Which of the following are divisible by 2, 3, and 8?

Question

I. 33,333,333,000
II. 2,334,455,921
III. 444,444,400

A. I​ only 
B. I and II​ only 
C. II only 
D. I​ and III only 
E. none of these

Abhishek009 · Answer

1. A number is divisible by 8 when the last 3 digits are divisible by 8 Using the said property we negate option II 2. A number is divisible by 2 when the units digit is even Using the said property we negate option II 3. A number is divisible by 3 when the sum of the digits is divisible by 3 I. 33,333,333,000 = sum of the digits is 8*3 = 24 divisible by 3 II. We will not waste our time checking this option sicne it has already been eliminated III. 444,444,400 = sum of the digits is 7*4 = 28 not divisible by 3 So option I is only possible number which is divisible by 2,3 & 8 :-D

Skywalker18 · Answer

2-If the last digit is even, the number is divisible by 2
3-If the sum of the digits is divisible by 3, the number is also divisible .
8- If the last three digits of a number are divisible by 8, then so is the whole number.

I. 33,333,333,000
Divisible by 2,3 and 8

II. 2,334,455,921
Odd number - not divisible by 2
No need to check further

III. 444,444,400
Sum of digits = 4*7= 28
Not divisible by 3

Answer A

Psiva00734 · Answer

a number is divisible by 2 if it is an even integer.

sum of number is divisbile by 3, then it is divisible.

last three digits of a number is divisbile by 8 then number is divisible by 8.

1, 33,333,333,000. 
it is a even integer

sum is divisible by 3.

last 3 digits is divsible by 8. so option 1 is divisble by 2,3, and 8.

2 2,334,455,921

it is an odd integer not divisibly by2 and 8 also.

sum of the numbers is 38. it is also not divisible by 3.

3 444,444,400.

even integer, last 3 digits 400 is divisibly by8. sum of digits is 28 not divisible by3

so option A is correct.

harvardgirl1 · Answer

This explanation is very clear thanks

ehsan090 · Answer

Even number cannot divide odd so option B and C ruled out straight away Left with 3 options now start from A A number can be divided by 3 if their sum of digit can be divided as well 33,333,333,000=24 so 3 qualify since number is even so 2 qualify as well for case of 8 last 3 digit should be divisible 000 can be divided

check last option now 444,444,400 sum=28/3 cannot be divided evenly ruled out without further checking So A is the correct choice

LeoN88 · Answer

II is not divisible by 2
III is not divisible by 3, sum of the digits is 28.

Let's see I.
Div. by 2,
Div. by 8, as last three digits are divisible by 8
Div. by 3 as sum of digits is 8*3

OA A, I only.

saurabh9gupta · Answer

I. Div by 2,3,8
2- because it is an even number
3- sum is divisible by 3
8 - last 3 digits are divisible by 8

II and III do not fit.

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Which of the following are divisible by 2, 3, and 8?

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Which of the following are divisible by 2, 3, and 8?

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