If n = (2^2)^22, then how many digits does n have?

Question

If  n = (2^2)^{22} , then how many digits does n have?

A. 12
B. 13
C. 14
D. 15
E. 16

Bunuel · Accepted Answer

n = (2^2)^{22}=

= 2^{2*22}=

= 2^{44}=

= 2^4*2^{40}=

= 16*(2^{10})^4=

= 16*(1024)^4 \approx

\approx 16*(10^3)^4 =

= 16*10^{12}

16*10^12 is 16 followed by 12 zeros, so it has fourteen digits.

Answer: C.

gmatophobia · Answer

Not sure if this is the best approach

n = (2^2)^{22}

n = (2)^{44}

n = (2)^{10} * (2)^{10} * (2)^{10} * (2)^{10} * 2^{4}

n = 1024 * 1024 * 1024 * 1024 * 16

n = (1000+24)^{2} * (1000+24)^{2} * 16

n = (10^{3}+24)^{2} * (10^{3}+24)^{2} * 16

n = (10^{6}+24^{2} + 48*10^{3})* (10^{3}+24)^{2} * 16

n = (10^{6}+48576)^{2} * 16

n = (10^{12}+48576^{2} + 48576 * 2 * 10^{6}) * 16

48576 is less than  10^6 , so the number of digits of its square will be less than 12. The same holds true for  48576 * 2 * 10^{6}

Hence the total number of digits will depend on the number of digits contributed by  10^{12} * 16  =  16 * 10^{12}  = 12 + 2 = 14

To test out let's consider a smaller value

16 * 10^1 = 160
16 * 10^2 = 1600

So the number of digits = 2 (for 16) + Power of 10

Option C

Kinshook · Answer

Asked:  If  n = (2^2)^{22} , then how many digits does x have?

n = (2^2)^{22}

log10 n = 22 log10 4 = 22*2*log10 2 = 22*2*.3010 = 44*.3010 = 13.244
Number of digits = 13 + 1 = 14

IMO C

gmatophobia · Answer

Trying to understand this method a bit better -

Say If I were to find the number of digits in  2^{10}

We know the answer is 4.

By above method wouldn't it be -

10	ext{log}_{10} 2 = 10 * 0.3010 = 3 ?

Kinshook · Answer

gmatophobia 
If x = 10 ; log10 10 = 1 ; Number of digits = 1+1 = 2
Please see my revised solution.

gmatophobia · Answer

Thanks for the reply, but isn't the highlighted portion still incorrect? Shouldn't the number of digits for 2^{10} = 4 and not 2. Am I missing something ?

Bunuel · Answer

P.S. Logarithms are not tested on the GMAT. Try solving this question without them.

appyankit · Answer

Gmat hard questions are like purposing A attitude girl, this is very tricky

Q.5680 * 5780 calculate the digit

Except last digit make all of them 0

5000*5000

Now 25,000,000 now calculate the digits and get the answer

Now back to the question

A => 2^2^22 => 2^44 => 2^10 * 2^10 * 2^10 * 2^4

=> 1024 * 1024 * 1024 * 1024 * 16

Make all them Zero except last digit

=> 1000 * 1000 * 1000 * 1000 * 10

Ans => 10,000,000,000,000

Posted from my mobile device

Kinshook · Answer

Asked: If  n = (2^2)^{22} , then how many digits does x have?

n = (2^2)^{22} = 2^44 = 2^40 * 2^4 = (1024)^4*16 
Number of digits = 3*4 + 2 = 14

IMO C

GiovanniMalfitano · Answer

2^3 = 8 less than ten, so it does not add a "number"
knowing that we can easily do 44/3 which is 14.66 but we need to take the integer so the answer is 14 (C)

GulfTube · Answer

how did you recognize that 1024 can be safely estimated to be 1000 without having any issues in mis counting the total digits?
 Bunuel

egmat · Answer

Hi GulfTube,

Your caution is right: rounding  can  wreck a digit count, so what you want is a test you can run  before  you trust the shortcut, not after.

Start with what a digit count actually depends on:  which two powers of 10 the number sits between . A  14 -digit number is anything from  10^13  up to just under  10^14 . So the only real question is whether the rounding can shove you across one of those walls.

Step 1 - note which way the error goes.

Replacing each  1024  with  1000  makes the number  smaller , so Bunuel's  16 x 10^12  is a  floor : the true n is at least that much. Written with a single leading digit, that floor is  1.6 x 10^13  - already  14  digits.

Step 2 - compare the error to the headroom.

To gain a  15th  digit, n would have to climb all the way to  10^14 , which is  6.25x  your floor. Now price the rounding: each swap costs a factor of  1024/1000 = 1.024 , and there are  four  of them, so  1.024^4 = 1.10  - about  10% . A  10%  correction cannot cover a  6.25x  gap, so  14  digits is locked in. (Exact value:  17,592,186,044,416  -  14  digits, as promised.)

That's the general test, and it's quick: write your estimate as  c x 10^k  with c between  1  and  10 , then ask whether the rounding error could push  c  past  10  (or below  1 ). Here c moves  1.6  to  1.76 . Nowhere near the wall. Even  ten  such roundings would only inflate it about  27% .

Where the shortcut does break:

Estimate  999^4  by rounding up to  1000^4 = 10^12 , and it looks like  13  digits. But  999^4 = 996,005,996,001  - only  12 . That estimate landed exactly  on  a power of  10 , so it had zero headroom, and a tiny error cost a whole digit.

Answer: C

If n = (2^2)^22, then how many digits does n have?

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If n = (2^2)^22, then how many digits does n have?

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