Which of the following CANNOT yield an integer when divided by 10 ?

SOLUTION

Which of the following CANNOT yield an integer when divided by 10 ?

(A) The sum of two odd integers
(B) An integer less than 10
(C) The product of two primes
(D) The sum of three consecutive integers
(E) An odd integer

In order an integer to be divisible by 10=2*5, the integer must be divisible by 2 (and 5), so it must be even, thus an odd integer is never divisible by 10 (or another way, an integer to be divisible by 10 its units digit must be 0, thus it must be even).

To discard other options:
(A) The sum of two odd integers. Not true: for example, 5+5=10 --> 10 IS divisible by 10.
(B) An integer less than 10. Not true: for example 0 IS divisible by 10.
(C) The product of two primes. Not true: 2*5=10 --> 10 IS divisible by 10.
(D) The sum of three consecutive integers. Not true: for example, -1+0+1=0 --> 0 IS divisible by 10.

Answer: E.
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IMO (E)
A) The sum of two odd integers - It could or could not yield integer: 3+5=8 or 25+5=30
B) An integer less than 10 - It could or could not yield integer: 3 or -30
C) The product of two primes -It could or could not yield integer: 2*3=6 or 5*2=10
D) The sum of three consecutive integers -It could or could not yield integer: 2+3+4=9 or 9+10+11=30
(E) An odd integer - it can't yield integer when divided by 10: Number should contain at least one 2 and 5 to yield an integer.

We can test examples as below:

A) 15+5 is divisible by 10
B) -10 is divisble by 10
C) 2*5 is divisible by 10
D) (19+20+21) is divisible by 10
E) All odd integers are not divisible by 10

Answer E.

(A) The sum of two odd integers
ODD+ODD= EVEN
e.g. 5+5=10
(B) An integer less than 10
Zero is a multiple of every integer.

(C) The product of two primes
2*5= 10

(D) The sum of three consecutive integers
a, a+1, a+2
a+a+1+a+2= 3(a+1)

If a= -1, then the consecutive integers would be -1,0,1
The answer would be zero which is a multiple of every integer.

(E) An odd integer
Correct answer. An odd integer can never be divisible by 10 as all the multiples of 10 end in 0 and thus will leave a remainder.

Just one tip: 0 is an integer!

Even though I chose the correct answer, initially it was not so clear for me to check the answer D.
(D) The sum of three consecutive integers

My approach as follows, we need to find the multiple of 10;
Hence, (First + Last)/2 = Aver, in our case Aver will be 10
(x + x + 2)/2 = 10
2x + 2 = 20
x = 9

The sum of three consecutive integers: x + x +1 x + 2 = 9, 10, 11 = 30/10 = Int.

The big takeaway from this one is to always look at all of the answers first before diving into test cases. E is such an obvious answer, but the test writers hide it in the last option as a trap to get you to waste time.

In order an integer to be divisible by 10=2*5, the integer must be divisible by 2 (and 5), so it must be even, thus an odd integer is never divisible by 10 (or another way, an integer to be divisible by 10 its units digit must be 0, thus it must be even).

Why can we read in this way 10 divided by odd number 5 gives 2 ?
either question framed in not good way or something is missing in answer choice.

Please help me in understanding Bunuel