GMAT 700+ Quant Question 21 | Arithmetic

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?

1. 21
2. 27
3. 9
4. 45
5. 12

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Video Explanation

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Explanatory Answer | GMAT Arithmetic Practice Question

Step 1: What are the factors of 36?

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.

Step 2: List Down Possibilities and Count

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

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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

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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

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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

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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

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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

Choice A is the correct answer.