When integer b is divided by 13, the remainder is 6. Which of the foll

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.

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

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

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

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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