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The product of digits of a positive three-digit

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Source: — Data Sufficiency |

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by [email protected] » Wed Dec 28, 2016 6:10 pm
Hi NandishSS,

We're told that the product of the digits in a positive 3-DIGIT number is 120. For that to be the case, one of the digits MUST be 5 and the other two digits MUST have a product of 24 (which means that the three digits will always be DISTINCT - and must be one of the following combinations: 3/5/8 or 4/5/6). Some examples include 385, 564, 853, etc. We're asked if K is a multiple of 5. This is a YES/NO question.

1) K is a multiple of 3.

IF...
K = 456 then the answer to the question is NO.
K = 465 then the answer to the question is YES.
Fact 1 is INSUFFICIENT

2) If we switch any digits in K, it will decrease.

Since the three digits will be distinct, the only way to guarantee that switching any 2 digits will DECREASE the value of K is if the three digits are in DESCENDING order... and that limits the possibilities to just two options:

IF....
K = 654 then the answer to the question is NO.
K = 853 then the answer to the question is NO.
The answer to the question is ALWAYS NO.
Fact 2 is SUFFICIENT

Final Answer: B

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by Jay@ManhattanReview » Wed Dec 28, 2016 10:45 pm
NandishSS wrote:The product of digits of a positive three-digit integer k is 120. Is k a multiple of 5?

(1) k is a multiple of 3.
(2) If we switch any digits in k, it will decrease.

OA:B

Source:800Score
We are given that the product of three digits of K = 120. For K to be a multiple of '5', it must have its unit digit equals to '5', thus the product of other two digits = 120/5 = 24. We can get the product 24 in two ways: 8*3 or 4*6.

=> The three digits are: either 5, 8, and 3 or 5, 4, and 6.

Since the order of the digits will decide whether K is divisible by 5, we must decide whether 5 is K's unit digit.

S1: K is a multiple of 3.

Divisibility rule of 3: Sum of all the digits is divisible by 3.

If the digits are 5, 4, and 6, the sum of the digits is divisible by 3. However, if the unit digit is 5, K is divisible by 5, else not. No unique answer! Insufficient.

S2: If we switch any digits in K, it will decrease.

The statement implies that if we order the digits: 5, 8, and 3 or 5, 4, and 6, the highest possible number would result. Thus, the possible numbers would 853 or 654. Whatever be the value of K: 653 or 654, none of them is divisible by 5. A unique answer. Sufficient.

OA: B

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by crackverbal » Thu Dec 29, 2016 2:16 am
Hi NandishSS,

Best way to solve these types of questions is to list down few possibilities then build upon with the given information.

Given: Product of digits of a three digit number is 120.

Generally any digit place can take up a integer value from 0 to 9(Except "0" cannot be in the first place of any digit number).

But in this question none of the digit places can take zero (because that will make the product as zero). Also since it's a multiple of 5, then one of the digits must be multiple of 5.

Let k = 5xy

5 * x * y = 120.

So x* y = 24, so possibilities could be 6 and 4 or 8 and 3.

So possibilities of the number could be 853 or 546 etc..

So we got a sense now, for this to be divisible by 5. We should know in certain that unit's place has to be 5.

Let's check the statements,

Statement I is insufficient:

K is a multiple of 3.

RULE: If the sum of all the digits of a number is multiple of 3, then the number is multiple or divisible by 3.

Here, we can't take 8, 5 and 3 combinations.

But even if we take the combinations of 5, 4 and 6 , still we are not sure whether the units place of k is 5 or not. It could be 654 or 465

So insufficient.

Statement II is sufficient:

To think of, we can write it from least to greatest from left to right,
853 or 654 etc.. In any case 5 cannot be in the units place. Otherwise it will contradict the given statement(because 5 is not the least digit in k).

So we can certainly say that k is not divisible by 5.

So sufficient.

So the answer is B.

Hope this helps.
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