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If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58401
If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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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

Based 10 notation, or decimal notation, is just a way of writing a number using 10 digits: 1, 2, 3, 4, 5, 6, 7, 8, and 0 (usual way), in contrast, for example, to binary numeral system (base-2 number system) notation.

$$10^{50}$$ has 51 digits: 1 followed by 50 zeros;
$$10^{50}-74$$ has 50 digits: 48 9's and 26 in the end;

So, the sum of the digits of $$10^{50}-74$$ equals to 48*9+2+6=440.

Hope it helps.
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Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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4
C. 440

10^x - 74 --> last 2 digits are always 2, 6.

10^2 - 74 = 26

10^3 - 74 = 926

10^4 - 74 = 9926 and so on....

If x > 1,
the sum of the digits --> (x-2) * 9 + 2 + 6. hence, (50-2) * 9 + 8 --> 440.
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Posts: 916
Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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solving for easier numbers ---> 10^3 and 74

(10^4) - 74 =

(10^4) - 10*7.4 =

10*[(10^3) - 7.4] =

10*[10^3 - 7.4] = 992.6*10 = 9926 = 9*2+2+6 = 26

Note that 10^4 will yield two nines a six and a two.

so solving for 10^50 and 74 will give 48 nines a six and a two:

10*[10^49 - 7.4] = 9*48+2+6 = 440

the answer is (C) VP  Status: mission completed!
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Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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dimitri92 wrote:
If 10^50 - 74 is written as an integer in a base 10 notation.What is the sum of the digits in that integer?

a. 424
b. 433
c. 440
d. 449
e. 467

C. 440
another approach is:
We know that 10^50 is ending 00, so 10^50-74=9....9926
total number of digits in 10^50-74 is 50, or 48 digits of 9 and two digits 2 and 6.
answer choice is 48*9+8=440

plugging numbers:
let represent the sum of the digits in that integer as Y, with the reminder 8, we can represent it in form Y=X*9+8, where X number of digits in 10^50-74 and 8=2+6.

Start with C and than move to B or D.

B. 433=X*9+1, X=48
C. 440=X*9+8, X=48 - correct as we have the reminder 8 and 48 number of digits (50-2), 2 digits are 26.
D. 449=X*9+8, X=49

Personally, I like NSP007's approach. My approaches are easy to comprehend.
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Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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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 $$9*48+2+6=440$$.

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Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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10^1 = 10
10^2 - 74 = 026
10^3 - 74 = 926
10^4 - 74 = 9926

Basically for 10^n , its 9999....(n-2)26.

So for 10^50-74, it is 99999....4826

48times9 + 2+6 = 440.
Intern  Joined: 04 Oct 2013
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Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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I don't understand why in the question it is mentioned "in base 10 notation"

Maybe its because English is not my mother tongue but that instruction really confused me. I thought I was looking for a number like "ten to the power of something".
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Joined: 02 Sep 2009
Posts: 58401
Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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Lobro wrote:
I don't understand why in the question it is mentioned "in base 10 notation"

Maybe its because English is not my mother tongue but that instruction really confused me. I thought I was looking for a number like "ten to the power of something".

Based 10 notation, or decimal notation, is just a way of writing a number using 10 digits: 1, 2, 3, 4, 5, 6, 7, 8, and 0 (usual way), in contrast, for example, to binary numeral system (base-2 number system) notation.

Similar questions to practice:
the-sum-of-the-digits-of-64-279-what-is-the-141460.html
the-sum-of-all-the-digits-of-the-positive-integer-q-is-equal-126388.html
10-25-560-is-divisible-by-all-of-the-following-except-126300.html
if-10-50-74-is-written-as-an-integer-in-base-10-notation-51062.html

Hope this helps.
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Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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100 - 74 = 26

The last 2 digits of the term would be 26; all else would be 9

99999999......26

Important rule:

Sum of ANY NUMBER added to 9 would give the SAME value of itself

For example; Consider number = 13

Sum of digits = 1+3 = 4

Adding 9 to 13 = 22 = 2+2 = 4

So the sum would always remain the same;

Back to our problem

99999999......26 = The sum of this number will add up to 2+6 = 8

From the options available, A & B can be discarded

9x1 = 9
9x2= 18
9x3= 27
9x4= 36
9x5= 45
9x6= 54
9x7= 63
9x8= 72 ........................................ 48th time
9x9= 81
9x10=90

99999999......26

$$10^{50}- 74$$ means 9 would be repeated 48 times; so last digit would be 2

Now we have 2+2+6 = 10 (Last digit is 0)

Only option C best fits = 440

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Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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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 $$9*48+2+6=440$$.

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" Senior SC Moderator V
Joined: 22 May 2016
Posts: 3548
If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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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 $$9*48+2+6=440$$.

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, but we will not see that clue until after we go further with the earlier pattern.
100 = 10$$^2$$. How many digits in $$10^2 -74?$$ TWO digits in the answer, 26

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 next place to the right and turn it into a 9.
So: (1) there are repeated 9s, and (2) there are 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:

Increasing powers of 10 (100, 1000, 10000) minus 74
produce an answer that has exactly the same number of digits as the exponent on the 10 has.
10$$^3$$ has 3 digits: a 9, a 2 and a 6.
We need to know the number of digits so that we know how many 9s there are.
[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) = 440

Hope 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?

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Re: If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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Bunuel 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

Based 10 notation, or decimal notation, is just a way of writing a number using 10 digits: 1, 2, 3, 4, 5, 6, 7, 8, and 0 (usual way), in contrast, for example, to binary numeral system (base-2 number system) notation.

$$10^{50}$$ has 51 digits: 1 followed by 50 zeros;
$$10^{50}-74$$ has 50 digits: 48 9's and 26 in the end;

So, the sum of the digits of $$10^{50}-74$$ equals to 48*9+2+6=440.

Hope it helps.

For once I did it exactly the way Bunuel did it. I came up here to check if what I did was the right way and not only did I get it right but it's exactly the way Bunuel explained it. This is the first time in my almost a year's time on Gmatclub!

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Help needed to get to a score of 700+ I'm at all ears!!
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If 10^50 – 74 is written as an integer in base 10 notation what is the  [#permalink]

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I had trouble understanding the prompt so I wrote out a short representation of the subtraction and it made sense pretty quickly:

1...00000
-_____74
=...99926

So, a bunch of 9s and 26 at the end.
Looking at answer choices, some are close to 450 which would be 50*9.
We remove 1 digit from 10^50 in carryover for subtraction so this makes sense.
But, we are also losing 9-2=7 and 9-6=3, or 10 since we replace two 9s with 26.
450-10 = 440, C. If 10^50 – 74 is written as an integer in base 10 notation what is the   [#permalink] 21 May 2019, 12:29
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