If x is a positive integer and 10^x – 74 in decimal notation

10^x is a (x+1)-digit number: 1 followed by x zeros.

10^x - 74 is a x-digit number: x-2 9's and 26 in the end. Thus the sum of the digits is (x-2)*9+2+6=440 --> x = 50.

Answer: E.

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Hope it helps.
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This might be a simpler method -

We know that since 10^x contains x number of zeros, therefore -
the last 0 will yield 10-4 = 6
the 2nd last zero will yield the number 9-7 = 2

Also, the rest will all be 9s. So all we need to do is calculate the number of 9s

let x be the number of 9s

9m + 6 + 2 = 440
m=48
Now we need to add 2 to the above number for the zeros for the last two digits that contributed to the subtraction
Therefore x= 48 + 2 = 50

Hope this helps

I think its E 50....My reasoning is as such:

Assuming its (10^x) - 74

I plugged in for x a 2 and 3

I saw that no matter what you put in x you get 10000alot of zeros - 74...Which always left you with a answer that had a 6 in the units and a 2 in the tens slot, and a bunch of 9s in the other slots...So the equation really is...

What number will give you 440? So it becomes 9(z) + 2 + 6 = 440....So how many 9s do I need in my answer. I solved for z and got 48 and then added 2 to it for the 2 and 6. OA please?

10^x – 74 = k....................k26
sum of k+.....+k+2+6 = 440
sum of k+.....+k = 440 - 8
sum of k+.....+k = 432

all k must be of integer 9.
so no of k = 432/9 = 48

so x has to be 50 (=48+2) to have k........k26 = 9..........926 = 10^x - 74.
so it is E.

Quite simple way..

GMATinsight's Solution


We need to find value of x for which sum of the digits of \(10^x – 74 = 440\)

For x = 2, \(10^x – 74 = 10^2 – 74 = 100-74 = 26\) i.e. Sum of digits = 0*9+2+6
For x = 3, \(10^x – 74 = 10^3 – 74 = 1000-74 = 936\) i.e. Sum of digits =1*9+2+6
For x = 4, \(10^x – 74 = 10^4 – 74 = 10000-74 = 9926\) i.e. Sum of digits =2*9+2+6
For x = 5, \(10^x – 74 = 10^5 – 74 = 100000-74 = 99926\) i.e. Sum of digits =2*9+2+6
For x = 6, \(10^x – 74 = 10^6 – 74 = 1000000-74 = 999926\) i.e. Sum of digits =4*9+2+6

i.e. Sum of the digits of \(10^x – 74\) will be \((x-2)*9+2+6 = 440\)

\((x-2)*9+2+6 = 440\)
i.e. \((x-2)*9 = 432\)
i.e. \((x-2) = 48\)
i.e. \((x) = 50\)

Answer: Option E





Nups1324

=> So, let's say x value is 2
10^2 = 100
10^2 -74 = 26 .
=> If x = 3
10^3-74 = 926
=> If x = 4
10^4-74 = 9926

You can see the pattern that after every power of 10 except 1&0 (because 10^1 = 10 & 10^0 = 1 would have made result in negative) result in same unit and ten digits.
We need to have the power of x, which gives number equivalent to the sum of 440.
In X = 2
10^2-74 = 26 , left digits sum = 2+6 = 8
In X = 3
10^3-74 = 926 , left digits sum = 9+2+6 = 17 and so on....

So we know out of 440 , 8 is the sum of ten and unit digit
440-8 = 432
When we divide 432 by 9 (because 10^x-74 will leave digits as 9999....................99926, we have to find the number of 9)
432/9 = 48

So the number of 9 before 26 is 48 (we are seeking value of x which leave a 50 digit value when 74 is deducted)
If you see the repeated pattern above 10^2-74 left 2 digit value
10^3 left 3 digit value
So 10^50 will leave 50 digit value where there are 48 9's till hundredth value 2 as ten and 6 as one's.
Hence answer is E
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10^x ends 0 in unit digit, tens digit, hundreds digit and son on. If so, then, deducting 74 from 10^x, results in 6 in unit digit, 2 in tens digit and 9 in rest of the digits. so:

440 = 2+6+(9x)
x = (440-8)/9
x = 432/9=48

lets see tha pattren:
10^2 - 74 = 100-74 = 26
10^3 - 74 = 1,000-74=926
10^4 - 74 = 10,000-74 = 9926
10^5 - 74 = 100,000-74 = 99926

so in 10^x, x has to be 50.

Right, once you realize that 9 x 48 is 430, you know that A-D can't work.

the last two digits will be 2&6. so
9(x-2)=400-(2+6)
=>x=50

Answer E = 50

We have (10^x)-74, where the sum of the digits is 440.

100-74 = 26, so we already have 2+6 = 8 from the 440.

Because we have a power of 10, we will have numbers like 100, 1000, 10000 and so on. This minus 74 rests 26 and a lot of 9s. E.g: 1000-74 = 9926.

So dividing the 432/9 = 48, that`s the number of 9s. and we have 2 other numbers (2 and 6) wich were 00 before the subtraction.

So we have 48 + 2 as an X = 50

Answer: E.
Reason: Rightmost 2 digits of pattern 10^x - 74 will be 26 (100 - 74 = 26, 1000 - 74 = 926...so on). One observation, the number of digits of the result of the pattern is equal to x (26 when x = 2, 926 when x = 3...so on). Now apart from last 2 digits 2 and 6, other digits in the result will be 9. We need to find n, where n*9 + (2+6) = 440 (given), which gives n = 48. And final answer is 48 + 2 (number of right hand 2 digits 2 and 6) = 50.

10^2 - 74=26
440-8(sum of 2 and 6)=432
432/9=48
so we need 10^48*10^2=10^50
the correct answer is E

The way I thought about this problem:

We know that we have some multiple of 10 that we're subtracting 74 from, so outside of 2+6 (left over from 10^n) we'll be dealing with 9's. The question is how many, because that will determine the value of n.

2+6=8 --> subtract from 440 --> we're left with 432

432/9=48

Now add 2 because we have two slots occupied by 2 and 6.

Thus, E is correct.

Can someone help me understand what decimal notation means in the context of this question? Thanks.

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.
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(10)^x - 74 = Integer whose Digits Add up to 440

x=2: 100 - 74 = 26 ---- 8

x=3: 1,000 - 74 = 926 ---- 17

x=4: 10,000 - 74 = 9,926 ---- 26

This is an Arithmetic Progression with Term 1 = 8 and the Common Difference d = 9

An = 8 + 9 *(n - 1)

we want to find which Term will give us a Sum of 440. An = 440.

440 = 8 + 9 * (n - 1)

440 = 8 + 9n - 9

441 = 9n

n = 49th term

for the 1st Term 8 ---- x = 2

thus, for the 49th Term 440 ---- x = 50

(10)^50 - 74 = Digits that SUM to 440

Answer E

Hi Bunuel,

I didn't understand this part,
"10^x - 74 is a x-digit number: x-2 9's and 26 in the end. Thus the sum of the digits is (x-2)*9+2+6=440 --> x = 50."

It will be great if you'd help me out with this.

Tagging others just in case Bunuel is not able to reply.
yashikaaggarwal chetan2u GMATinsight IanStewart nick1816 fskilnik.

Thank you

Posted from my mobile device

Wow.! That is so beautifully explained. This is a really nice concept. I honestly learned something new today.

Thank you so much yashikaaggarwal

100-74=26
1000-74=926
10000-74=9926
;
10^x-74=999...(x-2)times 26

SOD(10^x-74)=9(x-2)+2+6=9x-10=440
=>x=450/9=50