dave13 wrote:
Bunuel wrote:
udaymathapati wrote:
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
\(10^{50}\) has 51 digits (1 followed by 50 zeros). \(10^{50}-74\) has 50 digits: last 2 digits are 2 and 6 (26) and first 48 digits are 9's.
Like 1,000-74=926.
So the sum of the digits is \(9*48+2+6=440\).
Answer: C.
generis can you please explain ?
i dont understand how after \(10^{50}-74\) we have 50 digits
And how we get "last 2 digits are 2 and 6 (26) and first 48 digits are 9's"
dave13 , I've seen you use patterns. Good instinct. Use a pattern. (I think you missed the "1,000" pattern above.)
In other words, when exponents are huge, we can replicate the question with exponents that are manageable.
Make the powers of 10 smaller and test a few.
First we have to figure out what the digits ARE. That's just subtraction. Start with 100. (You could start with 1,000, which would be a little more accurate. 1,000 - 74 = 926. There is a 9. But, see below, 26 is always there.)
Given (100-74), what is the sum of the digits?*
100-74 = 26. Sum of the digits? (2+6)=8
How many digits in the answer? TWO. You wrote: "i dont understand how after \(10^{50}-74\) we have 50 digits"
The exponent, 50, gives us a clue. Back to the earlier pattern.
100 = 10\(^2\). How many digits in
\(10^2 -74?\) TWO digits in the answer, 26
But we have to be careful. If subtracting a positive integer (less than 100) from 10\(^2\), the possible number of digits in the answer is two OR one.
Two digits: (100-74) = 26
One digit: (100-94) = 6
The exponent is a clue only. Simple subtraction, with a few examples, will tell us how many digits. So let's go higher by powers of 10:
10\(^3\) = 1,000
10\(^4\) = 10,000
10\(^5\) = 100,000
Subtract 74 from each one. (Writing on paper really shows the pattern. Formatting here is hard):
(1,000 - 76) =
926
(10,000-76) =
9,926
100,000-76 =
99,926
\((10^3 - 74)\) has THREE digits. One 9, and 26
\((10^4 - 74)\) has FOUR digits. Two 9s, and 26
\((10^5 - 74)\) has FIVE digits. Three 9s, and 26
1) We are getting the same number of digits as the exponent on 10
2) The last two digits will always be 26
3) We have to borrow to move the initial 1 to the hundreds place. So there are repeated 9s. And only 9s until 26.
4) How many 9s? Exactly TWO fewer than 10's exponent (because 2 and 6 "use up" two of the digits)
Finally, SUM of the digits?
Back to the pattern:
After 100, increasing powers of 10 minus 74
produces an answer that has exactly the same number of digits as the exponent on the 10 has.
[Those digits will consist of varying quantities of the number 9, plus one numeral 2 and one numeral 6).
(1,000 - 76) =
926
(10,000-76) =
9,926
100,000-76 =
99,926
\(10^3 - 74 = ((1*9)+26)=(9+26)=35\)
\(10^4-74 = ((2*9)+26)) =(18+26)=44\)
\(10^5-74= ((3*9)+26))=(27+26)=53\)Try extrapolating from the pattern above to answer this question:
What is the sum of the digits of \(10^{50} - 74\)?
We get:
1) the number of 9s will be exactly two fewer than the exponent on the 10, so:
10\(^{50}\) = (50 - 2) = 48 instances of the number 9
2) there will also be one 2 and one 6
3)
the sum of the digits is
(48 * 9) =
432 (that part takes care of summing the
9s). Then add the
2:
(432 + 2) = 434. Then add the
6 and we are done.
(434 + 6) =
440Hope that helps.
*A fancy way to ask that question: If \(10^{2} - 74\) is written as an integer in base 10 notation, what is the sum of the digits in that integer?Does that help?
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