The three-digit positive integer x has the hundreds,

Question

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

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

Bunuel · Accepted Answer

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.

PareshGmat · Answer

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

Bunuel · Answer

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.

JoostGrijsen · Answer

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.

Bunuel · Answer

(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.

mcelroytutoring · Answer

Attached is a visual that should help.

Iam700 · Answer

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

FB2017 · Answer

(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.

Avinash_R1 · Answer

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!!

santro789 · Answer

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

dave13 · Answer

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 ?

pushpitkc · Answer

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!

onyx12102 · Answer

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

pushpitkc · Answer

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!

MHIKER · Answer

How are you coming to  x-y=210  after  2^{a-k}*3^{b-l}*5^{c-m}=2^2*3 ;

Bunuel · Answer

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

jpcappa · Answer

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.

jerinpeter · Answer

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.

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