH]
So c must equal 8. Let's check.

[MATH]48 * 6 = 288.[/MATH]

Thank you sir! but I am just wondering if there is an easier approach or steps for this? this question is for Grade 9 level or equivalent. I am just wondering
 
Thank you sir! but I am just wondering if there is an easier approach or steps for this? this question is for Grade 9 level or equivalent. I am just wondering
This is really a problem in logic for a Year 9 student. There is no need for algebra really if you are willing to use a bit of elbow grease.
Look at things like:
1. 4C x A = BCC so C x A must end in C, so as Dr P said back in post#3 what could C and A be? List them. The list is not that long.
2. Forty C x a single digit number. At max this could be 49x9 = 400-and-something, so B can't be bigger than 4.
3. A can't be 1 or 2, otherwise the product would not be a 3-digit number (B is not 0). So A>2.
This should lead you to the answer/s.
Try the options.
 
Thank you sir! but I am just wondering if there is an easier approach or steps for this? this question is for Grade 9 level or equivalent. I am just wondering
Actually, I would initially try to solve a problem like this just with common sense. If that did not quickly yield a solution, I would then create a notation that was likely to be useful.

[MATH]x = 100b + 10c + c = 100b + 11c,\ y = 40 + c, \text { and } y = \dfrac{x}{a},[/MATH]
while noting that a, b, and c are different positive integers less than 10.

Having created a notation, I'd play around a little and perhaps come up with

[MATH]y = \dfrac{x}{a} \implies ay = x \implies a(40 + c) = x \implies 40a + ac = x \implies[/MATH]
[MATH]40a + 10d + e = 100d + 10c + c \implies 10(4a + d) + e = 10(10b + c) + c \implies e = c \text { and } 4a + d = 10b + c.[/MATH]
The fact that ac = 10d + c is unusual. For example, 3 times 7 = 21 = 20 + 1, not 20 + 3. When do we get this unusual fact? If a is odd and c is 5, we get

[MATH]a = 2k - 1 = 2k - 2 + 1 = 2(k - 1) + 1 \implies ac = \{2(k - 1) + 1\} * 5 = 10(k - 1) + 5 = 10(k - 1) + c.[/MATH]
Now we can compute

[MATH]1 * 45 < 100,[/MATH]
[MATH]3 * 45 = 135 \ne 155,[/MATH]
[MATH]7 * 45 = 315 \ne 355, \text { and }[/MATH]
[MATH]9 * 45 = 405 \ne 455.[/MATH]
Any other possibilities?

If a = 6 and c is even

[MATH]c = 2k \implies ac = 6 * 2k = 12k = 10k + 2k = 10k + c.[/MATH]
[MATH]6 * 42 = 252 \ne 255,[/MATH]
[MATH]6 * 44 = 264 \ne 266, \text { and }[/MATH]
[MATH]6 * 48 = 288 \implies a = 6,\ b = 2, \text { and } c = 8.[/MATH]
In short, I'd use a mixture of good notation, understanding of place order, and trial and error. Of course, this is informal, but you can then make a more formal and exhaustive presentation.
 
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