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Manager  Joined: 03 Dec 2012
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The three-digit positive integer x has the hundreds,  [#permalink]

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

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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 $$c-m=0$$;

$$x-y=210$$.

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

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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

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

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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 $$c-m=0$$ --> $$x-y=210$$.

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

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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 $$c-m=0$$ --> $$x-y=210$$.

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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Posts: 58434
Re: The three-digit positive integer x has the hundreds,  [#permalink]

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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 $$c-m=0$$ --> $$x-y=210$$.

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:
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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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Attached is a visual that should help.
Attachments Screen Shot 2016-03-31 at 6.10.08 PM.png [ 137.01 KiB | Viewed 8437 times ]

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

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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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(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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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 $$c-m=0$$ --> $$x-y=210$$.

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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Can someone please explain me after this step --2a−k∗3b−l∗5c−m
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The three-digit positive integer x has the hundreds,  [#permalink]

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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 $$c-m=0$$ --> $$x-y=210$$.

ok i got it how you got$$2^2*3$$ but i still dont get how you got this part $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$  Hey pushpitkc are you there ? Senior PS Moderator V
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The three-digit positive integer x has the hundreds,  [#permalink]

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dave13 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 $$c-m=0$$ --> $$x-y=210$$.

ok i got it how you got$$2^2*3$$ but i still dont get how you got this part $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$  Hey pushpitkc are you there ? Hey dave13

I've always been here  We have $$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$ and can be re-written as $$\frac{(2^a)(3^b)(5^c)}{(2^k)(3^l)(5^m)} = 12$$ -> (1)

Now, since we know that $$\frac{2^a}{2^k} = 2^{a-k}$$ | $$\frac{3^b}{3^l} = 3^{b-l}$$ | $$\frac{5^c}{5^m} = 5^{c-m}$$. Also, 12 can be re-written as $$2^2*3^1*5^0$$

Equation (1) now becomes -> a - k = 2 | b - l = 1 | c - m = 0 and since we need the difference of
3 digit numbers abc(100a + 10b + c) and klm(100k + 10l + m) -> 2*100 + 1*10 + 0 = 210

Hope this clears your confusion!
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Re: The three-digit positive integer x has the hundreds,  [#permalink]

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I don't get why b-l = 1 etc, can someone please break this part down
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The three-digit positive integer x has the hundreds,  [#permalink]

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onyx12102 wrote:
I don't get why b-l = 1 etc, can someone please break this part down

Hey onyx12102

Please look at this part of my previous solution

LHS => $$\frac{2^a}{2^k} + \frac{3^b}{3^l} + \frac{5^c}{5^m}$$ = $$2^{a-k} +3^{b-l} + 5^{c-m}$$.
Also, RHS which is 12 can be re-written as $$2^2*3^1*5^0$$

Equating equal bases on both sides, we will get a - k = 3 , b - l = 1 , and c - m = 0

Hope this clears your confusion!
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You've got what it takes, but it will take everything you've got The three-digit positive integer x has the hundreds,   [#permalink] 28 Aug 2018, 02:21
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