What is the remainder when n is divided by 26, given that n divided by

Hi Brent, actually I wanted to place 2k+1 for St1+ST2, but what is with the case if our odd number=1 ? Can we always insert 2k+1 for an odd number ?

Yes, you can always insert 2k+1 for an odd integer. That is actually how we define an odd integer.

Cheers,
Brent

Hello!

wouldn't it be 2k-1? in that sense 13(2k-1) + 3 would not igual 16. I am confused about this because I read that the sequence of all positive odd integers is defined by An = 2N-1. Would this matter for the solution of the problem?

Thank you for your response.

Greetings.

2k - 1 is also an odd number for all integers k. However, if we use 2k-1 here, we need to do a little extra work at the end.

When we use 2k - 1, we get: n = 13(2k - 1) + 3
Simplify to get: n = 26k - 10
In other words, n is 10 LESS than some multiple of 26
Hmmm, what does this tell us about the remainder when n is divided by 26?
To find out, notice that we can take n = 26k - 10 and rewrite is as n = 26(k - 1 + 1) - 10
Or..... n = 26(k - 1) + 26 - 10
Simplify to get: n = 26(k-1) +16
So, n is 16 GREATER than some multiple of 26
So, we when we divide n by 26, we get a remainder of 16

PRO TIP: use 2k + 1 when you need a nice generic odd number

Cheers,
Brent

n= 13a +b

Statement 1= a is odd

There is no information about b

n= 13*1 +b
n= 13*3 +b
n= 13 *5 +b

Not sufficient.

Statement 2= b = 3
Multiple values possible.
n= 3
n= 16 (13*1 +3)
n= 29 (13*2+3)
n= 42 (13*3 +3)

Not sufficient.

Combining both statements:-
When a is 1 (odd) and b= 3, let's say 16

then when n/26 will have 16 as remainder

Similarly when n = 42 - 16 is remainder for n/26

C is the answer

N=16(a=1(Odd) and b=3) then Remainder=16

N=42(a=3 & B=3) Then remainder=8

N=68(a=5 and b=3) then Remainder=8

As we are getting different remainders how the answer is C?

I am wondering the same thing..

n = 13a + b

St 1 + 2) --> a = 1, 3, 5, 7 etc b = 3

n = 13a + 3
if a = 1, n = 16. 16/26 gives remainder 26! (because 26 does not go into 16, hence 16/26 is 0 remainder 26).
if a = 3, n = 42. 42/26 gives remainder 16.
if a = 5, n = 68. 68/26 gives remainder 16.

We see that this will hold for all "A" greater than 1 and also odd.. so from 3 onwards. However, a = 1 is a possibility and this yields a different remainder.

If someone could please provide an explanation that'd be great.

The highlighted part is wrong: the remainder can only be a non-negative integer that is less than the divisor. In the case of n=16, the divisor is 26. So it goes into it 0 times, thus the remainder is 16.
Think about it like fractions... 16/26 is less than 1... 0 + 16/26
Comparatively, a=3 gives 42/26 which is 1 + 16/26

Check: https://gmatclub.com/forum/remainders-144665.html
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... y-applied/

Official Solution

Courtesy: Veritas Prep

Solution: This means that out of “n” balls, if we make groups of 13, we will be able to make “a” groups and will have “b” balls leftover.

What happens when we try to combine two groups of 13 to make a group of 26? There are two possibilities: all groups of 13 will be used to make groups of 26 and “b” balls will be leftover (as before) or one group of 13 and “b” balls will be leftover.

What will decide whether a group of 13 will be leftover? If “a” is 2 i.e. we have two groups of 13, we will be able to make one group of 26 and no group of 13 will be leftover. If “a” is 3 i.e. we have three groups of 13, we will be able to make one group of 26 and one group of 13 will be leftover. If “a” is 4 i.e. we have four groups of 13, we will be able to make two groups of 26 and no group of 13 will be leftover. What do you conclude from these examples? If “a” is even, we will have no group of 13 leftover. If “a” is odd, we will have one group of 13 leftover. So the remainder when n is divided by 26 will depend on whether a is odd or even and the value of “b.” Let us look at the statements now:

Statement 1: a is odd.

If “a” is odd, then a group of 13 will be leftover. So the remainder will be 13 + b. But we do not know the value of “b.” So this statement alone is not sufficient.

Statement 2: b = 3

We now know that b = 3 but from this statement alone, we do not know whether a is odd or even. So this statement alone is not sufficient.

Taking both statements together, we know that remainder is 13+3 = 16. Hence both statements together are sufficient.

sorry I am unable to understand your explanation. since when we put say 45 , 68 both satisfies S1 N S2. but putting in original gives different valueof
R.. PLS guide.

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