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Let a “2-placer” be a terminating decimal, between 0 and 1, for which

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Let a “2-placer” be a terminating decimal, between 0 and 1, for which  [#permalink]

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Let a “2-placer” be a terminating decimal, between 0 and 1, for which the shortest possible representation has exactly 2 decimal places. For example, 0.09 and 0.87 are 2-placers, but 0.70 is not a 2-placer (because it can be written as 0.7, with only one decimal place). If m, n, and the product of m and n are each 2-placers, how many different values are possible for the sum m + n?

A) 40
B) 48
C) 60
D) 72
E) 197

Kudos for a correct solution.

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Re: Let a “2-placer” be a terminating decimal, between 0 and 1, for which  [#permalink]

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New post 11 May 2015, 05:13
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Bunuel wrote:
Let a “2-placer” be a terminating decimal, between 0 and 1, for which the shortest possible representation has exactly 2 decimal places. For example, 0.09 and 0.87 are 2-placers, but 0.70 is not a 2-placer (because it can be written as 0.7, with only one decimal place). If m, n, and the product of m and n are each 2-placers, how many different values are possible for the sum m + n?

A) 40
B) 48
C) 60
D) 72
E) 197

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

How can a pair of 2-placers multiply together to make another 2-placer?

Note that any 2-placer can be written as a fraction: an integer from 1 to 99, divided by 100. For example, the 2-placer 0.71 can be written as 71/100. The integers 10, 20, 30, …, 90 are excluded, since those create 1-placers rather than 2-placers.

Let m = M/100 and n = N/100, where M and N are the corresponding integers. Remember the restriction that M and N cannot be multiples of 10 (10, 20, 30, …).

The problem specifies that the product (M/100)(N/100) = MN/10,000 be a 2-placer as well. In other words, the fraction MN/10,000 must reduce to integer/100, but NOT to integer/10. This is equivalent to saying that 100 is a factor of MN, but 1000 is NOT a factor of MN.

Now consider how to select M and N to fit all of these restrictions. The product MN must contain at least the prime factors (2)(2)(5)(5) = 100. But neither M nor N alone can contain both a 2 and a 5, because neither M nor N may be a multiple of 10. Therefore, one of the two numbers must contain the prime factors 2*2 (and NOT 5), and the other must contain 5*5 (and NOT 2).

You’re ultimately interested in the sum of M and N, so it doesn’t matter which integer is called M and which is called N. Let’s say M is the one containing 2*2 and no 5, and N the one containing 5*5 and no 2.

M can be (2)(2) = 4 and any multiple that does not contain a 5: 8, 12, 16, 24, 28, 32, 36, …, 96. (Remember that M and N can only go up to 99.) Therefore, m can be any of 0.04, 0.08, 0.12, 0.16, 0.24, …, 0.96.

N could be (5)(5) = 25. It could also be (5)(5)(3) = 75. (Remember that N can’t contain a 2.) Therefore, n = 0.25 or 0.75.

Check that the product mn is a 2-placer as desired. For instance, if m = 0.96 and n = 0.75, then mn = 0.72. In each case, you’re taking either 25% or 75% of a multiple of 0.04, which you can think of as a multiple of 4. The result will always be a 2-placer.

Finally, consider the sum m + n:

- If n = 0.25, then there are twenty possible sums, corresponding to the twenty possible values of m:
m + n = 0.29, 0.33, 0.37, 0.41, 0.49, …, 1.21.

- If n = 0.75, then there are twenty more sums, corresponding to the twenty possible values of m:
m + n = 0.79, 0.83, 0.87, 0.91, 0.99, …, 1.71.

These two lists have no values in common. (Every decimal in the first list is one more than a multiple of 0.04; every decimal in the second list is one less than a multiple of 0.04.) There are thus 40 different sums of m and n.

The correct answer is (A).
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Re: Let a “2-placer” be a terminating decimal, between 0 and 1, for which  [#permalink]

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New post 05 May 2015, 00:00
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Bunuel wrote:
Let a “2-placer” be a terminating decimal, between 0 and 1, for which the shortest possible representation has exactly 2 decimal places. For example, 0.09 and 0.87 are 2-placers, but 0.70 is not a 2-placer (because it can be written as 0.7, with only one decimal place). If m, n, and the product of m and n are each 2-placers, how many different values are possible for the sum m + n?

A) 40
B) 48
C) 60
D) 72
E) 197

Kudos for a correct solution.


Since 'm' and 'n' are 2-Placer , product \(m*n\) can be rewritten as \(\frac{M*N}{10000}\). where \(\frac{M}{100} =m\) and \(\frac{n}{100}= N\)
In Order for product M*N/10000 to be 2-Placer there should be 2 '0' in product of M*N so , there should be 2*2*5*5 i.e. 100 in M*N .
Also , note that neither M nor N will be a multiple of 10 so M will be a multiple of 4 (without 5) and \(N\) will be a multiple of 25.

Possible values of M = 4,16,20,24,28,32, ...40...60...80..96 = 20 values.
Possible values of N = 25, 75 = 2 Values

Answer 2*20 = 40 = A
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Let a “2-placer” be a terminating decimal, between 0 and 1, for which  [#permalink]

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New post 24 Dec 2015, 19:04
Finally, consider the sum m + n:

- If n = 0.25, then there are twenty possible sums, corresponding to the twenty possible values of m:
m + n = 0.29, 0.33, 0.37, 0.41, 0.49, …, 1.21.

- If n = 0.75, then there are twenty more sums, corresponding to the twenty possible values of m:
m + n = 0.79, 0.83, 0.87, 0.91, 0.99, …, 1.71.

These two lists have no values in common. (Every decimal in the first list is one more than a multiple of 0.04; every decimal in the second list is one less than a multiple of 0.04.) There are thus 40 different sums of m and n.

The correct answer is (A).[/quote]

Bunuel
How u arrived on m+n (20 possible values)as highlighted above??
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Re: Let a “2-placer” be a terminating decimal, between 0 and 1, for which  [#permalink]

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Re: Let a “2-placer” be a terminating decimal, between 0 and 1, for which &nbs [#permalink] 11 Sep 2018, 12:26
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