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If k and p represent non-zero digits within the integers

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If k and p represent non-zero digits within the integers  [#permalink]

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New post 31 Jan 2012, 17:06
1
2
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

77% (01:26) correct 23% (00:59) wrong based on 192 sessions

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8k8
+ k88
--------
1,6p6

If k and p represent non-zero digits within the integers above, what is p?

A. 6
B. 7
C. 8
D. 9
E. 17

Anwer is A and I got it by putting the value of K as 7 as that's the only value that fits in the equation to give me p =6. Is there any other way to solve this question?

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Re: Non-Zero Digits  [#permalink]

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New post 31 Jan 2012, 17:49
1
8k8
+ k88
--------
1,6p6

If k and p represent non-zero digits within the integers above, what is p?


A. 6
B. 7
C. 8
D. 9
E. 17

8k8
k88
--------
16p6

Trial and error or just plug-in method might be the shortest way to solve this problem. Though you can narrow down the possible values of k to just two: 7 and 8 --> 8**+7**=16** or 8**+8**=16** (k can not be less than 7 or 9, as the result won't be 16**). After that it's easy to get that k=7 and p=6.

Answer: A.

Hope it's clear.
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Re: If k and p represent non-zero digits within the integers  [#permalink]

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New post 31 Jan 2012, 19:58
by solving I got k=7 and p=6
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find p  [#permalink]

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New post 01 Apr 2012, 06:38
1
Could you please help with this one (i'd like to see your logic - since mine was incorrect)
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Re: find p  [#permalink]

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New post 01 Apr 2012, 06:49
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Re: If k and p represent non-zero digits within the integers  [#permalink]

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New post 16 Apr 2012, 20:35
8k8= 800+10k+8
k88=100k+80+8
16p6=1000+600+10p+6

From the above we get 11k-p=71
Plugging in values and remembering both are non zero digits we have p=6
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Re: If k and p represent non-zero digits within the integers  [#permalink]

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New post 25 Jul 2014, 09:34
someonear wrote:
8k8= 800+10k+8
k88=100k+80+8
16p6=1000+600+10p+6

From the above we get 11k-p=71
Plugging in values and remembering both are non zero digits we have p=6


question : how exactly did you get 11k - p = 71 ,

if this person is no longer here, bunuel can you outline the steps?

Thanks
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Re: If k and p represent non-zero digits within the integers  [#permalink]

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New post 27 Jul 2014, 16:27
sagnik242 wrote:
someonear wrote:
8k8= 800+10k+8
k88=100k+80+8
16p6=1000+600+10p+6

From the above we get 11k-p=71
Plugging in values and remembering both are non zero digits we have p=6


question : how exactly did you get 11k - p = 71 ,

if this person is no longer here, bunuel can you outline the steps?

Thanks


(800 + 10k + 8) + (100k + 80 + 8) = 1000 + 600 + 10p + 6 --> 11k - p = 71.
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Re: If k and p represent non-zero digits within the integers  [#permalink]

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New post 29 Dec 2016, 00:35
Bunuel wrote:
sagnik242 wrote:
someonear wrote:
8k8= 800+10k+8
k88=100k+80+8
16p6=1000+600+10p+6

From the above we get 11k-p=71
Plugging in values and remembering both are non zero digits we have p=6


question : how exactly did you get 11k - p = 71 ,

if this person is no longer here, bunuel can you outline the steps?

Thanks


(800 + 10k + 8) + (100k + 80 + 8) = 1000 + 600 + 10p + 6 --> 11k - p = 71.



I tried to plug in the values but I am not getting the correct answer.
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Re: If k and p represent non-zero digits within the integers  [#permalink]

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New post 29 Dec 2016, 00:57
Bunuel wrote:
8k8
+ k88
--------
1,6p6

If k and p represent non-zero digits within the integers above, what is p?


A. 6
B. 7
C. 8
D. 9
E. 17

8k8
k88
--------
16p6

Trial and error or just plug-in method might be the shortest way to solve this problem. Though you can narrow down the possible values of k to just two: 7 and 8 --> 8**+7**=16** or 8**+8**=16** (k can not be less than 7 or 9, as the result won't be 16**). After that it's easy to get that k=7 and p=6.

Answer: A.

Hope it's clear.



The reason that we are not taking k as 8 is because it will be yield value of p as 17 and the resultant value will become 5 digit number?
Please correct me if my understanding is wrong.
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If k and p represent non-zero digits within the integers  [#permalink]

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New post 29 Dec 2016, 02:26
1
anairamitch1804 wrote:
Bunuel wrote:
8k8
+ k88
--------
1,6p6

If k and p represent non-zero digits within the integers above, what is p?


A. 6
B. 7
C. 8
D. 9
E. 17

8k8
k88
--------
16p6

Trial and error or just plug-in method might be the shortest way to solve this problem. Though you can narrow down the possible values of k to just two: 7 and 8 --> 8**+7**=16** or 8**+8**=16** (k can not be less than 7 or 9, as the result won't be 16**). After that it's easy to get that k=7 and p=6.

Answer: A.

Hope it's clear.



The reason that we are not taking k as 8 is because it will be yield value of p as 17 and the resultant value will become 5 digit number?
Please correct me if my understanding is wrong.





when k is 8

888---a
888---b
----
1776


Even if K is 8, p can not be 17. P can be 17 only if 8+k(left most digits of a and b)+carry from previous term (carry from k+8) yields the unit digit as 1 and sum of k+8(middle one)+1(carry) has unit digit as 7.

Because in addition, we consider only the unit digit of sum in each digit's addition until the last digits' sum come.

We are not considering 8 because it is violating the digits of the sum given in question

second digit from left should be 6 (as per question), however, in case of k=8, second digit from left is 7 which is not same. So, we can skip this choice.


hope it helps..


+1 Kudos if you like the post :)
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Re: If k and p represent non-zero digits within the integers  [#permalink]

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