Question

Each of the integers 334 and 419 have digits whose product is 36. How many 3-digit positive integers have digits whose product is 36?
(A) 21 (B) 15 (C) 18 (D) 24 (E) 12

Answer 1

$\large\overline{\underline{ \boxed{ \sf \red{\bigstar \: quesion}}}} </p><p>$

Each of the integers 334 and 419 have digits whose product is 36. How many 3-digit positive integers have digits whose product is 36?

$\large\overline{\underline{ \boxed{ \sf \red{\bigstar \: answer}}}} </p><p>$

Option (A)21

$\large\overline{\underline{ \boxed{ \sf \red{\bigstar \: explanation}}}} </p><p>$

For this question,

let’s start with the prime factorization of 36.

36 = 4 x 9 = (2)(2)(3)(3)

With this we can construct the following set of three numbers that multiply to 36.

They are

☞1, 4, 9

☞3, 3, 4

☞2, 2, 9

☞2, 3, 6

☞1, 6, 6

☞1, 2, 18

☞1, 3, 12

$\overline{\underline{ \boxed{ \sf {\ \: \: The \: last \: two \: sets \: will \: be \: ignored \: as \: they \ \: involve \: a \: two \: digit \: number.</p><p>}}}}$

With the other sets, let’s separate into two groups. Those with repeating numbers and those without.

Repeating:

(3, 3, 4), (2, 2, 9) and (1, 6, 6)

Non-repeating:

(1, 4, 9) and (2, 3, 6)

With each repeating set we have 3P3/2! since a digit repeats twice. This gets us 3!/2! = 3 for each set or 9 for all three sets.

With each non-repeating we have 3! which is 6 for each set, or 12 for both sets.

Adding them together,

☬we have 21 different 3 digit numbers whose product of its digits equal to 36.☬

Answer 2

Answer:

21 is your answer

Step-by-step explanation:

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