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

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

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New post 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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New post 15 Nov 2013, 07:13
1
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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Re: The three-digit positive integer x has the hundreds,  [#permalink]

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New post 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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New post 09 Dec 2014, 08:03
1
1
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 31 Mar 2016, 17:11
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Attached is a visual that should help.
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Screen Shot 2016-03-31 at 6.10.08 PM.png
Screen Shot 2016-03-31 at 6.10.08 PM.png [ 137.01 KiB | Viewed 6730 times ]


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

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New post 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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New post 11 Aug 2017, 01:48
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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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New post 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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New post 24 Aug 2017, 07:10
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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New post 27 Aug 2018, 05:38
1
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.




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 \(c-m=0\) --> \(x-y=210\) :hurt: :-)

Hey pushpitkc :) are you there ? :lol:
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The three-digit positive integer x has the hundreds,  [#permalink]

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New post 27 Aug 2018, 11:33
1
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 \(c-m=0\) --> \(x-y=210\).
Answer: C.




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 \(c-m=0\) --> \(x-y=210\) :hurt: :-)

Hey pushpitkc :) are you there ? :lol:


Hey dave13

I've always been here :) :lol:

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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New post 28 Aug 2018, 00:24
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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New post 28 Aug 2018, 01:21
1
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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The three-digit positive integer x has the hundreds, &nbs [#permalink] 28 Aug 2018, 01:21
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