The given question is a challenging GMAT 700+ level quant problem solving question combining concepts in permutation combination (counting methods) and elementary number properties.

Question 21: 149 is a 3-digit positive integer, product of whose digits is 1 × 4 × 9 = 36. How many 3-digit positive integers exist, product of whose digits is 36?

- 21
- 27
- 9
- 45
- 12

@ INR

1, 2, 3, 4, 6, 9, 12, 18, and 36.

Of these, 1, 2, 3, 4, 6, and 9 are single digit factors and can therefore, be digits of the 3 digit numbers.

**Possibility 1**: Let 9 be one of the 3 digits.

The product of the remaining 2 digits will, therefore, be 4.

Product of the remaining 2 digits | Possibilities for the 2 digits | Possible Rearrangements | Number of Outcomes |
---|---|---|---|

4 | 1 and 4 | {149, 194, 419, 491, 914, 941} | 6 |

4 | 2 and 2 | {229, 292, 922} | 3 |

**Possibility 2**: Let 6 be one of the 3 digits.

The product of the remaining 2 digits will, therefore, be 6.

Product of the remaining 2 digits | Possibilities for the 2 digits | Possible Rearrangements | Number of Outcomes |
---|---|---|---|

6 | 1 and 6 | {166, 616, 661} | 3 |

6 | 2 and 3 | {236, 263, 326, 362, 623, 632} | 6 |

**Possibility 3**: 4 is one of the three digits

The product of the remaining 2 digits is 9.

Product of the remaining 2 digits | Possibilities for the 2 digits | Possible Rearrangements | Number of Outcomes |
---|---|---|---|

9 | 3 and 3 | {334, 343, 433} | 3 |

9 | 1 and 4 | The 3 digits are 1, 4, and 9. | Counted in set 1 of possibility 1 |

**Possibility 4**: Let 3 be one of the three digits

The product of the remaining 2 digits is 12.

Product of the remaining 2 digits | Possibilities for the 2 digits | Possible Rearrangements | Number of Outcomes |
---|---|---|---|

12 | 2 and 6 | The 3 digits are 2, 3, and 6 | Counted in set 2 of possibility 2 |

12 | 3 and 4 | The 3 digits are 3, 3, and 4 | Counted in set 1 of possibility 3 |

**Possibility 5**: Let 2 be one of the three digits

The product of the remaining 2 digits is 18.

Product of the remaining 2 digits | Possibilities for the 2 digits | Possible Rearrangements | Number of Outcomes |
---|---|---|---|

18 | 2 and 9 | The 3 digits are 2, 2, and 9 | Counted in set 2 of possibility 1 |

18 | 3 and 6 | The digits are 2, 3, and 6 | Counted in set 2 of possibility 2 |

**Possibility 6**: Let 1 be one of the three digits

The product of the remaining 2 digits is 36.

Product of the remaining 2 digits | Possibilities for the 2 digits | Possible Rearrangements | Number of Outcomes |
---|---|---|---|

36 | 4 and 9 | The 3 digits are 1, 4, and 9 | Counted in set 1 of possibility 1 |

36 | 6 and 6 | The 3 digits are 1, 6, and 6 | Counted in set 1 of possibility 2 |

Number of such 3-digit positive integers is calculated by adding all the outcomes.

The outcomes from possibilities 4, 5, and 6 should not be counted because they have already been counted in the earlier possibilities.

the total Number of such 3-digit positive integers are 6 + 3 + 3 + 6 + 3 = 21 Numbers

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