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a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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07 Jan 2013, 06:59
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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
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Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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07 Jan 2013, 07:21
When a is divided by b remainder is 9 that means b is greater than or equals to 10, similarly d is greater than or equals to 6. b + d cannot be 15, hence E is the answer.



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Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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09 Jan 2013, 16:12
a/b gives reminder 9, hence \(b\geq{10}\) c/d gives reminder 5, hence \(d\geq{6}\) Add above inequalities: \((b+d)\geq{16}\) Among the answer choices, the only value that does NOT satisfy above constraint is 15. Hence choice(E) is the answer.
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Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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10 Jan 2013, 04:15
PraPon wrote: a/b gives reminder 9, hence \(b\geq{10}\) c/d gives reminder 5, hence \(d\geq{6}[/m]
Add above inequalities: [m](b+d)\geq{16}\)
Among the answer choices, the only value that does NOT satisfy above constraint is 15.
Hence choice(E) is the answer. Hi can u please explain highlighted part? I missing sumthing here..



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Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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10 Jan 2013, 04:31
bhavinshah5685 wrote: PraPon wrote: a/b gives reminder 9, hence \(b\geq{10}\) c/d gives reminder 5, hence \(d\geq{6}[/m]
Add above inequalities: [m](b+d)\geq{16}\)
Among the answer choices, the only value that does NOT satisfy above constraint is 15.
Hence choice(E) is the answer. 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 nonnegative integer and always less than divisor.For more check Remainders chapter of Math Book: remainders144665.htmla, 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.
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Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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06 Apr 2014, 10:27
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?
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Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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06 Apr 2014, 10:39



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Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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02 Feb 2015, 10:13
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=a9 and d=c5
=b+d =a9+c5 =a+c14
as a,b,c,d > 0 Int therefore a+b14 > 1 hence 15 is the only exception.
But Bunuel's explanation is more logical



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Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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15 Mar 2017, 13:51
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.
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Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]
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17 Mar 2017, 05:17
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.




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