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06 Sep 2024, 01:30
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C
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Difficulty:
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Question Stats:
59% (02:31) correct
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149 is a 3-digit positive integer, product of whose digits is 1x4x9 = 36. How many 3-digit positive integers exist whose product of digits is 36?
A. 9
B. 12
C. 21
D. 27
E. 45
A. 9
B. 12
C. 21
D. 27
E. 45
06 Sep 2024, 02:01
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The factor pairs of 36:
\(36*1\), \(18*2\), \(12*3\), \(9*4\) and \(6*6\)
Focusing on the first three as they have double digit values:
36*1: The 36*1 can be broken down to a product of three digits while keeping 1, in the following ways: \(9*4*1\) & \(6*6*1\).
\(9*4*1\) can be arranged in \(3! = 6\) different ways.
\(6*6*1\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
18*2: The 18*2 can be broken down to a product of three digits while keeping 2 in the following ways: \(9*2*2\) & \(6*3*2\).
\(9*2*2\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
\(6*3*2\) can be arranged in \(3! = 6\) different ways.
12*3: The 12*3 can be broken down to a product of three digits while keeping 3, in the following ways: \(9*4*1\) & \(4*3*3\).
\(4*3*3\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
Total Ways: \(6+3+3+6+3 = 21\)
ANSWER C
\(36*1\), \(18*2\), \(12*3\), \(9*4\) and \(6*6\)
Focusing on the first three as they have double digit values:
36*1: The 36*1 can be broken down to a product of three digits while keeping 1, in the following ways: \(9*4*1\) & \(6*6*1\).
\(9*4*1\) can be arranged in \(3! = 6\) different ways.
\(6*6*1\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
18*2: The 18*2 can be broken down to a product of three digits while keeping 2 in the following ways: \(9*2*2\) & \(6*3*2\).
\(9*2*2\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
\(6*3*2\) can be arranged in \(3! = 6\) different ways.
12*3: The 12*3 can be broken down to a product of three digits while keeping 3, in the following ways: \(9*4*1\) & \(4*3*3\).
\(4*3*3\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
Total Ways: \(6+3+3+6+3 = 21\)
ANSWER C
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Breaking 36 into its prime factor form => 2 * 2 * 3 * 3
We know have to fill 3 blanks _ _ _ with 3 digits whose product is 36. From the prime factor form we know we can form single digits by multiplying at most 2 prime factors.
Let us also consider 1 into the factored form; 36 = 1 * 2 * 2 * 3 * 3
1 _ _ => the other two blanks can have two 6's or one 4 and one 9; 3 ways to arrange 166 and 6 ways to arrange 149
2 _ _ => the other two blanks can have one 2 and one 9 or one 3 and one 6; 3 ways to arrange 229 and 6 ways to arrange 236
3 _ _ => the other two blanks can have one 3 and one 4 (case with one 6 and one 2 is already considered above); 3 ways to arrange 334
3 + 6 + 3 + 6 + 3 = 21 ways or 21 numbers
Answer C.
We know have to fill 3 blanks _ _ _ with 3 digits whose product is 36. From the prime factor form we know we can form single digits by multiplying at most 2 prime factors.
Let us also consider 1 into the factored form; 36 = 1 * 2 * 2 * 3 * 3
1 _ _ => the other two blanks can have two 6's or one 4 and one 9; 3 ways to arrange 166 and 6 ways to arrange 149
2 _ _ => the other two blanks can have one 2 and one 9 or one 3 and one 6; 3 ways to arrange 229 and 6 ways to arrange 236
3 _ _ => the other two blanks can have one 3 and one 4 (case with one 6 and one 2 is already considered above); 3 ways to arrange 334
3 + 6 + 3 + 6 + 3 = 21 ways or 21 numbers
Answer C.
