GMAT 700+ Quant Question 21 | Arithmetic

Counting Methods | Permutation Combination Practice | GMAT Sample Questions

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

Get to Q51 in GMAT Quant


Online GMAT Course
@ INR 3000


Video Explanation


GMAT Live Online Classes


Next Batch | December 3, 2020


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

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

Choice A is the correct answer.

GMAT Online Course - Quant
Try it free!

Register in 2 easy steps and
Start learning in 5 minutes!

★ Sign up for Free

Already have an Account?

★ Login to Continue

GMAT Live Online Classes

Next Batch Dec 3, 2020

★ GMAT Live Info

GMAT Coaching in Chennai

Next Integrated (Online + Offline) GMAT Batch
Dec 3, 2020

★ GMAT Class Info

GMAT Hard Math Videos On YouTube


GMAT Sample Questions | Topicwise GMAT Questions


Where is Wizako located?

Wizako - GMAT, GRE, SAT Prep
An Ascent Education Initiative
48/1 Ramagiri Nagar
Velachery Taramani Link Road.,
Velachery, Chennai 600 042. India

How to reach Wizako?

Phone: (91) 44 4500 8484
Mobile: (91) 95000 48484
WhatsApp: WhatsApp Now
Email: learn@wizako.com
Leave A Message