IF T = 1 / (2^9 * 5^3) IS EXPRESSED

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factor26: Senior | Next Rank: 100 Posts; Posts: 99; Joined: Sun Mar 13, 2011 12:42 pm

by factor26 » Sat Nov 12, 2011 4:35 pm

IF T = 1 / (2^9 * 5^3) IS EXPRESSED AS A TERMINATING DECIMAL, HOW MANY ZEROS WILL (T) HAVE BETWEEN THE DECIMAL POINT AND THE FIRST NONZERO DIGIT TO THE RIGHT OF THE DECIMAL POINT?

A - THREE

B - FOUR

C - FIVE

D - SIX

E - NINE

OG ANSWER IS B... I SOLVED THE QUESTION GOING THROUGH THE MATH THE LONG WAY ...

2 ^ 9 = 512
5 ^ 3 = 125
-----
64,000

1/64000 = .00001 ---> # OF 0'S BEFORE A NON-ZERO DIGIT = 4

CAN SOMEONE SHOW ME HOW TO SOLVE THIS AN EASIER WAY? THANKS!

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smackmartine: Legendary Member; Posts: 516; Joined: Fri Jul 31, 2009 3:22 pmFollowed by: 13 members; Thanked: 112 times

by smackmartine » Sat Nov 12, 2011 5:00 pm

IMO B
T = 1 / (2^9 * 5^3)

Notice that 2*5 = 10 , so let's try to leverage this fact.

T = 1/ (2)^6 (2*5)^3 = (1/2)^6* 1/(10)^3
Also, 1/2 = 0.5 or 5/10 so, let's re write the expression in terms of powers of 10.

T = (5)^6 * 1/(10)^(6+3) = (5)^6 /(10)^9

At this point you should remember that (5)^4 = 625 [since 25*25 = 625]
which gives 25* 625 = 15625 [relatively easier to calculate]

So , T = 0.000015625 ,so # of ZEROS WILL (T) HAVE BETWEEN THE DECIMAL POINT AND THE FIRST NONZERO DIGIT TO THE RIGHT OF THE DECIMAL POINT = 4

Take away: try to look for simple relation between the digits that can cut short long calculations.

Last edited by smackmartine on Sat Nov 12, 2011 6:28 pm, edited 1 time in total.

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Neo Anderson: Senior | Next Rank: 100 Posts; Posts: 92; Joined: Thu Oct 06, 2011 8:06 am; Thanked: 18 times

by Neo Anderson » Sat Nov 12, 2011 6:10 pm

T = 1/ (2)^5 (2*5)^3 = (1/2)^5* 1/(10)^3

Hey 'smackmartine'; I agree with your answer choice, however not sure how you got the above step,
IMO T = 1/ (2)^6 (2*5)^3 = (1/2)^6* 1/(10)^3

however the answer will be the same 'B'.

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Sat Nov 12, 2011 6:12 pm

1/(2^9 * 5^3) = 1/(10^3 * 2^6) = 1/1000 * 1/64

1/1000 - 3 0's
Now any number 'x' 10>x>100 in the denominator will have one '0', so 1/64 will have 1 '0'. Total = 4

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smackmartine: Legendary Member; Posts: 516; Joined: Fri Jul 31, 2009 3:22 pmFollowed by: 13 members; Thanked: 112 times

by smackmartine » Sat Nov 12, 2011 6:29 pm

Neo Anderson wrote:
T = 1/ (2)^5 (2*5)^3 = (1/2)^5* 1/(10)^3
Hey 'smackmartine'; I agree with your answer choice, however not sure how you got the above step,
IMO T = 1/ (2)^6 (2*5)^3 = (1/2)^6* 1/(10)^3

however the answer will be the same 'B'.

Edited..Thanks for the catch :)

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Nov 12, 2011 6:58 pm

factor26 wrote:IF T = 1 / (2^9 * 5^3) IS EXPRESSED AS A TERMINATING DECIMAL, HOW MANY ZEROS WILL (T) HAVE BETWEEN THE DECIMAL POINT AND THE FIRST NONZERO DIGIT TO THE RIGHT OF THE DECIMAL POINT?

A - THREE

B - FOUR

C - FIVE

D - SIX

E - NINE

OG ANSWER IS B... I SOLVED THE QUESTION GOING THROUGH THE MATH THE LONG WAY ...

2 ^ 9 = 512
5 ^ 3 = 125
-----
64,000

1/64000 = .00001 ---> # OF 0'S BEFORE A NON-ZERO DIGIT = 4

CAN SOMEONE SHOW ME HOW TO SOLVE THIS AN EASIER WAY? THANKS!

The number of 0's depends on how many powers of 10 are in the denominator.

1/(2â�¹*5Â³) = 1/(2â�¶)(2Â³*5Â³) = 1/(64*10Â³) = 1/64,000.

1/10,000 = .0001.
1/100,000 = .00001.
Since the denominator (64,000) is between 10,000 and 100,000, four 0's will appear to the right of the decimal point.

The correct answer is B.

Only when the denominator reaches 1,000,000 will a fifth 0 be needed:
1/1,000,000 = .000001.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8138; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Fri Jun 26, 2015 9:09 am

factor26 wrote:IF T = 1 / (2^9 * 5^3) IS EXPRESSED AS A TERMINATING DECIMAL, HOW MANY ZEROS WILL (T) HAVE BETWEEN THE DECIMAL POINT AND THE FIRST NONZERO DIGIT TO THE RIGHT OF THE DECIMAL POINT?

A - THREE

B - FOUR

C - FIVE

D - SIX

E - NINE

OG ANSWER IS B... I SOLVED THE QUESTION GOING THROUGH THE MATH THE LONG WAY ...

2 ^ 9 = 512
5 ^ 3 = 125
-----
64,000

1/64000 = .00001 ---> # OF 0'S BEFORE A NON-ZERO DIGIT = 4

CAN SOMEONE SHOW ME HOW TO SOLVE THIS AN EASIER WAY? THANKS!

Solution:

We use the term "leading zeroes" to describe the zeroes between the decimal point and the first nonzero decimal digit lying to the right of the decimal point. We can use the following rule to determine the number of leading zeroes in a fraction when it is converted to a decimal:

If X is an integer with k digits, then 1/X will have k - 1 leading zeroes unless X is a perfect power of 10, in which case there will be k - 2 leading zeroes.

We see that t is in the form 1/X. Because the denominator X has more twos than fives, we know X is not a perfect power of 10. Before considering the fraction as a whole, we first must determine the number of digits in the denominator.

Rewriting the denominator, we get 2^9 x 5^3 = (2^6 x 2^3) x 5^3 = 2^6 x (2^3 x 5^3) = 2^6 x 10^3 = 64 x (1,000) = 64,000, which is a 5-digit integer. Thus, k = 5.

Using our rule, we see that the fraction t has 5 - 1 = 4 leading zeroes.

Answer: B

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nikhilgmat31: Legendary Member; Posts: 518; Joined: Tue May 12, 2015 8:25 pm; Thanked: 10 times

by nikhilgmat31 » Wed Jul 01, 2015 11:01 pm

1/64000
can be simplified as 1/64 * .00015

1/64 equals 0.01

so 0.01 *0.00015 = 0.000015

so 4 zeros

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