What is the smallest number which when increased by 3 is divisible by

Question

What is the smallest number which when increased by 3 is divisible by 27, 35, 25 and 21?

(A) 3272

(B) 3572

(C) 4122

(D) 4722

(E) 4922

Skywalker18 · Answer

Prime factorization of numbers
27 = 3^3
35 = 5*7
25 = 5^2
21 = 3*7

Smallest number divisble by above 4 numbers will be LCM = 3^3 * 5^2 * 7 = 4725
Now , the number we need is 3 less than 4725 , therefore the number is 4722

Answer D

Abhishek009 · Answer

The problem can be reduced if we add 3 to the answer options and check divisibility rule of 3 ( Since 27 is a multiple of 3)

I am deliberately not involving in checking the divisibility rule for 5 and 7 since all the answer options have units digit as 5 and divisibility by 7 is a cumbersome 3 step process, where as divisibility by 3 is simply the sum of all the digits of the number.

(A) 3272 -> 3272+3 = 3275 Sum of the digits is 17 which is not divisible by 3.

(B) 3572 -> 3572+3 = 3575 Sum of the digits is 20 which is not divisible by 3.

(C) 4122 -> 4122+3 = 4125 Sum of the digits is 12 which is divisible by 3 { Keep this option for now }

(D) 4722 -> 4722+3 = 4725 Sum of the digits is 18 which is divisible by 3 { Keep this option for now }

(E) 4922 -> 4922+3 = 4925 Sum of the digits is 20 which is not divisible by 3

Now check options (C) and (D)

C) 4125 / 27 = Not completely divisible , so this is not the answer.

D) 4725 / 27 = 175 Completely divisible , so this is our answer.

HKD1710 · Answer

We need the highest power of each prime from all the numbers given.

3 comes 3 times in 27

5 and 7 comes once in 35

5 comes twice in 25

3 and 7 comes once in 21.

so we need three 3's, two 5's and one 7 for a number to be divisible by 27, 35, 25 and 21.

number = 3*3*3*5*5*7 = 4725

correct answer choice would be 3 less than this number because in question it give that "smallest number which when increased by 3"

Hence 4722 i.e.  "D"  is the correct answer.

laddaboy · Answer

I used the divisibility test for 9 ( any number which is divisible by 9 has it's sum of digits equal to 9 ) and found out only Option D satisfies this condition.

ScottTargetTestPrep · Answer

Let’s first determine the LCM of 27, 35, 25, and 21

27 = 3^3

35 = 5 x 7

25 = 5^2

21 = 3 x 7

So the LCM is 3^3 x 5^2 x 7 = 4,725. Since the number sought needs to be increased by 3 to become 4,725, so that number is 4,725 - 3 = 4,722.

Answer: D

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What is the smallest number which when increased by 3 is divisible by

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