The three-digit positive integer x has the hundreds,

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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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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Hope this helps.
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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.

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

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

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

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Hope it helps.
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Attached is a visual that should help.

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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

(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.

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!!

Can someone please explain me after this step --2a−k∗3b−l∗5c−m

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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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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How are you coming to \(x-y=210\) after \(2^{a-k}*3^{b-l}*5^{c-m}=2^2*3\);
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We got that:
\(a-k=2\) - the difference between the hundreds digits of x and y is 2;
\(b-l=1\) - the difference between the tens digits of x and y is 1;
\(c-m=0\) - the difference between the units digits of x and y is 0;

abc
-
klm
______
210
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Not sure if it helps anyone but I solved by plugging in numbers.
1. To make it as simple as possible, start by pretending that abc and klm are both the values 111.
2. Using the formula \((2^a\))(\(3^b\))(\(5^c\)) = 12(\(2^k\))(\(3^l\))(\(5^m\)) and plugging in my starter numbers, 30 obviously doesn’t equal 360. So I need to manipulate the left side variables to get a factor of 12 to balance it out.
3. To get 12, you can do prime factorization to get “2 x 2 x 3”. In the given equation, this is effectively cubing the 2 and squaring the 3.
4. So I am left with abc = 321 and klm 111.
5. abc – klm = 210

Maybe more confusing, but my brain hurts trying to do it the other way.

First, let us simplify the exponential equation:
(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)
(2^a)(3^b)(5^c) = (3)(4)(2^k)(3^l)(5^m)
(2^a)(3^b)(5^c) = (3^1)(2^2)(2^k)(3^l)(5^m)
(2^a)(3^b)(5^c) = (2^k+2)(3^l + 1)(5^m)

When the bases on both sides of an equation are equal, and the bases are prime numbers, the exponents of the respective bases must also be equal: a = k + 2; b = l + 1; and c = m. Now recall that a, b, and c represent the hundreds, tens, and units digits of the three-digit integer x; similarly, k, l, and m represent the hundreds, tens, and units digits of the three-digit integer y.

Therefore, the hundreds digit of x is 2 greater than the hundreds digit of y; the tens digit of x is 1 greater than the tens digit of y; finally, the units digit of x equals the units digit of y. Using this information, we can set up our subtraction problem and find the value of (x – y):
a - k = 2; b - l =1; and c - m = 0
abc – klm = 210

The correct answer is C.