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# The three-digit positive integer x has the hundreds,

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The three-digit positive integer x has the hundreds, [#permalink]

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15 Nov 2013, 07:00
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The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?

A) 21
B) 200
C) 210
D) 300
E) 310
[Reveal] Spoiler: OA

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Re: The three-digit positive integer x has the hundreds, [#permalink]

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15 Nov 2013, 07:09
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mohnish104 wrote:
The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?

A) 21
B) 200
C) 210
D) 300
E) 310

x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively.
y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively.

$$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$ --> $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$ --> $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$.

Answer: C.
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Re: The three-digit positive integer x has the hundreds, [#permalink]

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15 Nov 2013, 07:13
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Bunuel wrote:
mohnish104 wrote:
The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?

A) 21
B) 200
C) 210
D) 300
E) 310

x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively.
y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively.

$$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$ --> $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$ --> $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$.

Answer: C.

Similar questions to practice:
k-and-l-are-each-four-digit-positive-integers-with-thousands-126646.html
the-function-f-is-defined-for-each-positive-three-digit-100847.html
for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html

Hope this helps.
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The three-digit positive integer x has the hundreds, [#permalink]

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02 Dec 2014, 01:00
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x = 100a + 10b + c ........... (1)

y = 100k + 10l + m ............ (2)

We require to find x-y

(1) - (2)

$$2^a* 3^b * 5^c = 12 * 2^k* 3^l * 5^m$$

$$2^a * 3^b * 5^c = 2^{k+2} * 3^{l+1} * 5^m$$

Equating powers

a = k+2; b = l+1; c = m

Substituting the above values in equation (1)

100k + 200 + 10l + 10 + c - (100k + 10l + m) = 210

Answer = C
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Re: The three-digit positive integer x has the hundreds, [#permalink]

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09 Dec 2014, 07:58
Bunuel wrote:
mohnish104 wrote:
The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?

A) 21
B) 200
C) 210
D) 300
E) 310

x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively.
y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively.

$$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$ --> $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$ --> $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$.

Answer: C.

I am sorry but i dont understand the slightest thing of what's going on here. How does that suddenly translate to 2^2*3? really confused here.

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Re: The three-digit positive integer x has the hundreds, [#permalink]

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09 Dec 2014, 08:03
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JoostGrijsen wrote:
Bunuel wrote:
mohnish104 wrote:
The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?

A) 21
B) 200
C) 210
D) 300
E) 310

x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively.
y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively.

$$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$ --> $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$ --> $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$.

Answer: C.

I am sorry but i dont understand the slightest thing of what's going on here. How does that suddenly translate to 2^2*3? really confused here.

$$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$;

$$\frac{(2^a)(3^b)(5^c)}{(2^k)(3^l)(5^m)} = 12$$;

$$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$.

Similar questions to practice:
k-and-l-are-each-four-digit-positive-integers-with-thousands-91004.html
for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html
k-and-l-are-each-four-digit-positive-integers-with-thousands-126646.html
the-function-f-is-defined-for-each-positive-three-digit-100847.html
for-a-three-digit-number-xyz-where-x-y-and-z-are-the-59284.html

Hope it helps.
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Re: The three-digit positive integer x has the hundreds, [#permalink]

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31 Mar 2016, 17:11
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Attached is a visual that should help.
Attachments

Screen Shot 2016-03-31 at 6.10.08 PM.png [ 137.01 KiB | Viewed 3598 times ]

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Re: The three-digit positive integer x has the hundreds, [#permalink]

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28 Apr 2017, 03:57
what is the next step after this 2a−k∗3b−l∗5c−m=22∗32a−k∗3b−l∗5c−m=22∗3 ? I cant figure it out. can someone break this down well for me please

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Re: The three-digit positive integer x has the hundreds, [#permalink]

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11 Aug 2017, 01:48
(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)
can be re written as (2^a-k)(3^b-l)(5^c-m)=2^2.3^1.5^0
a-k=2
b-l=1
c-m=0
The above 3 are derived from laws of indices, which in turn points to the subtraction of each of the unit's,ten's and hundred's place of x & y.
Option C.

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Re: The three-digit positive integer x has the hundreds, [#permalink]

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15 Aug 2017, 16:57
Bunuel wrote:
mohnish104 wrote:
The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y?

A) 21
B) 200
C) 210
D) 300
E) 310

x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively.
y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively.

$$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$ --> $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$ --> $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$.

Answer: C.

please clarify my doubt:

does the wording sound good? The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. it would be better if it was as

am i right? previous statement can be interpreted as there is a positive integer a, whose hundreds digit is the hundred digit of x. please clarify!!

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Re: The three-digit positive integer x has the hundreds, [#permalink]

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24 Aug 2017, 07:10
Can someone please explain me after this step --2a−k∗3b−l∗5c−m

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Re: The three-digit positive integer x has the hundreds,   [#permalink] 24 Aug 2017, 07:10
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# The three-digit positive integer x has the hundreds,

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