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habil
Is there another way to approach this problem

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According to Bunuel's solution, since "b" is not divisible by 13, "b" will not be divisible by any integer which has 13 as factor so that b/26 (i.e., 2 * 13) cannot result in an integer.

For option A, 13b/52, even though the denominator has 13 as factor, this is cancelled out by the 13 in the numerator, resulting in an integer.

Alternatively, you can also plug in numbers:

b
6
19 (= 6 + 13)
32 (= 19 + 13)
45 (= 32 + 13)
58 (= 45 + 13)
71 (= 58 + 13)
84 (= 71 + 13)
97 (= 84 + 13)
110 (= 97 + 13)
123 (= 110 + 13)
136 (= 123 + 13)
149 (= 136 + 13)
.
.
. etc

You realise that 6 and 84 are divisible by 6; 84 is divisible by 12; 136 is divisible by 17; but none of the numbers for integer "b" are divisible by 13 or an integer with 13 as a factor!

So b/26 will not result in an integer.

Hope this helps.
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Is there another way to approach this problem

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Hi

The logic is same as suggested by Bunuel..
Let the number be b as given..
b =13y+6....
Now substitute b as 13y +6 in all choices and you will get the ans
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Bunuel
When integer b is divided by 13, the remainder is 6. Which of the following cannot be an integer?

A. 13b/12
B. b/26
C. b/17
D. b/12
E. b/6

Assume the number to be 13b + 6

Of the given numbers, only option B has a multiple of 13 in the denominator
But our number can never be a multiple of 13

Hence b/26 cannot be an integer.

Correct Option: B
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Bunuel
When integer b is divided by 13, the remainder is 6. Which of the following cannot be an integer?

A. 13b/12
B. b/26
C. b/17
D. b/12
E. b/6

If, when b is divided by 13, the remainder is 6, then that means b = 13q + 6 for some integer q. Let’s analyze each answer choice to see whether the given expression can produce an integer.

A) 13b/12

We need to see if 13(13q + 6)/12 could equal an integer for some integer value of q. We can choose q = 6. If q = 6, then 13q + 6 = 84, which is is divisible by 12; hence 13(84)/12 is an integer.

B) b/26

Could (13q + 6)/26 result in an integer for some integer value of q? Notice that 26 is exactly 2 times 13. So, for any integer value of q, 13q will either be divisible by 26 (if q is even) or produce a remainder of 13 (if q is odd). Adding 6 to 13q, the expression will either produce a remainder of 6 or 19, but will never produce a remainder of zero. Therefore, (13q + 6)/26 = b/26 can never equal an integer.

Answer: B
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chetan2u
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Is there another way to approach this problem

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Hi

The logic is same as suggested by Bunuel..
Let the number be b as given..
b =13y+6....
Now substitute b as 13y +6 in all choices and you will get the ans

Hi, I didnot understand. I know b= 13y+6, when I substitute the value of b in the options I still get a fraction. How do I conclude b is an integer?? Can you please assist?
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Kezia9
chetan2u
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Is there another way to approach this problem

Posted from my mobile device

Hi

The logic is same as suggested by Bunuel..
Let the number be b as given..
b =13y+6....
Now substitute b as 13y +6 in all choices and you will get the ans

Hi, I didnot understand. I know b= 13y+6, when I substitute the value of b in the options I still get a fraction. How do I conclude b is an integer?? Can you please assist?

Hi...
When you substitute b as 13y+6, you get INTEGER in following way..
1) b/6....(13y+6)/6=13y/6 + 6/6... So when y is 0, 13y/6 and 6/6 both will give you 0 as Remainder so it will be integer at b=6
2) b/12...(13y+6)/12.. when y is 6.. 13y+6=13*6+6=6(13+1)=6*14=6*2*7=12*7.. so div by 12
3) 13b/52...13(13y+6)/52..whenever y is multiple of 2 but not 4..
13(13*2+6)/52=13*(32)/52=13*4*8/52=52*8/52=8
4) b/17...(13y+6)/17..when y is 10..(13*10+6)/17=136/17=8..

But you may not require all this if you know why b/26 is not an integer..
b/26=(13y+6)/26=13y/26 +6/26..
Now 13y/26 will leave a Remainder of 13 when y is odd and 0 when y is even..
But 6/26 will always leave 6 as Remainder.
Total remainder can be two
1) 13+6=19
2) 0+6=6
Thus there will always be a Remainder of 19 or 6, hence b/26 will never be an integer

Hope this helps
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