GMAT Problem Solving- Find The Smallest Positive 4-Digit Number Which, When Increased by 8, is Divisible by 12, 18, 30, and 45

Question: Find the smallest positive 4-digit number which, when increased by 8, is divisible by 12, 18, 30 and 45

1. 1072
2. 1,080
3. 1,088
4. 1,096
5. 1,12

“Find the smallest positive 4-digit number which, when increased by 8, is divisible by 12, 18, 30 and 45”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. GMAT Quant section consists of 31 questions in total. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

It is asked in the question to find out the smallest positive four-digit number which when increased by 8, is divisible by 12,8, 30,45.
A number is divisible by all the numbers 12, 8, 30, and 45, so the number must be the LCM of these numbers.
LCM(12 ,8, 30, 45)
The LCM is 180
If the number is divisible by 180 then automatically the given number will be automatically divisible by 12, 8, 30, 45
Our goal is to find the minimum four-digit number which is divisible by 180.
To find the minimum four-digit number which is divisible by 180 we have to divide 1000 by 180 because 1000 is the minimum four-digit number.
1000/ 180 = 5.555
Now let the factor be 5 and 6
180 * 5 = 900 (not suitable as it is 3 digit number)
180 * 6 = 1080 ( it is 4 digit number)
Therefore 1080 is the smallest four-digit number which is divisible by 180
But it should be noted that in the question it is asked to find out the number which when added with 8 will be divisible by 12,8, 30, and 45.
Therefore the correct answer will be 8 less than the number.
1080 - 8 = 1072
So 1072 is the smallest 4-digit number which when added with 8 will be divisible by 12,8, 30, and 45.

Correct Answer: A

Approach Solution 2:

The answer will be a number that will either end in 0 or 5 when 8 is added, while also being divisible by 3. This is because of the fact that 12, 18, 30,45 are all divisible by 3, while all the numbers that 45 will go into will end in 0 or 5
From the given options, the only number that when added with 8 will end in 0 or 5 is 1072.
Also to double-check 1072 + 8 = 1080
1080 is divisible by 3.

Correct Answer: A

Approach Solution 3:

The question asks for the lowest positive four-digit integer that, when multiplied by 8, divides into 12, 8, 30, and 45.
A number must be the LCM of the integers 12, 8, 30, and 45 since it is divisible by each of these numbers.
LCM(12 ,8, 30, 45) (12 ,8, 30, 45)
180 is the LCM.
The given integer will automatically be divisible by 12, 8, 30, and 45 if the value is divisible by 180.
The least four-digit number that can be divided by 180 is what we are looking for.
Since 1000 is the smallest four-digit number, we must divide it by 180 to obtain the minimal four-digit number that is divisible by 180.
1000/ 180 = 5.555

Allow factors 5 and 6 now.
180 * 5 = 900 (not appropriate as it is 3 digit number) (not suitable as it is 3 digit number)
180 * 6 = 1080 ( that is 4 digit number) ( it is 4 digit number)
So, the smallest four-digit number that can be divided by 180 is 1080.
However, it should be noted that the question asks you to identify the number that, when added to 8, would divide equally into 12, 8, 30, and 45.
As a result, the right response will be 8 less than the given number.
1080 - 8 = 1072
Thus, 1072 is the smallest 4-digit number that can be divided into 12, 8, 30, and 45 when added to 8.

Correct Answer: A

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