I know GMAT put the following questions on the test with no intention of you actually solving them (no messy math),..so I was wondering if there's an easy way to solve questions like these two? Insights are appreciated!
Question #1:
If 10^50 – 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?
A. 424
B. 433
C. 440
D. 449
E. 467
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Question #2:
In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 16th, 17th, and 18th terms in the sequence?
A. 2^18
B. 3(2^17)
C. 7(2^16)
D. 3(2^16)
E. 7(2^15)
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What's the quick and dirty way of solving Qs like these 2?
This topic has expert replies
Q1:
10^50 - 74 = (9999....-48 times)26.
for example 10^3 = 1000-74 = 926, (9 digit appears once i.e 3-2) . similarly 10^50 -74 ( 9 digit will appear 48 times i.e 50-2)
Therefore sum would be 48*9 + 2 + 6 = 440 (Ans: C)
Q2:
the seris can be written as
2^0,2^1,2^2,2^3,2^4 and so on.
nth term = 2^(n-1)
Therefore 16,17 and 18 terms will be:
2^15,2^16,2^17
Sum would be :
2^15+2^16+2^17 = 2^15(1+2^1+2^2)
=> 7(2^15)
(Ans: E)
10^50 - 74 = (9999....-48 times)26.
for example 10^3 = 1000-74 = 926, (9 digit appears once i.e 3-2) . similarly 10^50 -74 ( 9 digit will appear 48 times i.e 50-2)
Therefore sum would be 48*9 + 2 + 6 = 440 (Ans: C)
Q2:
the seris can be written as
2^0,2^1,2^2,2^3,2^4 and so on.
nth term = 2^(n-1)
Therefore 16,17 and 18 terms will be:
2^15,2^16,2^17
Sum would be :
2^15+2^16+2^17 = 2^15(1+2^1+2^2)
=> 7(2^15)
(Ans: E)
