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06 Dec 2021, 08:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
58% (02:43) correct
42%
(02:18)
wrong
based on 170
sessions
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The product (8)(888...8), where the second factor has k digits, is an integer whose digits have a sum of 1000. What is k?
(A) 901
(B) 911
(C) 919
(D) 991
(E) 999
Are You Up For the Challenge: 700 Level Questions
(A) 901
(B) 911
(C) 919
(D) 991
(E) 999
Are You Up For the Challenge: 700 Level Questions
06 Dec 2021, 17:18
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Bunuel
While there are often much slicker ways to handle problems like this (and I look forward to the discussion below), an approach I often fall back on in situations involving calculations with improbably large numbers is to start the calculation with smaller versions of the numbers in question to see if a pattern emerges. In this case, the approach would go..
8 x (8) = 64
8 x (88) = 704
8 x (888) = 7104
8 x (8888) = 71104
8 x (88888) = 711104...
To be clear, the above math is partly for illustration. In doing the problem, I probably wouldn't calculate all of these numbers; rather, I would start with a pairing like 8 x (88888) and see the pattern that emerges while doing the math. In this case, doing the math lands you with a number that starts following the pattern 711...1104, in which the ellipsis represents a number of 1s that grows depending how many digits the second factor in the original multiplication has. The question now becomes how many 1s we'll need for the digits to add up to 1000 and what that means for the variable in question, k. The number that results from this calculation will always have a 7 as its lead digit, a 0 as its tens digit, and a 4 as its units digit. Because these digits themselves add up to 11, to get to a digit-sum of 1000, the product will need 989 1s.
It can be tempting to go to the answers looking for 989 at that point, but always return to the question first. What does k have to be in order for us to get 989 1s? Let's return to the calculation above: when there were four 8s in the second factor, there were two 1s in the final product. When there were five 8s, we got three 1s. The number of 1s in the final product is always two fewer than the number of 8s in the second factor. Following this pattern, in order to get 989 1s in the final product, we'll need 991 8s in the second factor. Therefore, the answer is D.
General Discussion
06 Dec 2021, 22:22
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Bunuel
8 * (88...8) = 64 * (11...1)
There are k 1s in the second term.
64 * (11...1) = 64 (100000... + ... + 1000 + 100 + 10 + 1)
Hence, to get the product, we are essentially adding
6400000...
..640000...
....64000...
......6400...
....
..........640
............64 +
______________
........1104
______________
For every digit after the units digit and tens digit, we will get 11 out of which we will carry 1. So the sum will be 71111...11104. If sum of all digits here is 1000, we must have 989 1s.
Now how is it related to k?
If k = 1, Sum = 64
If k = 2, Sum 704
If k = 3, Sum = 7104
If k = 4, Sum = 71104
Hence, k = Number of 1s + 2 = 989 + 2 = 991
Answer (D)
