a, b, c, and d are positive integers. If the remainder is 9

Hi can u please explain highlighted part? I missing sumthing here..

If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since \(15 = 6*2 + 3\).

Notice that \(0\leq{r}<x\) means that remainder is a non-negative integer and always less than divisor.

For more check Remainders chapter of Math Book: remainders-144665.html

a, b, c, and d are positive integers. If the remainder is 9 when a is divided by b, and the remainder is 5 when c is divided by d, which of the following is NOT a possible value for b + d?

(A) 20
(B) 19
(C) 18
(D) 16
(E) 15

According to the above, since the remainder is 9 when a is divided by b, then b (divisor) must be greater than 9 (remainder). So, the least value of b is 10.

Similarly, since he remainder is 5 when c is divided by d, then d must be greater than 5. So, the least value of d is 6.

Hence, the least value of b + d is 10 + 6 = 16. Therefore 15 (option E) is NOT a possible value for b + d.

Answer: E.

Hope it's clear.

what if a = 1 and b= 9...then wouldn't 1/9 still have a remainder of 9? doesn't the rule that b must be greater than or equal to 10 not hold in this case?


Posted from my mobile device

No.

Let me ask you a question: how many leftover apples would you have if you had 1 apple and wanted to distribute in 9 baskets evenly? Each basket would get 0 apples and 1 apple would be leftover (remainder).

When a divisor is more than dividend, then the remainder equals to the dividend, for example:
3 divided by 4 yields the reminder of 3: \(3=4*0+3\);
9 divided by 14 yields the reminder of 9: \(9=14*0+9\);
1 divided by 9 yields the reminder of 1: \(1=9*0+1\).

Given
a,b,c,d > 0 Int

b+d != ? (! = not)


a=bq+9 (q=1,2,3.....)
c=dr+5 (r=1,2,3.....)

when q=r=1

a=b+9 amd c=d+5
b=a-9 and d=c-5

=b+d
=a-9+c-5
=a+c-14


as a,b,c,d > 0 Int
therefore a+b-14 > 1
hence 15 is the only exception.

But Bunuel's explanation is more logical

To answer, we must recognize an important rule: the divisor must be greater than the remainder. Let's look at a few examples:
10 ÷ 4 = 2 remainder 2 (divisor 4 is greater than remainder 2)
23 ÷ 6 = 3 remainder 5 (divisor 6 is greater than remainder 5)

If the divisor weren't greater than the remainder, the divisor would be able to divide into the dividend at least one more time. Let's take an incorrect example to illustrate:
23 ÷ 6 = 2 remainder 11.

In this case, we've framed the operation such that the divisor is LESS than the remainder (6 is less than 11). The error is that 6 actually goes into 23 three times. The remainder is what is left over when the divisor has been divided into the dividend as many times as possible. Therefore, if a divided by b gives a remainder of 9, we can conclude that b is greater than 9: b ≥ 10.

Likewise, if c divided by d gives a remainder of 5, we can conclude that d is greater than 5: d ≥ 6. Therefore, we can determine a minimum for the sum: b + d ≥ 16.
Only 15 is too small.

The correct answer is E.

a/b = Q (9), so 'b' cant be 9 or less than 9. It has to be greater than 9.
c/d = Q (5), so 'd' cant be d or less than 5. It has to be greater than 5.

Now by adding the least values of b & c,
b+c = 10+6 = 16. (Considering only the least values.)

Among the answer choices, the only value that does NOT satisfy above constraint is option E i.e. 15.

for the given condition the least values of a,b,c,d is
9,6,5,10 ; therefore b+d ; 6+10 ; 16
least value is 16 so not possible is 15
option E

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.