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

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a, b, c, and d are positive integers. If the remainder is 9 [#permalink]  07 Jan 2013, 05:59
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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
[Reveal] Spoiler: OA

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Kudos [?]: 289 [3] , given: 22

Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]  07 Jan 2013, 06:21
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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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Kudos [?]: 505 [0], given: 182

Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]  09 Jan 2013, 15:12
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a/b gives reminder 9, hence $$b\geq{10}$$
c/d gives reminder 5, hence $$d\geq{6}$$

$$(b+d)\geq{16}$$

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

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Kudos [?]: 67 [0], given: 15

Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]  10 Jan 2013, 03:15
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.

Hi can u please explain highlighted part? I missing sumthing here..
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Kudos [?]: 50073 [0], given: 7527

Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]  10 Jan 2013, 03:31
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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.

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.

Hope it's clear.
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Kudos [?]: 9 [0], given: 11

Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]  06 Apr 2014, 09:27
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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Kudos [?]: 50073 [1] , given: 7527

Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]  06 Apr 2014, 09:39
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Expert's post
HCalum11 wrote:
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: $$1=9*0+1$$.

Theory on remainders problems: remainders-144665.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199

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Kudos [?]: 66 [0], given: 14

Re: a, b, c, and d are positive integers. If the remainder is 9 [#permalink]  02 Feb 2015, 09:13
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=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
Re: a, b, c, and d are positive integers. If the remainder is 9   [#permalink] 02 Feb 2015, 09:13
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