Solution to Problem 6

See for the problem here: Click here

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Correct Answer:

(1) The following are the only numbers: 19, 29, 39, 49, 59, 69, 79, 89, 99.

(2) There exists no such three digit number \(abc\) satisfying

\(a+b+c+(a\times b\times c) = abc\).

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Correct answers for (1) were given by Manjunatha M R., Abhishek M. N. and Darshan A. Correct answer to (2) was given only by Manjunatha M. R. We present below the common solution to (1) and two solutions to (2), one by Manjunatha M. R. and the other solution by the poser.

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Solution to (1): We have the following condition as has been mentioned in the comment section of the problem by Darshan

\(a+b+(a\times b) = 10a+b\).

This immediatley implies

\(9a = a\times b\) .

As \(a\neq 0\), we must have

\(b=9\),

irresepctive of whatever \(a\) may be. Thus, with \(b=9\), we have the following numbers, for which, one can easily verify that the given conditions are satisfied

\(19, 29, 39, 49, 59, 69, 79, 89\) and \(99\).

Solution to 2: Solution by Manjunatha M. R.

If \(xyz\) is athree digit number, then \(xyz = 100x+10y+z\), where \(x\in (0,9]\) and \(y, z\in [0,9]\).

Let \(x \in (0,9]\), \(y\in(0,9]\) and \(z\in \mathbb{R}\). Consider

\(x+y+z+(x\times y \times z) = 100x+10y+z\).

The above implies

\(x\times y \times z = 99x+9y\).

Or, equivalently

\(z = \frac{99}{y}+\frac{9}{x}\).

Now, \(x\leq 9\) and \(y \leq 9\). This implies

\(\frac{1}{x}\geq \frac{1}{9} \) and \(\frac{1}{y} \geq \frac{1}{9} \).

This implies,

\(\frac{9}{x} \geq 1\) and \(\frac{99}{y}\geq 11\). \(\therefore \frac{9}{x}+\frac{99}{y} \geq 12, \) \(\forall \,\, x, y\in (0,9]\).

Therfore, we must have that

\(z \geq 12\),

which cannot happen as \(z\) belongs to \([0,9]\).

Now, the only left out case is \(y=0\). In such a case, we must have that \(99x + 9y = 0\), which simply cannot happen unless \(x=0\). But we must want \(x\neq 0\). Thus the proof that no such number exists.

Original solution:

If \(abc\) is a three digit number sastifying

\(a+b+c + (a\times b \times c) = abc = 100a+10b+c\), then, we have \(a\times b \times c = 99a + 9b\). This implies, \(a(b\times c-99) = 9b\).

Notice that for any two numbers \(b, c \in [0, 9]\), the maximum value of \(b\times c) = 81\). Thus \(b\times c -99\) must be a negative number in the above equation. While, \(9b\) is non negative. LHS is always negative (it cannot even be zero) and RHS is non negative. This contradiction leads us to the conclusion that no such three digit number exists.