#### nanaseailie

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I am always clueless in doing this kind of problem

how are you supposed to intelligently guess the numbers?

- Thread starter nanaseailie
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I am always clueless in doing this kind of problem

how are you supposed to intelligently guess the numbers?

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It might be easier to see if you write it as a multiplication. ie 4C x A = BCC

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It helps to simplify the notation a bit, and not much trial and error is needed. (Perhaps none is, but I used some.)

I am always clueless in doing this kind of problem

how are you supposed to intelligently guess the numbers?

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

\(\displaystyle x = 100b + 10c + c = 100b + 11c \text { and } y = 40 + c \implies 122 \le x \in \mathbb Z,\ y \in \mathbb Z, \text { and } 41 \le y \le 49.\)

It is given that \(\displaystyle y = \dfrac{x}{a} \implies ay = x \implies x \le 9 * 49 = 441 \implies 122 \le x \le 433.\)

\(\displaystyle \text {If } a = 5, \text { integer } y = \dfrac{x}{5} = \dfrac{100b + 11c}{5} = 20b + 2c + \dfrac{c}{5}.\)

But no permissible value of c will cancel 5 so we have a contradiction. Thus, a is not 5.

Playing around with

\(\displaystyle 41a \le ay \le 49a \text { and } ay = x = 100b + 11c \ge 122 \implies\) we find the following:

\(\displaystyle a \ge 3,\)

\(\displaystyle b = 1 \iff a = 3 \text { or } 4,\)

\(\displaystyle a = 6 \implies b = 2,\)

\(\displaystyle a = 7 \implies b = 2 \text { or } 3,\)

\(\displaystyle a = 8 \implies b = 3,\text { and}\)

\(\displaystyle a = 9 \implies b = 3 \text { or } 4,\)

\(\displaystyle \text {If } a = 4, \text { integer } y = \dfrac{x}{4} = \dfrac{100b + 11c}{4} = 25b + 2c + \dfrac{3c}{4}.\)

The only permissible value of c that will cancel 4 is 8, which entails that

\(\displaystyle y = 40 + 8 = 48 \text { and } x = 100 * 1 + 11 * 8 = 188, \text { but } 4 * 48 = 192 \ne 18\)\(\displaystyle 8\).

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

\(\displaystyle a(40 + c) = ay = x = 100b + 10c + c \implies ac = 10d + c.\)

Now we run through the multiplication tables and find

\(\displaystyle a = 3 \implies c = 5 \implies x = 155 \text { and } y = 45, \text { but } 3 * 45 = 135 \ne 155,\)

So a is not 3.

\(\displaystyle a = 7 \implies c = 5 \implies x = 355 \text { and } y = 45, \text { but } 7 * 45 = 315 \ne 355,\)

So a is not 7. And a = 8 does not work at all.

\(\displaystyle a = 9 \implies c = 5 \implies x = 455 \text { and } y = 45, \text { but } 9 * 45 = 405 \ne 455,\)

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.

\(\displaystyle 44 * 6 = 264 \ne 244.\)

So c must equal 8. Let's check.

\(\displaystyle 48 * 6 = 288.\)

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It might be easier to see if you write it as a multiplication. ie 4C x A = BCC

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No because BCC doesn't mean BxCxC here.

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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 wonderingIt 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.

\(\displaystyle x = 100b + 10c + c = 100b + 11c \text { and } y = 40 + c \implies 122 \le x \in \mathbb Z,\ y \in \mathbb Z, \text { and } 41 \le y \le 49.\)

It is given that \(\displaystyle y = \dfrac{x}{a} \implies ay = x \implies x \le 9 * 49 = 441 \implies 122 \le x \le 433.\)

\(\displaystyle \text {If } a = 5, \text { integer } y = \dfrac{x}{5} = \dfrac{100b + 11c}{5} = 20b + 2c + \dfrac{c}{5}.\)

But no permissible value of c will cancel 5 so we have a contradiction. Thus, a is not 5.

Playing around with

\(\displaystyle 41a \le ay \le 49a \text { and } ay = x = 100b + 11c \ge 122 \implies\) we find the following:

\(\displaystyle a \ge 3,\)

\(\displaystyle b = 1 \iff a = 3 \text { or } 4,\)

\(\displaystyle a = 6 \implies b = 2,\)

\(\displaystyle a = 7 \implies b = 2 \text { or } 3,\)

\(\displaystyle a = 8 \implies b = 3,\text { and}\)

\(\displaystyle a = 9 \implies b = 3 \text { or } 4,\)

\(\displaystyle \text {If } a = 4, \text { integer } y = \dfrac{x}{4} = \dfrac{100b + 11c}{4} = 25b + 2c + \dfrac{3c}{4}.\)

The only permissible value of c that will cancel 4 is 8, which entails that

\(\displaystyle y = 40 + 8 = 48 \text { and } x = 100 * 1 + 11 * 8 = 188, \text { but } 4 * 48 = 192 \ne 18\)\(\displaystyle 8\).

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

\(\displaystyle a(40 + c) = ay = x = 100b + 10c + c \implies ac = 10d + c.\)

Now we run through the multiplication tables and find

\(\displaystyle a = 3 \implies c = 5 \implies x = 155 \text { and } y = 45, \text { but } 3 * 45 = 135 \ne 155,\)

So a is not 3.

\(\displaystyle a = 7 \implies c = 5 \implies x = 355 \text { and } y = 45, \text { but } 7 * 45 = 315 \ne 355,\)

So a is not 7. And a = 8 does not work at all.

\(\displaystyle a = 9 \implies c = 5 \implies x = 455 \text { and } y = 45, \text { but } 9 * 45 = 405 \ne 455,\)

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.

\(\displaystyle 44 * 6 = 264 \ne 244.\)

So c must equal 8. Let's check.

\(\displaystyle 48 * 6 = 288.\)

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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.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

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.

Actually, I would initially try to solve a problem like this just with common sense. If that did notThank 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

\(\displaystyle x = 100b + 10c + c = 100b + 11c,\ y = 40 + c, \text { and } y = \dfrac{x}{a},\)

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

\(\displaystyle y = \dfrac{x}{a} \implies ay = x \implies a(40 + c) = x \implies 40a + ac = x \implies\)

\(\displaystyle 40a + 10d + e = 100d + 10c + c \implies 10(4a + d) + e = 10(10b + c) + c \implies e = c \text { and } 4a + d = 10b + c.\)

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

\(\displaystyle a = 2k - 1 = 2k - 2 + 1 = 2(k - 1) + 1 \implies ac = \{2(k - 1) + 1\} * 5 = 10(k - 1) + 5 = 10(k - 1) + c.\)

Now we can compute

\(\displaystyle 1 * 45 < 100,\)

\(\displaystyle 3 * 45 = 135 \ne 155,\)

\(\displaystyle 7 * 45 = 315 \ne 355, \text { and }\)

\(\displaystyle 9 * 45 = 405 \ne 455.\)

Any other possibilities?

If a = 6 and c is even

\(\displaystyle c = 2k \implies ac = 6 * 2k = 12k = 10k + 2k = 10k + c.\)

\(\displaystyle 6 * 42 = 252 \ne 255,\)

\(\displaystyle 6 * 44 = 264 \ne 266, \text { and }\)

\(\displaystyle 6 * 48 = 288 \implies a = 6,\ b = 2, \text { and } c = 8.\)

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.