How many digits are there in the product 2^23*5^24*7^3?

Question

A. 24
B. 25
C. 26
D. 27
E. 28

kdatt1991 · Accepted Answer

Seems like a tricky question, but I hope that I have been able to crack it! Here's my solution: (2^{23})*(5^{24})*(7^{3}) = (2^{23})*(5^{23})*(5)*(7)*(7)*(7) (2^{23})*(5^{23})*(5)*(7)*(7)*(7) = ((2*5)^{23})*(5)*(7)*(7)*(7) ((2*5)^{23})*(5)*(7)*(7)*(7) = (10^{23})*(35)*(49) ...... From this step onwards it is probably possible to estimate the number of digits by approximating (10^{23})*(35)*(49) to (10^{23})*(35)*(50) ! But, just to make sure: (10^{23})*(35)*(49) = (10^{23})*(35)*(50-1) .... Therefore (10^{23})*(1750 - 35) which is can be simplified to (10^{23})*(1715) (10^{23})*(1715) should have exactly 27 digits! I think the answer is D! Please consider giving me KUDOS if you felt this post was helpful and correct! or please enlighten me (in case my answer's incorrect) so that I can learn and improve from my mistakes! Thanks.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

The key to this problem is rearranging the math to play to your strengths. You should feel comfortable multiplying 2s by 5s to get 10s, so if you extract 2^23*5^23, you can visualize that number: 10^23, which is a 1 followed by 23 zeroes. Then you're left with 5^1*7^3, which you could either multiply out (not fun but not impossible, either) or again repackage to (5)(7) * (7)(7), which is 35 * 49. That is close enough to 35 * 50 that you can quickly see that that number will have 4 digits, so your final number will be those 4 digits followed by 23 zeroes for a total of 27 digits.

zerosleep · Answer

Well, 2^23 * 5^24 8 7^3 can be simplified 
5 * 7^3 * 10^23

Now either we can multiply 5 and 343 (7^3 = 343) and check or intuitively we can easily see that -
7^3 will be definitely greater than 200 (7*7 = 49 and we have one more 7 to multiply roughly would tak eus to 280+ 
if you dont remember 7^3 =343)... 
what matter here is it will surely give me a -> 3 digit number <-
which when multiplied by 5 will give me no more than a 4 digit number. (we already saw that no is greater than 200 so definitely 4 digit number and not 3)

Hence we can say that on simplifiction we get (4 digit number) * 10^23 
This will give me 4 digit number followed by 23 zeroes and hence no of digits will be 27
Ans- :D

TudorM · Answer

Hi Bunuel,

We can simplify to: 2^23*5^23*5*7^3= (2*5)^23*5*343=10^23*1715

We can see a pattern in the powers of 10 =>
10^1 has 2 digits
10^2 has 3 digits
....
10^23 has 24 digits

IF we simplify 10^1*1715= 17150 =>1715 adds 3 digits to any number, power of 10 =>

(a number with 24 digits, from 10^23) * ( a number that adds 3 digits) = 27 digits

CORRECT RESPONSE D

Manick9 · Answer

Great Explanation.
+1 Kudos

CounterSniper · Answer

2^23*5^23*5*7^3
or,
10^23*5*343
10^23*1715
=27 digits

salaxzar · Answer

To find no of digits, first thing comes in our mind is -
either we do multiplication and see for ourselves. But here, looking at the mammoth factors (2^23 * 5^24 8 7^3), we know it is not posssible or at least very time consuming.

SO, next thing comes in our mind is - we know that 10^2 = 100 = 3 digits
10^3 = 1000 = 4 digits
In fact, 10^ n will have n+1 digits.

Now, I see this because I see a lot of 2's and 5's in the option. Hence, let us try to simplify the given equation - 
2^23 * 5^24 * 7^3 = (2*5)^23 * 5 * 7^3
= 10^23 * 5 * 7^3 *
= 24 digits + whatever we get from rest

Let us solve the rest, 
7^3 will surely give me 3 digit no, if you do not know 7^3 = 343
Multiplying by 5 will give me a 4 digit number.

This will give me 4 digit number followed by 23 zeroes and hence no of digits will be 27. 
Ans D

Madhavi1990 · Answer

I did a slightly more lengthy approach (but for someone still struggling with Quant; maybe useful):
- intial amount: 2^23 5^24 7^3
- least number of digits - 2*3*5 = 70 --> 2 digits; leaving me with 70 x 2^22 x 5 23 x 7 ^2 
- next number -- 70 x 70 x 2^21 x 5^22 x7
- next number -- 343000 x 2^20 x 5 ^21 (now we can understand from this step only zeroes would get added on)
- we have now 3430000 (7 digits) x 2^19 x 5^20 --> we have now 19 zeroes that get added on; leaving aside 5 to power 1
- so we have 7 + 19 = 25 zeroes and 343*5 = 2 additional digits (1715) that get added on
in total 27 digits
Hope this was helpful to anyone that read this :)

ScottTargetTestPrep · Answer

Solution:
 
Since 2^23 x 5^24 x 7^3 = (2 x 5)^23 x 5 x 7^3 = 10^23 x 1,715, we see that the product is the number 1,715 followed by 23 zeros. Therefore, there are 4 + 23 = 27 digits in the product.

Answer: D

HarishKumar · Answer

We have  2^{23}*5^{24}*7^3

which can be rewritten as

2^{23}*5^{23}*5^1*7^3

which gives us

(2*5)^{23}*5^1*7^3

(10)^{23}*5^1*7^3

Now,  10^{23}  is going to give us  1  followed by  23  zeroes, in total  24  digits.

Now coming to  5^1*7^3 .

7^3 = 343  (I happened to know this, but it is better to memorize all cubes up till 10 and squares up till 20 or at least 15 for the GMAT)

Now  5^1 * 343 = 5*343  and we can be sure it won't cross 4 digits. It would be approximately  1500 (5*300) + 200 (5*40) = 1700 , though exactly it will  1715 .

Now  1715  multiplied by  1  is going to be  1715 , thus giving us  4  digits in addition to the  23  zeroes.

Thus total number of digits is going to be  4 + 23 = 27  digits.

Answer is  Choice D .

Maria240895chile · Answer

2^{23}*5^{24}*7^3

the easiest way is to multiply everything by 2

2^{23}*2*5^{24}*7^3 
 =2^{24}*5^{24}*7^3 
 =(2*5)^{24}*7^3 
 =10^{24}*7^3

so  10^{24}  will have a 1 and 24 ceros, so 25 digits and  7^3=343  so 3 digits

because we multiplied by 2 we have to divide by 2 which means one cero less(ex:100/2=50, 1000/2=500...and so on), leaving 24 digits+3 digits=27

Prashant3112 · Answer

2^23*5^24*7^3 =7*7*7*2*2*2*5*5*5*5*10^20 =7*7*7*5*10^24 =1715*10^24. (28 digits - D)

How many digits are there in the product 2^235^247^3?

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How many digits are there in the product 2^235^247^3?