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..
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