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Is the three-digit number n less than 550?

(1) The product of the digits in n is 30 --> \(30=1*2*3*5\) --> n can be any combination of digits from the following two sets: {1, 6(=2*3), 5} or {2, 3, 5}. n can be lees than 550 (eg. 165, 156, ...) OR more than 550 (eg 615, 651). Not sufficient.

(2) The sum of digits in n is 10 --> Clearly not sufficient. Many combinations are possible 109, 901, ....

(1)+(2) We have the set {2, 3, 5} (as only in this set the sum of the digits is 10). From this set we can not form three digit number more than 550 (max=532). Sufficient.

Answer: C.
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Bunuel
Is the three-digit number n less than 550?

(1) The product of the digits in n is 30 --> \(30=1*2*3*5\) --> n can be any combination of digits from the following two sets: {1,6(=2*3),5} or {2,3,5}. n can be lees than 550 (eg. 165, 156, ...) OR more than 550 (eg 615, 651). Not sufficient.

(2) The sum of digits in n is 10 --> Clearly not sufficient. Many combinations are possible 109, 901, ....

(1)+(2) We have the set {2,3,5} (as only in this set the sum of the digits is 10). From this set we can not form three digit number more than 550 (max=532). Sufficient.

Answer: C.

Excellent Bunuel. I missed the factor "1" when checking the factors of "30". And if you miss this "1" as the factor the answer you end up with is "A", which indeed is wrong.
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ha....I also forgot "1" as a factor....thanks for reminding.....
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GGRRRRRRR.....Made the same mistake...Forgot to include 1 and chose the answer as A.
I should not make similar mistakes in the real exam.
Thanks BUnnel.
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Isnt there a better way ...?

Any three digit number is 100x + 10y + z

From 1 we have xyz = 30
From 2 we have x+y+z = 10

We have three variables and three equations ..

Without solving, cant we guess that solution to these three equations will give us an answer ... ?
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Is the three-digit number n less than 550?

(1) The product of the digits in n is 30.
(2) The sum of the digits in n is 10.
\(100 \leqslant \,\,n = \left\langle {ABC} \right\rangle \,\, \leqslant 999\)

\(\left\langle {ABC} \right\rangle \,\,\mathop < \limits^? \,\,550\)


\(\left( 1 \right)\,\,\,A \cdot B \cdot C = 30\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {1,5,6} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {5,6,1} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)


\(\left( 2 \right)\,\,\,A + B + C = 10\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {1,2,7} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\\\
\,{\text{Take}}\,\,\left( {A,B,C} \right) = \left( {5,5,0} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)


\(\left( {1 + 2} \right)\,\,\,\left\langle {ABC} \right\rangle \,\,\, \geqslant \,\,\,550\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}\\
\,A = 5\,\,,\,\,B \geqslant 5\,\,\,\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,B = 5,\,\,C = 0\,\,\, \Rightarrow \,\,\,\left( 1 \right)\,\,{\text{contradicted}} \hfill \\\\
\,\,\,{\text{or}} \hfill \\\\
\,A \geqslant 6\,\,\,\mathop \Rightarrow \limits_{\left( 1 \right)}^{\left( * \right)} \,\,\,A = 6\,\,\,\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,B \cdot C = 5\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,A + B + C = 12\,\,\, \Rightarrow \,\,\,\left( 2 \right)\,\,{\text{contradicted}} \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)

\(\left( * \right)\,\,\,\left( 1 \right)\,\,\, \Rightarrow \,\,\,A,B,C\,\,{\text{are}}\,\,\left( {{\text{positive}}} \right)\,\,{\text{divisors}}\,\,{\text{of}}\,\,30\,\,\, \Rightarrow \,\,\,A \notin \left\{ {7,8,9} \right\}\)

\(\left( {**} \right)\,\,\,B \cdot C = 5\,\,\, \Rightarrow \,\,\,B,C\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{pair}}\,\,{\text{of}}\,\,\left( {{\text{positive}}} \right)\,\,{\text{divisors}}\,\,{\text{of}}\,\,5\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}\\
\,\left( {B,C} \right) = \left( {1,5} \right) \hfill \\\\
\,\,{\text{or}} \hfill \\\\
\,\left( {B,C} \right) = \left( {5,1} \right) \hfill \\ \\
\end{gathered} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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wait,

1) The question says the product of the digits in n is 30.

2) n is a three-digit number

3) Those three digits are 2,3 and 5.

1 should not be included, only 2,3, and 5. The question does not say "the product of the factors of n" it says the products of the digits in n,

Without 1, the correct answer is A.
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PabloOli
wait,

1) The question says the product of the digits in n is 30.

2) n is a three-digit number

3) Those three digits are 2,3 and 5.

1 should not be included, only 2,3, and 5. The question does not say "the product of the factors of n" it says the products of the digits in n,

Without 1, the correct answer is A.

The correct answer is C, not A. For (1): n can be lees than 550 (eg. 165, 156, ...) OR more than 550 (eg 615, 651). Not sufficient.
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